# Drawing Sorgenfrey continuous functions

The Sorgenfrey line is a well known topological space. It is the real number line with open intervals defined as sets of the form $[a,b)$. Though this is a seemingly small tweak, it generates a vastly different space than the usual real number line. In this post, we look at the Sorgenfrey line from the continuous function perspective, in particular, the continuous functions that map the Sorgenfrey line into the real number line. In the process, we obtain insight into the space of continuous functions on the Sorgenfrey line.

The Sorgenfrey Line

Let $\mathbb{R}$ denote the real number line. The usual open intervals are of the form $(a,b)=\left\{x \in \mathbb{R}: a. The union of such open intervals is called an open set. If more than one topologies are considered on the real line, these open sets are referred to as the usual open sets or Euclidean open sets (on the real line). The open intervals $(a,b)$ form a base for the usual topology on the real line. One important fact abut the usual open sets is that the usual open sets can be generated by the intervals $(a,b)$ where both end points are rational numbers. Thus the usual topology on the real line is said to have a countable base.

Now tweak the usual topology by calling sets of the form $[a,b)=\left\{x \in \mathbb{R}: a \le x open intervals. Then form open sets by taking unions of all such open intervals. The collection of such open sets is called the Sorgenfrey topology (on the real line). The real number line $\mathbb{R}$ with the Sorgenfry topology is called the Sorgenfrey line, denoted by $\mathbb{S}$. The Sorgenfrey line has been discussed in this blog, starting with this post. This post examines continuous functions from $\mathbb{S}$ into the real line. In the process, we gain insight on the space of continuous functions defined on $\mathbb{S}$.

Note that any usual open interval $(a,b)$ is the union of intervals of the form $[c,d)$. Thus any usual (Euclidean) open set is an open set in the Sorgenfrey line. Thus the usual topology (on the real line) is contained in the Sorgenfrey topology, i.e. the usual topology is a weaker (coarser) topology.

Let $C(\mathbb{R})$ be the set of all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ where the domain is the real number line with the usual topology. Let $C(\mathbb{S})$ be the set of all continuous functions $f:\mathbb{S} \rightarrow \mathbb{R}$ where the domain is the Sorgenfrey line. In both cases, the range is always the number line with the usual topology. Based on the preceding paragraph, any continuous function $f:\mathbb{R} \rightarrow \mathbb{R}$ is also continuous with respect to the Sorgenfrey line, i.e. $C(\mathbb{R}) \subset C(\mathbb{S})$.

Pictures of Continuous Functions

Consider the following two continuous functions.

Figure 1 – CDF of the standard normal distribution

Figure 2 – CDF of the uniform distribution

The first one (Figure 1) is the cumulative distribution function (CDF) of the standard normal distribution. The second one (Figure 2) is the CDF of the uniform distribution on the interval $(0,a)$ where $a>0$. Both of these are continuous in the usual Euclidean topology (in the domain). Such graphs would make regular appearance in a course on probability and statistics. They also show up in a calculus course as an everywhere differentiable curve (Figure 1) and as a differentiable curve except at finitely many points (Figure 2). Both of these functions can also be regarded as continuous functions on the Sorgenfrey line.

Consider a function that is continuous in the Sorgenfrey line but not continuous in the usual topology.

Figure 3 – Right continuous function

Figure 3 is a function that maps the interval $(-\infty,0)$ to -1 and maps the interval $[0,\infty)$ to 1. It is not continuous in the usual topology because of the jump at $x=0$. But it is a continuous function when the domain is considered to be the Sorgenfrey line. Because of the open intervals being $[a,b)$, continuous functions defined on the Sorgenfrey line are right continuous.

The cumulative distribution function of a discrete probability distribution is always right continuous, hence continuous in the Sorgenfrey line. Here’s an example.

Figure 4 – CDF of a discrete uniform distribution

Figure 4 is the CDF of the uniform distribution on the finite set $\left\{0,1,2,3,4 \right\}$, where each point has probability 0.2. There is a jump of height 0.2 at each of the points from 0 to 4. Figure 3 and Figure 4 are step functions. As long as the left point of a step is solid and the right point is hollow, the step functions are continuous on the Sorgenfrey line.

The take away from the last four figures is that the real-valued continuous functions defined on the Sorgenfrey line are right continuous and that step functions (with the left point solid and the right point hollow) are Sorgenfrey continuous.

A Family of Sorgenfrey Continuous Functions

The four examples of continuous functions shown above are excellent examples to illustrate the Sorgenfrey topology. We now introduce a family of continuous functions $f_a:\mathbb{S} \rightarrow \mathbb{R}$ for $0. These continuous functions will lead to additional insight on the function space whose domain space is the Sorgenfrey line.

For any $0, the following gives the definition and the graph of the function $f_a$.

$\displaystyle f_a(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ -\infty

Figure 5 – a family of Sorgenfrey continuous functions

Function Space on the Sorgenfrey Line

This is the place where we switch the focus to function space. The set $C(\mathbb{S})$ is a subset of the product space $\mathbb{R}^\mathbb{R}$. So we can consider $C(\mathbb{S})$ as a topological space endowed with the topology inherited as a subspace of $\mathbb{R}^\mathbb{R}$. This topology on $C(\mathbb{S})$ is called the pointwise convergence topology and $C(\mathbb{S})$ with the product subspace topology is denoted by $C_p(\mathbb{S})$. See here for comments on how to work with the pointwise convergence topology.

For the present discussion, all we need is some notation on a base for $C_p(\mathbb{S})$. For $x \in \mathbb{S}$, and for any open interval $(a,b)$ (open in the usual topology of the real number line), let $[x,(a,b)]=\left\{h \in C_p(\mathbb{S}): h(x) \in (a, b) \right\}$. Then the collection of intersections of finitely many $[x,(a,b)]$ would form a base for $C_p(\mathbb{S})$.

The following is the main fact we wish to establish.

The function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum. In particular, the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

The above result will derive several facts on the function space $C_p(\mathbb{S})$, which are discussed in a section below. More interestingly, the proof of the fact that $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$ is based purely on the definition of the functions $f_a$ and the Sorgenfrey topology. The proof given below does not use any deep or high powered results from function space theory. So it should be a nice exercise on the Sorgenfrey topology.

I invite readers to either verify the fact independently of the proof given here or follow the proof closely. Lots of drawing of the functions $f_a$ on paper will be helpful in going over the proof. In this one instance at least, drawing continuous functions can help gain insight on function spaces.

Working out the Proof

The following diagram was helpful to me as I worked out the different cases in showing the discreteness of the family $F=\left\{f_a: 0. The diagram is a valuable aid in convincing myself that a given case is correct.

Figure 6 – A comparison of three Sorgenfrey continuous functions

Now the proof. First, $F$ is relatively discrete in $C_p(\mathbb{S})$. We show that for each $a$, there is an open set $O$ containing $f_a$ such that $O$ does not contain $f_w$ for any $w \ne a$. To this end, let $O=[a,V_1] \cap [-a,V_2]$ where $V_1$ and $V_2$ are the open intervals $V_1=(0.9,1.1)$ and $V_2=(-0.1,0.1)$. With Figure 6 as an aid, it follows that for $0, $f_b \notin O$ and for $a, $f_c \notin O$.

The open set $O=[a,V_1] \cap [-a,V_2]$ contains $f_a$, the function in the middle of Figure 6. Note that for $0, $f_b(-a)=1$ and $f_b(-a) \notin V_2$. Thus $f_b \notin O$. On the other hand, for $a, $f_c(a)=0$ and $f_c(a) \notin V_1$. Thus $f_c \notin O$. This proves that the set $F$ is a discrete subspace of $C_p(\mathbb{S})$ relative to $F$ itself.

Now we show that $F$ is closed in $C_p(\mathbb{S})$. To this end, we show that

for each $g \in C_p(\mathbb{S})$, there is an open set $U$ containing $g$ such that $U$ contains at most one point of $F$.

Actually, this has already been done above with points $g$ that are in $F$. One thing to point out is that the range of $f_a$ is $\left\{0,1 \right\}$. As we consider $g \in C_p(\mathbb{S})$, we only need to consider $g$ that maps into $\left\{0,1 \right\}$. Let $g \in C_p(\mathbb{S})$. The argument is given in two cases regarding the function $g$.

Case 1. There exists some $a \in (0,1)$ such that $g(a) \ne g(-a)$.

We assume that $g(a)=0$ and $g(-a)=1$. Then for all $0, $f_b(a)=1$ and for all $a, $f_c(-a)=0$. Let $U=[a,(-0.1,0.1)] \cap [-a,(0.9,1.1)]$. Then $g \in U$ and $U$ contains no $f_b$ for any $0 and $f_c$ for any $a. To help see this argument, use Figure 6 as a guide. The case that $g(a)=1$ and $g(-a)=0$ has a similar argument.

Case 2. For every $a \in (0,1)$, we have $g(a) = g(-a)$.

Claim. The function $g$ is constant on the interval $(-1,1)$. Suppose not. Let $0 such that $g(a) \ne g(b)$. Suppose that $0=g(b) < g(a)=1$. Consider $W=\left\{w. Clearly the number $a$ is an upper bound of $W$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 1 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=0$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=1$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=1$ and $g(b)=g(-b)=0$. Since $g(x_n)=0$ for all $n$, $g(-x_n)=0$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=0$, contradicting $g(-u)=1$. Thus we cannot have $0=g(b) < g(a)=1$.

Now suppose we have $1=g(b) > g(a)=0$ where $0. Consider $W=\left\{w. Clearly $W$ has an upper bound, namely the number $a$. Let $u \le a$ be a least upper bound of $W$. The function $g$ has value 0 on the interval $(u,a)$. Otherwise, $u$ would not be the least upper bound of the set $W$. There is a sequence of points $\left\{x_n \right\}$ in the interval $(b,u)$ such that $x_n \rightarrow u$ from the left such that $g(x_n)=1$ for all $n$. Otherwise, $u$ would not be the least upper bound of the set $W$.

It follows that $g(u)=0$. Otherwise, the function $g$ is not continuous at $u$. Now consider the 6 points $-a<-u<-b. By the assumption in Case 2, $g(u)=g(-u)=0$ and $g(b)=g(-b)=1$. Since $g(x_n)=1$ for all $n$, $g(-x_n)=1$ for all $n$. Note that $-x_n \rightarrow -u$ from the right. Since $g$ is right continuous, $g(-u)=1$, contradicting $g(-u)=0$. Thus we cannot have $1=g(b) > g(a)=0$.

The claim that the function $g$ is constant on the interval $(-1,1)$ is established. To wrap up, first assume that the function $g$ is 1 on the interval $(-1,1)$. Let $U=[0,(0.9,1.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$. Now assume that the function $g$ is 0 on the interval $(-1,1)$. Since $g$ is Sorgenfrey continuous, it follows that $g(-1)=0$. Let $U=[-1,(-0.1,0.1)]$. It is clear that $g \in U$. It is also clear from Figure 5 that $U$ contains no $f_a$.

We have established that the set $F=\left\{f_a: 0 is a closed and discrete subspace of $C_p(\mathbb{S})$.

What does it Mean?

The above argument shows that the set $F$ is a closed an discrete subspace of the function space $C_p(\mathbb{S})$. We have the following three facts.

 Three Results $C_p(\mathbb{S})$ is separable. $C_p(\mathbb{S})$ is not hereditarily separable. $C_p(\mathbb{S})$ is not a normal space.

To show that $C_p(\mathbb{S})$ is separable, let’s look at one basic helpful fact on $C_p(X)$. If $X$ is a separable metric space, e.g. $X=\mathbb{R}$, then $C_p(X)$ has quite a few nice properties (discussed here). One is that $C_p(X)$ is hereditarily separable. Thus $C_p(\mathbb{R})$, the space of real-valued continuous functions defined on the number line with the pointwise convergence topology, is hereditarily separable and thus separable. Recall that continuous functions in $C_p(\mathbb{R})$ are also Soregenfrey line continuous. Thus $C_p(\mathbb{R})$ is a subspace of $C_p(\mathbb{S})$. The space $C_p(\mathbb{R})$ is also a dense subspace of $C_p(\mathbb{S})$. Thus the space $C_p(\mathbb{S})$ contains a dense separable subspace. It means that $C_p(\mathbb{S})$ is separable.

Secondly, $C_p(\mathbb{S})$ is not hereditarily separable since the subspace $F=\left\{f_a: 0 is a closed and discrete subspace.

Thirdly, $C_p(\mathbb{S})$ is not a normal space. According to Jones’ lemma, any separable space with a closed and discrete subspace of cardinality of continuum is not a normal space (see Corollary 1 here). The subspace $F=\left\{f_a: 0 is a closed and discrete subspace of the separable space $C_p(\mathbb{S})$. Thus $C_p(\mathbb{S})$ is not normal.

Remarks

The topology of the Sorgenfrey line is vastly different from the usual topology on the real line even though the the Sorgenfrey topology is obtained by a seemingly small tweak from the usual topology. The real line is a metric space while the Sorgenfrey line is not metrizable. The real number line is connected while the Sorgenfrey line is not. The countable power of the real number line is a metric space and thus a normal space. On the other hand, the Sorgenfrey line is a classic example of a normal space whose square is not normal. See here for a basic discussion of the Sorgenfrey line.

The pictures of Sorgenfrey continuous functions demonstrated here show that the real number line and the Sorgenfrey line are also very different from a function space perspective. The function space $C_p(\mathbb{R})$ has a whole host of nice properties: normal, Lindelof (hence paracompact and collectionwise normal), hereditarily Lindelof (hence hereditarily normal), hereditarily separable, and perfectly normal (discussed here).

Though separable, the function space $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum, making it not hereditarily separable and not normal.

For more information about $C_p(X)$ in general and $C_p(\mathbb{S})$ in particular, see [1] and [2]. A different proof that $C_p(\mathbb{S})$ contains a closed and discrete subspace of cardinality continuum can be found in Problem 165 in [2].

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

# Normality in Cp(X)

Any collectionwise normal space is a normal space. Any perfectly normal space is a hereditarily normal space. In general these two implications are not reversible. In function spaces $C_p(X)$, the two implications are reversible. There is a normal space that is not countably paracompact (such a space is called a Dowker space). If a function space $C_p(X)$ is normal, it is countably paracompact. Thus normality in $C_p(X)$ is a strong property. This post draws on Dowker’s theorem and other results, some of them are previously discussed in this blog, to discuss this remarkable aspect of the function spaces $C_p(X)$.

Since we are discussing function spaces, the domain space $X$ has to have sufficient quantity of real-valued continuous functions, e.g. there should be enough continuous functions to separate the points from closed sets. The ideal setting is the class of completely regular spaces (also called Tychonoff spaces). See here for a discussion on completely regular spaces in relation to function spaces.

Let $X$ be a completely regular space. Let $C(X)$ be the set of all continuous functions from $X$ into the real line $\mathbb{R}$. When $C(X)$ is endowed with the pointwise convergence topology, the space is denoted by $C_p(X)$ (see here for further comments on the definition of the pointwise convergence topology).

When Function Spaces are Normal

Let $X$ be a completely regular space. We discuss these four facts of $C_p(X)$:

1. If the function space $C_p(X)$ is normal, then $C_p(X)$ is countably paracompact.
2. If the function space $C_p(X)$ is hereditarily normal, then $C_p(X)$ is perfectly normal.
3. If the function space $C_p(X)$ is normal, then $C_p(X)$ is collectionwise normal.
4. Let $X$ be a normal space. If $C_p(X)$ is normal, then $X$ has countable extent, i.e. every closed and discrete subset of $X$ is countable, implying that $X$ is collectionwise normal.

Fact #1 and Fact #2 rely on a representation of $C_p(X)$ as a product space with one of the factors being the real line. For $x \in X$, let $Y_x=\left\{f \in C_p(X): f(x)=0 \right\}$. Then $C_p(X) \cong Y_x \times \mathbb{R}$. This representation is discussed here.

Another useful tool is Dowker’s theorem, which essentially states that for any normal space $W$, the space $W$ is countably paracompact if and only if $W \times C$ is normal for all compact metric space $C$ if and only if $W \times [0,1]$ is normal. For the full statement of the theorem, see Theorem 1 in this previous post, which has links to the proofs and other discussion.

To show Fact #1, suppose that $C_p(X)$ is normal. Immediately we make use of the representation $C_p(X) \cong Y_x \times \mathbb{R}$ where $x \in X$. Since $Y_x \times \mathbb{R}$ is normal, $Y_x \times [0,1]$ is also normal. By Dowker’s theorem, $Y_x$ is countably paracompact. Note that $Y_x$ is a closed subspace of the normal $C_p(X)$. Thus $Y_x$ is also normal.

One more helpful tool is Theorem 5 in in this previous post, which is like an extension of Dowker’s theorem, which states that a normal space $W$ is countably paracompact if and only if $W \times T$ is normal for any $\sigma$-compact metric space $T$. This means that $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal.

We want to show $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact. Since $Y_x \times \mathbb{R} \times \mathbb{R}$ is normal (based on the argument in the preceding paragraph), $(Y_x \times \mathbb{R}) \times [0,1]$ is normal. Thus according to Dowker’s theorem, $C_p(X) \cong Y_x \times \mathbb{R}$ is countably paracompact.

For Fact #2, a helpful tool is Katetov’s theorem (stated and proved here), which states that for any hereditarily normal $X \times Y$, one of the factors is perfectly normal or every countable subset of the other factor is closed (in that factor).

To show Fact #2, suppose that $C_p(X)$ is hereditarily normal. With $C_p(X) \cong Y_x \times \mathbb{R}$ and according to Katetov’s theorem, $Y_x$ must be perfectly normal. The product of a perfectly normal space and any metric space is perfectly normal (a proof is found here). Thus $C_p(X) \cong Y_x \times \mathbb{R}$ is perfectly normal.

The proof of Fact #3 is found in Problems 294 and 295 of [2]. The key to the proof is a theorem by Reznichenko, which states that any dense convex normal subspace of $[0,1]^X$ has countable extent, hence is collectionwise normal (problem 294). See here for a proof that any normal space with countable extent is collectionwise normal (see Theorem 2). The function space $C_p(X)$ is a dense convex subspace of $[0,1]^X$ (problem 295). Thus if $C_p(X)$ is normal, then it has countable extent and hence collectionwise normal.

Fact #4 says that normality of the function space imposes countable extent on the domain. This result is discussed in this previous post (see Corollary 3 and Corollary 5).

Remarks

The facts discussed here give a flavor of what function spaces are like when they are normal spaces. For further and deeper results, see [1] and [2].

Fact #1 is essentially driven by Dowker’s theorem. It follows from the theorem that whenever the product space $X \times Y$ is normal, one of the factor must be countably paracompact if the other factor has a non-trivial convergent sequence (see Theorem 2 in this previous post). As a result, there is no Dowker space that is a $C_p(X)$. No pathology can be found in $C_p(X)$ with respect to finding a Dowker space. In fact, not only $C_p(X) \times C$ is normal for any compact metric space $C$, it is also true that $C_p(X) \times T$ is normal for any $\sigma$-compact metric space $T$ when $C_p(X)$ is normal.

The driving force behind Fact #2 is Katetov’s theorem, which basically says that the hereditarily normality of $X \times Y$ is a strong statement. Coupled with the fact that $C_p(X)$ is of the form $Y_x \times \mathbb{R}$, Katetov’s theorem implies that $Y_x \times \mathbb{R}$ is perfectly normal. The argument also uses the basic fact that perfectly normality is preserved when taking product with metric spaces.

There are examples of normal but not collectionwise normal spaces (e.g. Bing’s Example G). Resolution of the question of whether normal but not collectionwise normal Moore space exists took extensive research that spanned decades in the 20th century (the normal Moore space conjecture). The function $C_p(X)$ is outside of the scope of the normal Moore space conjecture. The function space $C_p(X)$ is usually not a Moore space. It can be a Moore space only if the domain $X$ is countable but then $C_p(X)$ would be a metric space. However, it is still a powerful fact that if $C_p(X)$ is normal, then it is collectionwise normal.

On the other hand, a more interesting point is on the normality of $X$. Suppose that $X$ is a normal Moore space. If $C_p(X)$ happens to be normal, then Fact #4 says that $X$ would have to be collectionwise normal, which means $X$ is metrizable. If the goal is to find a normal Moore space $X$ that is not collectionwise normal, the normality of $C_p(X)$ would kill the possibility of $X$ being the example.

Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Tkachuk V. V., A $C_p$-Theory Problem Book, Topological and Function Spaces, Springer, New York, 2011.

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$\copyright$ 2017 – Dan Ma

# Compact metrizable scattered spaces

A scattered space is one in which there are isolated points found in every subspace. Specifically, a space $X$ is a scattered space if every non-empty subspace $Y$ of $X$ has a point $y \in Y$ such that $y$ is an isolated point in $Y$, i.e. the singleton set $\left\{y \right\}$ is open in the subspace $Y$. A handy example is a space consisting of ordinals. Note that in a space of ordinals, every non-empty subset has an isolated point (e.g. its least element). In this post, we discuss scattered spaces that are compact metrizable spaces.

Here’s what led the author to think of such spaces. Consider Theorem III.1.2 found on page 91 of Arhangelskii’s book on topological function space [1], which is Theorem 1 stated below:

Thereom 1
For any compact space $X$, the following conditions are equivalent:

• The function space $C_p(X)$ is a Frechet-Urysohn space.
• The function space $C_p(X)$ is a k space.
• $X$ is a scattered space.

Let’s put aside the Frechet-Urysohn property and the k space property for the moment. For any Hausdorff space $X$, let $C(X)$ be the set of all continuous real-valued functions defined on the space $X$. Since $C(X)$ is a subspace of the product space $\mathbb{R}^X$, a natural topology that can be given to $C(X)$ is the subspace topology inherited from the product space $\mathbb{R}^X$. Then $C_p(X)$ is simply the set $C(X)$ with the product subspace topology (also called the pointwise convergence topology).

Let’s say the compact space $X$ is countable and infinite. Then the function space $C_p(X)$ is metrizable since it is a subspace of $\mathbb{R}^X$, a product of countably many lines. Thus the function space $C_p(X)$ has the Frechet-Urysohn property (being metrizable implies Frechet-Urysohn). This means that the compact space $X$ is scattered. The observation just made is a proof that any infinite compact space that is countable in cardinality must be scattered. In particular, every infinite compact and countable space must have an isolated point. There must be a more direct proof of this same fact without taking the route of a function space. The indirect argument does not reveal the essential nature of compact metric spaces. The essential fact is that any uncountable compact metrizable space contains a Cantor set, which is as unscattered as any space can be. Thus the only scattered compact metrizable spaces are the countable ones.

The main part of the proof is the construction of a Cantor set in a compact metrizable space (Theorem 3). The main result is Theorem 4. In many settings, the construction of a Cantor set is done in the real number line (e.g. the middle third Cantor set). The construction here is in a more general setting. But the idea is still the same binary division process – the splitting of a small open set with compact closure into two open sets with disjoint compact closure. We also use that fact that any compact metric space is hereditarily Lindelof (Theorem 2).

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Compact metrizable spaces

We first define some notions before looking at compact metrizable spaces in more details. Let $X$ be a space. Let $A \subset X$. Let $p \in X$. We say that $p$ is a limit point of $A$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. So the notion of limit point here is from a topology perspective and not from a metric perspective. In a topological space, a limit point does not necessarily mean that it is the limit of a convergent sequence (however, it does in a metric space). The proof of the following theorem is straightforward.

Theorem 2
Let $X$ be a hereditarily Lindelof space (i.e. every subspace of $X$ is Lindelof). Then for any uncountable subset $A$ of $X$, all but countably many points of $A$ are limit points of $A$.

We now discuss the main result.

Theorem 3
Let $X$ be a compact metrizable space such that every point of $X$ is a limit point of $X$. Then there exists an uncountable closed subset $C$ of $X$ such that every point of $C$ is a limit point of $C$.

Proof of Theorem 3
Note that any compact metrizable space is a complete metric space. Consider a complete metric $\rho$ on the space $X$. One fact that we will use is that if there is a sequence of closed sets $X \supset H_1 \supset H_2 \supset H_3 \supset \cdots$ such that the diameters of the sets $H$ (based on the complete metric $\rho$) decrease to zero, then the sets $H_n$ collapse to one point.

The uncountable closed set $C$ we wish to define is a Cantor set, which is constructed from a binary division process. To start, pick two points $p_0,p_1 \in X$ such that $p_0 \ne p_1$. By assumption, both points are limit points of the space $X$. Choose open sets $U_0,U_1 \subset X$ such that

• $p_0 \in U_0$,
• $p_1 \in U_1$,
• $K_0=\overline{U_0}$ and $K_1=\overline{U_1}$,
• $K_0 \cap K_1 = \varnothing$,
• the diameters for $K_0$ and $K_1$ with respect to $\rho$ are less than 0.5.

Note that each of these open sets contains infinitely many points of $X$. Then we can pick two points in each of $U_0$ and $U_1$ in the same manner. Before continuing, we set some notation. If $\sigma$ is an ordered string of 0’s and 1’s of length $n$ (e.g. 01101 is a string of length 5), then we can always extend it by tagging on a 0 and a 1. Thus $\sigma$ is extended as $\sigma 0$ and $\sigma 1$ (e.g. 01101 is extended by 011010 and 011011).

Suppose that the construction at the $n$th stage where $n \ge 1$ is completed. This means that the points $p_\sigma$ and the open sets $U_\sigma$ have been chosen such that $p_\sigma \in U_\sigma$ for each length $n$ string of 0’s and 1’s $\sigma$. Now we continue the picking for the $(n+1)$st stage. For each $\sigma$, an $n$-length string of 0’s and 1’s, choose two points $p_{\sigma 0}$ and $p_{\sigma 1}$ and choose two open sets $U_{\sigma 0}$ and $U_{\sigma 1}$ such that

• $p_{\sigma 0} \in U_{\sigma 0}$,
• $p_{\sigma 1} \in U_{\sigma 1}$,
• $K_{\sigma 0}=\overline{U_{\sigma 0}} \subset U_{\sigma}$ and $K_{\sigma 1}=\overline{U_{\sigma 1}} \subset U_{\sigma}$,
• $K_{\sigma 0} \cap K_{\sigma 1} = \varnothing$,
• the diameters for $K_{\sigma 0}$ and $K_{\sigma 1}$ with respect to $\rho$ are less than $0.5^{n+1}$.

For each positive integer $m$, let $C_m$ be the union of all $K_\sigma$ over all $\sigma$ that are $m$-length strings of 0’s and 1’s. Each $C_m$ is a union of finitely many compact sets and is thus compact. Furthermore, $C_1 \supset C_2 \supset C_3 \supset \cdots$. Thus $C=\bigcap \limits_{m=1}^\infty C_m$ is non-empty. To complete the proof, we need to show that

• $C$ is uncountable (in fact of cardinality continuum),
• every point of $C$ is a limit point of $C$.

To show the first point, we define a one-to-one function $f: \left\{0,1 \right\}^N \rightarrow C$ where $N=\left\{1,2,3,\cdots \right\}$. Note that each element of $\left\{0,1 \right\}^N$ is a countably infinite string of 0’s and 1’s. For each $\tau \in \left\{0,1 \right\}^N$, let $\tau \upharpoonright n$ denote the string of the first $n$ digits of $\tau$. For each $\tau \in \left\{0,1 \right\}^N$, let $f(\tau)$ be the unique point in the following intersection:

$\displaystyle \bigcap \limits_{n=1}^\infty K_{\tau \upharpoonright n} = \left\{f(\tau) \right\}$

This mapping is uniquely defined. Simply conceptually trace through the induction steps. For example, if $\tau$ are 01011010…., then consider $K_0 \supset K_{01} \supset K_{010} \supset \cdots$. At each next step, always pick the $K_{\tau \upharpoonright n}$ that matches the next digit of $\tau$. Since the sets $K_{\tau \upharpoonright n}$ are chosen to have diameters decreasing to zero, the intersection must have a unique element. This is because we are working in a complete metric space.

It is clear that the map $f$ is one-to-one. If $\tau$ and $\gamma$ are two different strings of 0’s and 1’s, then they must differ at some coordinate, then from the way the induction is done, the strings would lead to two different points. It is also clear to see that the map $f$ is reversible. Pick any point $x \in C$. Then the point $x$ must belong to a nested sequence of sets $K$‘s. This maps to a unique infinite string of 0’s and 1’s. Thus the set $C$ has the same cardinality as the set $\left\{0,1 \right\}^N$, which has cardinality continuum.

To see the second point, pick $x \in C$. Suppose $x=f(\tau)$ where $\tau \in \left\{0,1 \right\}^N$. Consider the open sets $U_{\tau \upharpoonright n}$ for all positive integers $n$. Note that $x \in U_{\tau \upharpoonright n}$ for each $n$. Based on the induction process described earlier, observe these two facts. This sequence of open sets has diameters decreasing to zero. Each open set $U_{\tau \upharpoonright n}$ contains infinitely many other points of $C$ (this is because of all the open sets $U_{\tau \upharpoonright k}$ that are subsets of $U_{\tau \upharpoonright n}$ where $k \ge n$). Because the diameters are decreasing to zero, the sequence of $U_{\tau \upharpoonright n}$ is a local base at the point $x$. Thus, the point $x$ is a limit point of $C$. This completes the proof. $\blacksquare$

Theorem 4
Let $X$ be a compact metrizable space. It follows that $X$ is scattered if and only if $X$ is countable.

Proof of Theorem 4
$\Longleftarrow$
In this direction, we show that if $X$ is countable, then $X$ is scattered (the fact that can be shown using the function space argument pointed out earlier). Here, we show the contrapositive: if $X$ is not scattered, then $X$ is uncountable. Suppose $X$ is not scattered. Then every point of $X$ is a limit point of $X$. By Theorem 3, $X$ would contain a Cantor set $C$ of cardinality continuum.

$\Longrightarrow$
In this direction, we show that if $X$ is scattered, then $X$ is countable. We also show the contrapositive: if $X$ is uncountable, then $X$ is not scattered. Suppose $X$ is uncountable. By Theorem 2, all but countably many points of $X$ are limit points of $X$. After discarding these countably many isolated points, we still have a compact space. So we can just assume that every point of $X$ is a limit point of $X$. Then by Theorem 3, $X$ contains an uncountable closed set $C$ such that every point of $C$ is a limit point of $C$. This means that $X$ is not scattered. $\blacksquare$

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Remarks

A corollary to the above discussion is that the cardinality for any compact metrizable space is either countable (including finite) or continuum (the cardinality of the real line). There is nothing in between or higher than continuum. To see this, the cardinality of any Lindelof first countable space is at most continuum according to a theorem in this previous post (any compact metric space is one such). So continuum is an upper bound on the cardinality of compact metric spaces. Theorem 3 above implies that any uncountable compact metrizable space has to contain a Cantor set, hence has cardinality continuum. So the cardinality of a compact metrizable space can be one of two possibilities – countable or continuum. Even under the assumption of the negation of the continuum hypothesis, there will be no uncountable compact metric space of cardinality less than continuum. On the other hand, there is only one possibility for the cardinality of a scattered compact metrizable, which is countable.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2015 \text{ by Dan Ma}$

# Comparing two function spaces

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. It is a well known fact that any continuous real-valued function $f$ defined on either $\omega_1$ or $\omega_1+1$ is eventually constant, i.e., there exists some $\alpha<\omega_1$ such that the function $f$ is constant on the ordinals beyond $\alpha$. Now consider the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$. Thus individually, elements of these two function spaces appear identical. Any $f \in C_p(\omega_1)$ matches a function $f^* \in C_p(\omega_1+1)$ where $f^*$ is the result of adding the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. This fact may give the impression that the function spaces $C_p(\omega_1)$ and $C_p(\omega_1+1)$ are identical topologically. The goal in this post is to demonstrate that this is not the case. We compare the two function spaces with respect to some convergence properties (countably tightness and Frechet-Urysohn property) as well as normality.

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Tightness

One topological property that is different between $C_p(\omega_1)$ and $C_p(\omega_1+1)$ is that of tightness. The function space $C_p(\omega_1+1)$ is countably tight, while $C_p(\omega_1)$ is not countably tight.

Let $X$ be a space. The tightness of $X$, denoted by $t(X)$, is the least infinite cardinal $\kappa$ such that for any $A \subset X$ and for any $x \in X$ with $x \in \overline{A}$, there exists $B \subset A$ for which $\lvert B \lvert \le \kappa$ and $x \in \overline{B}$. When $t(X)=\omega$, we say that $X$ has countable tightness or is countably tight. When $t(X)>\omega$, we say that $X$ has uncountable tightness or is uncountably tight.

First, we show that the tightness of $C_p(\omega_1)$ is greater than $\omega$. For each $\alpha<\omega_1$, define $f_\alpha: \omega_1 \rightarrow \left\{0,1 \right\}$ such that $f_\alpha(\beta)=0$ for all $\beta \le \alpha$ and $f_\alpha(\beta)=1$ for all $\beta>\alpha$. Let $g \in C_p(\omega_1)$ be the function that is identically zero. Then $g \in \overline{F}$ where $F$ is defined by $F=\left\{f_\alpha: \alpha<\omega_1 \right\}$. It is clear that for any countable $B \subset F$, $g \notin \overline{B}$. Thus $C_p(\omega_1)$ cannot be countably tight.

The space $\omega_1+1$ is a compact space. The fact that $C_p(\omega_1+1)$ is countably tight follows from the following theorem.

Theorem 1
Let $X$ be a completely regular space. Then the function space $C_p(X)$ is countably tight if and only if $X^n$ is Lindelof for each $n=1,2,3,\cdots$.

Theorem 1 is a special case of Theorem I.4.1 on page 33 of [1] (the countable case). One direction of Theorem 1 is proved in this previous post, the direction that will give us the desired result for $C_p(\omega_1+1)$.

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The Frechet-Urysohn property

In fact, $C_p(\omega_1+1)$ has a property that is stronger than countable tightness. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space (see this previous post). Of course, $C_p(\omega_1)$ not being countably tight means that it is not a Frechet-Urysohn space.

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Normality

The function space $C_p(\omega_1+1)$ is not normal. If $C_p(\omega_1+1)$ is normal, then $C_p(\omega_1+1)$ would have countable extent. However, there exists an uncountable closed and discrete subset of $C_p(\omega_1+1)$ (see this previous post). On the other hand, $C_p(\omega_1)$ is Lindelof. The fact that $C_p(\omega_1)$ is Lindelof is highly non-trivial and follows from [2]. The author in [2] showed that if $X$ is a space consisting of ordinals such that $X$ is first countable and countably compact, then $C_p(X)$ is Lindelof.

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Embedding one function space into the other

The two function space $C_p(\omega_1+1)$ and $C_p(\omega_1)$ are very different topologically. However, one of them can be embedded into the other one. The space $\omega_1+1$ is the continuous image of $\omega_1$. Let $g: \omega_1 \longrightarrow \omega_1+1$ be a continuous surjection. Define a map $\psi: C_p(\omega_1+1) \longrightarrow C_p(\omega_1)$ by letting $\psi(f)=f \circ g$. It is shown in this previous post that $\psi$ is a homeomorphism. Thus $C_p(\omega_1+1)$ is homeomorphic to the image $\psi(C_p(\omega_1+1))$ in $C_p(\omega_1)$. The map $g$ is also defined in this previous post.

The homeomposhism $\psi$ tells us that the function space $C_p(\omega_1)$, though Lindelof, is not hereditarily normal.

On the other hand, the function space $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. Note that $C_p(\omega_1+1)$ is countably tight, which is a hereditary property.

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Remark

There is a mapping that is alluded to at the beginning of the post. Each $f \in C_p(\omega_1)$ is associated with $f^* \in C_p(\omega_1+1)$ which is obtained by appending the point $(\omega_1,a)$ to $f$ where $a$ is the eventual constant real value of $f$. It may be tempting to think of the mapping $f \rightarrow f^*$ as a candidate for a homeomorphism between the two function spaces. The discussion in this post shows that this particular map is not a homeomorphism. In fact, no other one-to-one map from one of these function spaces onto the other function space can be a homeomorphism.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.
2. Buzyakova, R. Z., In search of Lindelof $C_p$‘s, Comment. Math. Univ. Carolinae, 45 (1), 145-151, 2004.

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$\copyright \ 2014 \text{ by Dan Ma}$

# Cp(omega 1 + 1) is monolithic and Frechet-Urysohn

This is another post that discusses what $C_p(X)$ is like when $X$ is a compact space. In this post, we discuss the example $C_p(\omega_1+1)$ where $\omega_1+1$ is the first compact uncountable ordinal. Note that $\omega_1+1$ is the successor to $\omega_1$, which is the first (or least) uncountable ordinal. The function space $C_p(\omega_1+1)$ is monolithic and is a Frechet-Urysohn space. Interestingly, the first property is possessed by $C_p(X)$ for all compact spaces $X$. The second property is possessed by all compact scattered spaces. After we discuss $C_p(\omega_1+1)$, we discuss briefly the general results for $C_p(X)$.

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Initial discussion

The function space $C_p(\omega_1+1)$ is a dense subspace of the product space $\mathbb{R}^{\omega_1}$. In fact, $C_p(\omega_1+1)$ is homeomorphic to a subspace of the following subspace of $\mathbb{R}^{\omega_1}$:

$\Sigma(\omega_1)=\left\{x \in \mathbb{R}^{\omega_1}: x_\alpha \ne 0 \text{ for at most countably many } \alpha < \omega_1 \right\}$

The subspace $\Sigma(\omega_1)$ is the $\Sigma$-product of $\omega_1$ many copies of the real line $\mathbb{R}$. The $\Sigma$-product of separable metric spaces is monolithic (see here). The $\Sigma$-product of first countable spaces is Frechet-Urysohn (see here). Thus $\Sigma(\omega_1)$ has both of these properties. Since the properties of monolithicity and being Frechet-Urysohn are carried over to subspaces, the function space $C_p(\omega_1+1)$ has both of these properties. The key to the discussion is then to show that $C_p(\omega_1+1)$ is homeopmophic to a subspace of the $\Sigma$-product $\Sigma(\omega_1)$.

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Connection to $\Sigma$-product

We show that the function space $C_p(\omega_1+1)$ is homeomorphic to a subspace of the $\Sigma$-product of $\omega_1$ many copies of the real lines. Let $Y_0$ be the following subspace of $C_p(\omega_1+1)$:

$Y_0=\left\{f \in C_p(\omega_1+1): f(\omega_1)=0 \right\}$

Every function in $Y_0$ has non-zero values at only countably points of $\omega_1+1$. Thus $Y_0$ can be regarded as a subspace of the $\Sigma$-product $\Sigma(\omega_1)$.

By Theorem 1 in this previous post, $C_p(\omega_1+1) \cong Y_0 \times \mathbb{R}$, i.e, the function space $C_p(\omega_1+1)$ is homeomorphic to the product space $Y_0 \times \mathbb{R}$. On the other hand, the product $Y_0 \times \mathbb{R}$ can also be regarded as a subspace of the $\Sigma$-product $\Sigma(\omega_1)$. Basically adding one additional factor of the real line to $Y_0$ still results in a subspace of the $\Sigma$-product. Thus we have:

$C_p(\omega_1+1) \cong Y_0 \times \mathbb{R} \subset \Sigma(\omega_1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

Thus $C_p(\omega_1+1)$ possesses all the hereditary properties of $\Sigma(\omega_1)$. Another observation we can make is that $\Sigma(\omega_1)$ is not hereditarily normal. The function space $C_p(\omega_1+1)$ is not normal (see here). The $\Sigma$-product $\Sigma(\omega_1)$ is normal (see here). Thus $\Sigma(\omega_1)$ is not hereditarily normal.

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A closer look at $C_p(\omega_1+1)$

In fact $C_p(\omega_1+1)$ has a stronger property that being monolithic. It is strongly monolithic. We use homeomorphic relation in (1) above to get some insight. Let $h$ be a homeomorphism from $C_p(\omega_1+1)$ onto $Y_0 \times \mathbb{R}$. For each $\alpha<\omega_1$, let $H_\alpha$ be defined as follows:

$H_\alpha=\left\{f \in C_p(\omega_1+1): f(\gamma)=0 \ \forall \ \alpha<\gamma<\omega_1 \right\}$

Clearly $H_\alpha \subset Y_0$. Furthermore $H_\alpha$ can be considered as a subspace of $\mathbb{R}^\omega$ and is thus metrizable. Let $A$ be a countable subset of $C_p(\omega_1+1)$. Then $h(A) \subset H_\alpha \times \mathbb{R}$ for some $\alpha<\omega_1$. The set $H_\alpha \times \mathbb{R}$ is metrizable. The set $H_\alpha \times \mathbb{R}$ is also a closed subset of $Y_0 \times \mathbb{R}$. Then $\overline{A}$ is contained in $H_\alpha \times \mathbb{R}$ and is therefore metrizable. We have shown that the closure of every countable subspace of $C_p(\omega_1+1)$ is metrizable. In other words, every separable subspace of $C_p(\omega_1+1)$ is metrizable. This property follows from the fact that $C_p(\omega_1+1)$ is strongly monolithic.

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Monolithicity and Frechet-Urysohn property

As indicated at the beginning, the $\Sigma$-product $\Sigma(\omega_1)$ is monolithic (in fact strongly monolithic; see here) and is a Frechet-Urysohn space (see here). Thus the function space $C_p(\omega_1+1)$ is both strongly monolithic and Frechet-Urysohn.

Let $\tau$ be an infinite cardinal. A space $X$ is $\tau$-monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$, we have $nw(\overline{A}) \le \tau$. A space $X$ is monolithic if it is $\tau$-monolithic for all infinite cardinal $\tau$. It is straightforward to show that $X$ is monolithic if and only of for every subspace $Y$ of $X$, the density of $Y$ equals to the network weight of $Y$, i.e., $d(Y)=nw(Y)$. A longer discussion of the definition of monolithicity is found here.

A space $X$ is strongly $\tau$-monolithic if for any $A \subset X$ with $\lvert A \lvert \le \tau$, we have $w(\overline{A}) \le \tau$. A space $X$ is strongly monolithic if it is strongly $\tau$-monolithic for all infinite cardinal $\tau$. It is straightforward to show that $X$ is strongly monolithic if and only if for every subspace $Y$ of $X$, the density of $Y$ equals to the weight of $Y$, i.e., $d(Y)=w(Y)$.

In any monolithic space, the density and the network weight coincide for any subspace, and in particular, any subspace that is separable has a countable network. As a result, any separable monolithic space has a countable network. Thus any separable space with no countable network is not monolithic, e.g., the Sorgenfrey line. On the other hand, any space that has a countable network is monolithic.

In any strongly monolithic space, the density and the weight coincide for any subspace, and in particular any separable subspace is metrizable. Thus being separable is an indicator of metrizability among the subspaces of a strongly monolithic space. As a result, any separable strongly monolithic space is metrizable. Any separable space that is not metrizable is not strongly monolithic. Thus any non-metrizable space that has a countable network is an example of a monolithic space that is not strongly monolithic, e.g., the function space $C_p([0,1])$. It is clear that all metrizable spaces are strongly monolithic.

The function space $C_p(\omega_1+1)$ is not separable. Since it is strongly monolithic, every separable subspace of $C_p(\omega_1+1)$ is metrizable. We can see this by knowing that $C_p(\omega_1+1)$ is a subspace of the $\Sigma$-product $\Sigma(\omega_1)$, or by using the homeomorphism $h$ as in the previous section.

For any compact space $X$, $C_p(X)$ is countably tight (see this previous post). In the case of the compact uncountable ordinal $\omega_1+1$, $C_p(\omega_1+1)$ has the stronger property of being Frechet-Urysohn. A space $Y$ is said to be a Frechet-Urysohn space (also called a Frechet space) if for each $y \in Y$ and for each $M \subset Y$, if $y \in \overline{M}$, then there exists a sequence $\left\{y_n \in M: n=1,2,3,\cdots \right\}$ such that the sequence converges to $y$. As we shall see below, $C_p(X)$ is rarely Frechet-Urysohn.

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General discussion

For any compact space $X$, $C_p(X)$ is monolithic but does not have to be strongly monolithic. The monolithicity of $C_p(X)$ follows from the following theorem, which is Theorem II.6.8 in [1].

Theorem 1
Then the function space $C_p(X)$ is monolithic if and only if $X$ is a stable space.

See chapter 3 section 6 of [1] for a discussion of stable spaces. We give the definition here. A space $X$ is stable if for any continuous image $Y$ of $X$, the weak weight of $Y$, denoted by $ww(Y)$, coincides with the network weight of $Y$, denoted by $nw(Y)$. In [1], $ww(Y)$ is notated by $iw(Y)$. The cardinal function $ww(Y)$ is the minimum cardinality of all $w(T)$, the weight of $T$, for which there exists a continuous bijection from $Y$ onto $T$.

All compact spaces are stable. Let $X$ be compact. For any continuous image $Y$ of $X$, $Y$ is also compact and $ww(Y)=w(Y)$, since any continuous bijection from $Y$ onto any space $T$ is a homeomorphism. Note that $ww(Y) \le nw(Y) \le w(Y)$ always holds. Thus $ww(Y)=w(Y)$ implies that $ww(Y)=nw(Y)$. Thus we have:

Corollary 2
Let $X$ be a compact space. Then the function space $C_p(X)$ is monolithic.

However, the strong monolithicity of $C_p(\omega_1+1)$ does not hold in general for $C_p(X)$ for compact $X$. As indicated above, $C_p([0,1])$ is monolithic but not strongly monolithic. The following theorem is Theorem II.7.9 in [1] and characterizes the strong monolithicity of $C_p(X)$.

Theorem 3
Let $X$ be a space. Then $C_p(X)$ is strongly monolithic if and only if $X$ is simple.

A space $X$ is $\tau$-simple if whenever $Y$ is a continuous image of $X$, if the weight of $Y$ $\le \tau$, then the cardinality of $Y$ $\le \tau$. A space $X$ is simple if it is $\tau$-simple for all infinite cardinal numbers $\tau$. Interestingly, any separable metric space that is uncountable is not $\omega$-simple. Thus $[0,1]$ is not $\omega$-simple and $C_p([0,1])$ is not strongly monolithic, according to Theorem 3.

For compact spaces $X$, $C_p(X)$ is rarely a Frechet-Urysohn space as evidenced by the following theorem, which is Theorem III.1.2 in [1].

Theorem 4
Let $X$ be a compact space. Then the following conditions are equivalent.

1. $C_p(X)$ is a Frechet-Urysohn space.
2. $C_p(X)$ is a k-space.
3. The compact space $X$ is a scattered space.

A space $X$ is a scattered space if for every non-empty subspace $Y$ of $X$, there exists an isolated point of $Y$ (relative to the topology of $Y$). Any space of ordinals is scattered since every non-empty subset has a least element. Thus $\omega_1+1$ is a scattered space. On the other hand, the unit interval $[0,1]$ with the Euclidean topology is not scattered. According to this theorem, $C_p([0,1])$ cannot be a Frechet-Urysohn space.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A useful representation of Cp(X)

Let $X$ be a completely regular space. The space $C_p(X)$ is the space of all real-valued continuous functions defined on $X$ endowed with the pointwise convergence topology. In this post, we show that $C_p(X)$ can be represented as the product of a subspace of $C_p(X)$ with the real line $\mathbb{R}$. We prove the following theorem. See here for an application of this theorem.

Theorem 1
Let $X$ be a completely regular space. Let $x \in X$. Let $Y$ be defined by:

$Y=\left\{f \in C_p(X): f(x)=0 \right\}$

Then $C_p(X)$ is homeomorphic to $Y \times \mathbb{R}$.

The above theorem can be found in [1] (see Theorem I.5.4 on p. 37). In [1], the homeomorphism is stated without proof. For the sake of completeness, we provide a detailed proof of Theorem 1.

Proof of Theorem 1
Define $h: C_p(X) \rightarrow Y \times \mathbb{R}$ by $h(f)=(f-f(x),f(x))$ for any $f \in C_p(X)$. The map $h$ is a homeomorphism.

The map is one-to-one

First, we show that it is a one-to-one map. Let $f,g \in C_p(X)$ where $f \ne g$. Assume that $f(x) \ne g(x)$. Then $h(f) \ne h(g)$. So assume that $f(x)=g(x)$. Then the functions $f-f(x)$ and $g-g(x)$ are different, which means $h(f) \ne h(g)$.

The map is onto

Now we show $h$ maps $C_p(X)$ onto $Y \times \mathbb{R}$. Let $(g,t) \in Y \times \mathbb{R}$. Let $f=g+t$. Note that $f(x)=g(x)+t=t$. Then $f-f(x)=g$. We have $h(f)=(g,t)$.

Note. Showing the continuity of $h$ and $h^{-1}$ is a matter of working with the basic open sets in the function space carefully (e.g. making the necessary shifting). Some authors just skip the details and declare them continuous, e.g. [1]. Readers are welcome to work out enough of the details to see the key idea.

The map is continuous

Show that $h$ is continuous. Let $f \in C_p(X)$. Let $U \times V$ be an open set in $Y \times \mathbb{R}$ such that $h(f) \in U \times V$ and,

$U=\left\{g \in Y: \forall \ i=1,\cdots,n, g(x_i) \in U_i \right\}$

$\forall \ i=1,\cdots,n, \ U_i=(f(x_i)-f(x)-\frac{1}{k},f(x_i)-f(x)+\frac{1}{k})$

$V=(f(x)-\frac{1}{k},f(x)+\frac{1}{k})$

where $x_1,\cdots,x_n$ are arbitrary points in $X$ and $k$ is some large positive integer. Define the following:

$\forall \ i=1,\cdots,n, \ W_i=(f(x_i)-\frac{1}{2k},f(x_i)+\frac{1}{2k})$

$W_{n+1}=(f(x)-\frac{1}{2k},f(x)+\frac{1}{2k})$

$x_{n+1}=x$

Then define the open set $W$ as follows:

$W=\left\{q \in C_p(X): \forall \ i=1,\cdots,n,n+1, q(x_i) \in W_i \right\}$

Clearly $f \in W$. We need to show $h(W) \subset U \times V$. Let $q \in W$. Then $h(q)=(q-q(x),q(x))$. We need to show that $q-q(x) \in U$ and $q(x) \in V$. Note that $q(x_{n+1})=q(x) \in W_{n+1}$. For each $i=1,\cdots,n$, $q(x_i) \in W_i$. So we have the following:

$f(x_i)-\frac{1}{2k}

$f(x)-\frac{1}{2k}

Subtracting the above two inequalities, we have the following:

$f(x_i)-f(x)-\frac{1}{k}

The above inequality shows that for each $i=1,\cdots,n$, $q(x_i) -q(x) \in U_i$. Hence $q-q(x) \in U$. It is clear that $q(x) \in V$. This completes the proof that the map $h$ is continuous.

The inverse is continuous

We now show that $h^{-1}$ is continuous. Let $(g,t) \in Y \times \mathbb{R}$. Note that $h^{-1}(g,t)=g+t$. Let $M$ be an open set in $C_p(X)$ such that $g+t \in M$ and

$M=\left\{f \in C_p(X): \forall \ i=1,\cdots,n+1, f(x_i) \in M_i \right\}$

$\forall \ i=1,\cdots,n, \ M_i=(g(x_i)+t-\frac{1}{m},g(x_i)+t+\frac{1}{m})$

$x_{n+1}=x$

$M_{n+1}=(t-\frac{1}{m},t+\frac{1}{m})$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and $m$ is some large positive integer. Now define an open subset $G \times T$ of $Y \times \mathbb{R}$ such that $(g,t) \in G \times T$ and

$G=\left\{q \in Y: \forall \ i=1,\cdots,n+1, q(x_i) \in G_i \right\}$

$\forall \ i=1,\cdots,n, \ G_i=(g(x_i)-\frac{1}{2m},g(x_i)+\frac{1}{2m})$

$T=(t-\frac{1}{2m},t+\frac{1}{2m})$

We need to show that $h^{-1}(G \times T) \subset M$. Let $(q,a) \in G \times T$. We then have the following inequalities.

$\forall \ i=1,\cdots,n, \ g(x_i)-\frac{1}{2m}

$t-\frac{1}{2m}

Adding the above two inequalities, we obtain:

$\forall \ i=1,\cdots,n, \ g(x_i)+t-\frac{1}{m}

The above implies that $\forall \ i=1,\cdots,n$, $q(x_i)+a \in M_i$. It is clear that $q(x_{n+1})+a=q(x)+a=a \in M_{n+1}$. Thus $q+a \in M$. This completes the proof that $h^{-1}$ is continuous.

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Reference

1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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$\copyright \ 2014 \text{ by Dan Ma}$

# A useful embedding for Cp(X)

Let $X$ be a Tychonoff space (also called completely regular space). By $C_p(X)$ we mean the space of all continuous real-valued functions defined on $X$ endowed with the pointwise convergence topology. In this post we discuss a scenario in which a function space can be embedded into another function space. We prove the following theorem. An example follows the proof.

Theorem 1
Suppose that the space $Y$ is a continuous image of the space $X$. Then $C_p(Y)$ can be embedded into $C_p(X)$.

Proof of Theorem 1
Let $t:X \rightarrow Y$ be a continuous surjection, i.e., $t$ is a continuous function from $X$ onto $Y$. Define the map $\psi: C_p(Y) \rightarrow C_p(X)$ by $\psi(f)=f \circ t$ for all $f \in C_p(Y)$. We show that $\psi$ is a homeomorphism from $C_p(Y)$ into $C_p(X)$.

First we show $\psi$ is a one-to-one map. Let $f,g \in C_p(Y)$ with $f \ne g$. There exists some $y \in Y$ such that $f(y) \ne g(y)$. Choose some $x \in X$ such that $t(x)=y$. Then $f \circ t \ne g \circ t$ since $(f \circ t)(x)=f(t(x))=f(y)$ and $(g \circ t)(x)=g(t(x))=g(y)$.

Next we show that $\psi$ is continuous. Let $f \in C_p(Y)$. Let $U$ be open in $C_p(X)$ with $\psi(f) \in U$ such that

$U=\left\{q \in C_p(X): \forall \ i=1,\cdots,n, \ q(x_i) \in U_i \right\}$

where $x_1,\cdots,x_n$ are arbitrary points of $X$ and each $U_i$ is an open interval of the real line $\mathbb{R}$. Note that for each $i$, $f(t(x_i)) \in U_i$. Now consider the open set $V$ defined by:

$V=\left\{r \in C_p(Y): \forall \ i=1,\cdots,n, \ r(t(x_i)) \in U_i \right\}$

Clearly $f \in V$. It follows that $\psi(V) \subset U$ since for each $r \in V$, it is clear that $\psi(r)=r \circ t \in U$.

Now we show that $\psi^{-1}: \psi(C_p(Y)) \rightarrow C_p(Y)$ is continuous. Let $\psi(f)=f \circ t \in \psi(C_p(Y))$ where $f \in C_p(Y)$. Let $G$ be open with $\psi^{-1}(f \circ t)=f \in G$ such that

$G=\left\{r \in C_p(Y): \forall \ i=1,\cdots,m, \ r(y_i) \in G_i \right\}$

where $y_1,\cdots,y_m$ are arbitrary points of $Y$ and each $G_i$ is an open interval of $\mathbb{R}$. Choose $x_1,\cdots,x_m \in X$ such that $t(x_i)=y_i$ for each $i$. We have $f(t(x_i)) \in G_i$ for each $i$. Define the open set $H$ by:

$H=\left\{q \in \psi(C_p(Y)) \subset C_p(X): \forall \ i=1,\cdots,m, \ q(x_i) \in G_i \right\}$

Clearly $f \circ t \in H$. Note that $\psi^{-1}(H) \subset G$. To see this, let $r \circ t \in H$ where $r \in C_p(Y)$. Now $r(t(x_i))=r(y_i) \in G_i$ for each $i$. Thus $\psi^{-1}(r \circ t)=r \in G$. It follows that $\psi^{-1}$ is continuous. The proof of the theorem is now complete. $\blacksquare$

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Example

The proof of Theorem 1 is not difficult. It is a matter of notating carefully the open sets in both function spaces. However, the embedding makes it easy in some cases to understand certain function spaces and in some cases to relate certain function spaces.

Let $\omega_1$ be the first uncountable ordinal, and let $\omega_1+1$ be the successor ordinal to $\omega_1$. Furthermore consider these ordinals as topological spaces endowed with the order topology. As an application of Theorem 1, we show that $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. Define a continuous surjection $g:\omega_1 \rightarrow \omega_1+1$ as follows:

$g(\gamma) = \begin{cases} \omega_1 & \mbox{if } \ \gamma =0 \\ \gamma-1 & \mbox{if } \ 1 \le \gamma < \omega \\ \gamma & \mbox{if } \ \omega \le \gamma < \omega_1 \end{cases}$

The map $g$ is continuous from $\omega_1$ onto $\omega_1+1$. By Theorem 1, $C_p(\omega_1+1)$ can be embedded as a subspace of $C_p(\omega_1)$. On the other hand, $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. The function space $C_p(\omega_1+1)$ is a Frechet-Urysohn space, which is a property that is carried over to any subspace. The function $C_p(\omega_1)$ is not Frechet-Urysohn. Thus $C_p(\omega_1)$ cannot be embedded in $C_p(\omega_1+1)$. A further comparison of these two function spaces is found in this subsequent post.

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$\copyright \ 2014 \text{ by Dan Ma}$