# Sequentially compact spaces, I

All spaces under consideration are Hausdorff. Countably compactness and sequentially compactness are notions related to compactness. A countably compact space is one in which every counable open cover has a finite subcover, or equivalently, every countably infinite subset has a limit point. The limit points contemplated here are from the topological point of view, i.e. the point $p \in X$ is a limit point of $A \subset X$ if every open subset of $X$ containing $p$ contains a point of $A$ distinct from $p$. On the other hand, a space $X$ is sequentially compact if every sequence $\left\{x_n:n=1,2,3,\cdots\right\}$ of points of $X$ has a subsequence that converges. We present examples showing that the notion of sequentially compactness is different from compactness.

Let $\omega_1$ be the first uncountable ordinal. The space of all countable ordinals $W=[0,\omega_1)$ with the ordered topology is sequentially compact and not compact. Let $\left\{w_n\right\}$ be a sequence of points in $W$. Let $A=\left\{w_n:n=1,2,3,\cdots\right\}$. If $A$ is finite, then the sequence $\left\{w_n\right\}$ is eventually constant and thus has a convergent subsequence. So assume $A$ is an infinite set. Then we can choose an increasing sequence of integers $n(1),n(2),n(3),\cdots$ such that $w_{n(1)}. Let $\alpha<\omega_1$ be the least upper bound of all $w_{n(j)}$. Then subsequence $w_{n(j)}$ converges to $\alpha$.

The notion of sequentially compactness is not to be confused with the notion of being a sequential space. The space $[0,\omega_1]=\omega_1+1$, the space of countable ordinals with one additional point $\omega_1$ at the end, is a sequentially compact space for the same reason that $[0,\omega_1)$ is sequentially compact. However, $[0,\omega_1]$ is not sequential. Note that $[0,\omega_1)$ is a sequentially closed set but not closed in $[0,\omega_1]$. On the other hand, being a sequential space does not imply compactness or sequentially compactness, e.g. the real line $\mathbb{R}$.

For discussion of sequential spaces, see Sequential spaces, I, Sequential spaces, II and Sequential spaces, III.

We now present an example of a compact space that is not sequentially compact. Let $I=[0,1]$ be the unit interval. Let $2=\left\{0,1\right\}$, the two-point discrete space. Let $X=2^{I}$ be the product space of uncountably many copies of $2=\left\{0,1\right\}$ indexed by $I$. We show that $X$ is not sequentially compact. To this end, we define a sequence $\left\{f_n\right\}$ that has not convergent subsequence.

For any $y \in \mathbb{R}$, let $[y]$ be the greatest integer less than or equal to $y$. For each $t \in I$ and for each $n=1,2,3,\cdots$, let $t_n$ be:

$t_n=10^n t-[10^n t]$.

For example, if $q=\frac{1}{\sqrt{2}}=0.7071067811 \cdots$, then $q_1=0.071067811 \cdots$, $q_2=0.71067811 \cdots$ and $q_3=0.1067811 \cdots$. For each $n=1,2,3,\cdots$, define $f_n:I \mapsto 2$ by the following:

$\displaystyle f_n(t)=\left\{\begin{matrix}0&\thinspace t_n <0.5 \\{1}&\thinspace t_n \ge 0.5 \end{matrix}\right.$

With the above example, $f_1(q)=0$, $f_2(q)=1$, $f_3(q)=0$, $f_4(q)=0$ and so on. In general, if the $(n+1)^{st}$ decimal place of the number $t$ is less then $5$, then $f_n(t)=0$. Otherwise $f_n(t)=1$.

Let’s observe that if $g_n \in X=2^{I}$ converges to $g \in X$, then $g_n(t) \in X=2^{I}$ converges to $g(t) \in X$ for each $t \in I$ (hence the product topology is called the topology of pointwise convergence). We claim that the sequence $\left\{f_n\right\}$ has no convergence subsequence. To this end, we show that each subsequence of $\left\{f_n\right\}$ does not converge at some $t \in I$.

Let $n(1) be any increasing sequence of positive integers. We define $t \in I$ such that $f_{n(1)}(t),f_{n(2)}(t),f_{n(3)}(t),\cdots$ is an alternating sequence of zeros and ones. Consider $t \in I$ satisfying the following:

For each $j \le n(1)$, the $j^{th}$ decimal place of $t$ is $9$,

For each $n(1), the $j^{th}$ decimal place of $t$ is $1$,

For each $n(2), the $j^{th}$ decimal place of $t$ is $9$ and so on.

For example, if $n(1)=2$, $n(2)=5$ and $n(3)=9$, then let $t=0.9911199991 \cdots$. With this in mind, $f_{n(1)}(t),f_{n(2)}(t),f_{n(3)}(t),\cdots$ is an alternating sequence of zeros and ones.

Reference

1. Engelking, R. General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# A Note On The Sorgenfrey Line

The Sorgenfrey Line is a topological space whose underlying space is the real line. The topology is generated by the basis of the half open intervals $[a, b)$ where $a$ and $b$ are real numbers. For students of topology, the Sorgenfrey Line is a handy example of (1) “Lindelof x Lindelof” does not have to be Lindelof, (2) “normal x normal” does not have to be normal, (3) “paracompact x paracompact” does not have to be paracompact, (4) “perfectly normal x perfectly normal” does not have to be perfectly normal (does not even have to be normal). In other words, these four properties are not preserved by taking Cartesian product. The goal of this note is to prove these and a few other results about the Sorgenfrey Line. In this note, $S$ is to denote the Sorgenfrey Line.

We will show these results:

A
$S$ is Lindelof (thus is normal and paracompact).
B
$S$ is hereditarily Lindelof.
C
Compact subsets of $S$ are countable. Thus $S$ is an example of a space that is Lindelof and not $\sigma$-compact.
D
$S \times S$ is not Lindelof.
E
$S \times S$ is not normal.
F
$S \times S$ is not paramcompact.
G
$S$ is not second countable, thus not metrizable.
H
$S$ is perfectly normal.

In proving C, we will use the following lemma (Lemma 1), which was proved in a previous post. This is a special countable extent property of the real line. Note that a space $X$ has extent of cardinality $\mathcal{K}$ if $\mathcal{K}$ is the least upper bound on the sizes of all closed and discrete subsets of $X$. In proving F, the Jones Lemma will be used. This lemma is essentially saying that the extent of a separable normal space cannot be the cardinality continuum or greater. A space $X$  is perfectly normal if $X$ is normal and every closed subset is a $G_{\delta}$ set (equivalently every open subset is an $F_{\sigma}$ set). These two lemmas are stated below.

Lemma 1
Every uncountable subset of $\mathbb{R}$ has a two-sided limit point.
Lemma 2 (Jones’ Lemma)
If $X$ is a separable normal space, then it has no closed and discrete subset of cardinality continuum.

See this post for a proof of Jones’ Lemma.

Proof of A. Let $\mathcal{A}$ be an open cover of $S$ consisting of open intervals of the form $[a, b)$.

Let $T=\cup\lbrace{(a, b): [a, b)\epsilon\mathcal{A}}\rbrace$. We claim that $U=S - T$ is a countable set. Suppose that $U$ is uncountable. By Lemma 1, there exists a real number $p$ that is a two-sided limit point of $U$. This means that for every open interval $(s,t)$ (open interval in the usual topology on the real line) with $p \in (s,t)$, the interval $(s,t)$ contains points of $U$ on the left side of $p$ as well as on the right side of $p$.

Choose some $[a,b) \in \mathcal{A}$ such that $p \in [a,b)$. Note that $(a,b)$ is a subset of $T$ and so should not contain points of $U=S-T$. Because $p$ is a two-sided limit point of $U$, $(a,b)$ will contain points of $U$, a contradiction. So $U$ must be countable.

Note that $T$ is an open set in the usual topology, which is Lindelof. So we can find countably many $[c,d) \in \mathcal{A}$ such that the union of all such $(c,d)$ covers $T$. Then find countably many $[c,d) \in \mathcal{A}$ that cover the countably many points in $U=S-T$. Combining both sets of $[c,d)$, we see that $\mathcal{A}$ has a countable subcover. $\blacksquare$

Proof of B. Take any uncountable subspace of $S$, we can apply the same proof as in A.

Proof of C. Let $A\subset{S}$ be a compact subspace that is uncoutable. By Lemma 1, $A$ has a two-sided limit point $y$ (i.e. it is both a left-sided limit point and a right-sided limit point in the usual topology). Let $\lbrace{y_n}\rbrace$ be a sequence of points in $A$ that converges to $y$ from the left. Then $\lbrace{(\infty,y_n)}\rbrace$  and $[y,\infty)$ form an open cover of $A$ that has no finite subcover. Thus all compact subspaces of the Sorgenfrey Line are countable. Furthermore $S$ is an example of a Lindelof but not $\sigma$-compact space.

Proof of D. If $S \times S$ is Lindelof, then it would be a separable normal space and cannot have closed and discrete subset of cardinality continuum or greater (according to Jones’ Lemma). Note that $D=\lbrace{(x,-x): x\epsilon{S}}\rbrace$ is a closed and discrete subset of $S \times S$, which has cardinality continuum. This shows that $S \times S$ cannot be Lindeloff. $\blacksquare$

Proof of E and F. For E, use the same argument as in the proof of D. For F, note that paracompactness implies normality. $\blacksquare$

Proof of G. If $S$ is second countable (has a countable base), then it would be a separable metrizable space and $S \times S$ would also be a separable metrizable space. But $S \times S$ is not even Lindeloff. Thus the Sorgenfrey Line cannot be second countable. $\blacksquare$

Proof of H. Let $W$ be an open subset of $S$. Let $O$ be the interor of $W$ in the usual topology. By a similar argument in the proof of A above, we can show that $W-O$ is countable. Since $O$ is an open set in the usual topology, it is an $F_{\sigma}$ set in the usual topology (and thus in the Sorgenfrey topology). It follows that $O$ plus countably many points would form an $F_{\sigma}$ set. Thus every open subset of the Sorgenfrey line $S$ is a $G_\delta$-set. Since the Sorgenfrey line is Lindelof, it is normal. Thus the Sorgenfrey line is a normal space in which every open set is a $G_\delta$-set (i.e. it is perfectly normal). $\blacksquare$

# A Countable Extent Property Unique to the Real Line

Being a separable metrizable space, the real line with the usual topology has countable extent. A space $X$ has countable extent if all closed and discrete subsets of $X$ are countable. In general, the extent of a space is the least cardinality of a closed and discrete set in that space. In this brief note, we show that the real line has another countable extent property that is unique to the real line. We show that every uncountable subset of the real line has a two-sided limit point. As an application, it follows from this fact that the Sorgenfrey line is Lindelof and that all compact subspaces in the Sorgenfrey Line are countable (see the blog post A Note On The Sorgenfrey Line).

Given a space $X$ and given $p \in {X}$, the point $p$ is a limit point of a subset $A\subset{X}$ if every open set containing $p$ contains a point of $A$ distinct from $p$. Equivalently, we can replace open set in this definition with members of a base (basic open sets). Thus, the limit here is from a topological perspective and not from a metric space perspective. A space $X$ has countable extent if all closed and discrete subsets are countable. This is equivalent to the statement that every uncountable subset has a limit point.

In the real line, we say that the point $p$ is a limit point of $A\subset{\mathbb{R}}$ if every open interval $(a, b)$ containing $p$ contains a point of $A$ distinct from $p$. We say $p$ is a two-sided limit point of $A\subset{\mathbb{R}}$ if every open interval $(a, b)$ containing $p$ contains points of $A$ on both sides of $p$ (i.e. both intervals $(a, p)$ and $(p, b)$ contain points of $A$). Likewise, one-sided (left-sided, right-sided) limit point means every open interval containing $p$ contains points of $A$ on one side (left side, right side, respectively).

Lemma
Every uncountable subset of $\mathbb{R}$ has a two-sided limit point.

Proof. Let $A\subset{\mathbb{R}}$ be uncountable. The set $A$ as a topological space is a Lindelof space. Thus $A$ has a limit point (in fact, has uncountably many limit points).

Suppose that $A$ has no two-sided limit points. Then the uncountably many limit points of $A$ must be either left-sided or right-sided limit points. We assume the case that $A$ has uncountably many right sided limit points. The proof for the left-sided case is similar. Let $B$ be the set of all right-sided limit points of $A$.

Since points in $B$ are right-sided limits of $A$, for each $x \in {B}$, there is a rational number $a_x$ such that $(a_x, x)\cap{A}=\phi$. Matching up the rational numbers with uncountably many points in $B$, there is a rational number $r$ such that the following set $C$ is uncountable.

$C=\lbrace{x \in {B}: r=a_x}\rbrace$

The uncountable set $C$ has a limit point $y$. Note that $y$ is a limit point of $A$. For the rational number $r$ indicated above, it must be the case $r < y$. Note that all points of $C$ are to the right of $r$. If $r \ge y$, then there would no points of $C$ in an open interval of $y$, which goes against the fact that $y$ is a limit point of $C$.

By a similar argument as in the preceding paragraph, there are no points of $C$ on the right side of $y$. If there is some $x \in {C}$ and $y < x$ then $(a_x, x)=(r, x)$ is an open interval containing the point $y$ that contains no points of $A$, contradicting the fact that $y$ is a limit point of $A$. Thus $y$ is only a left-sided limit point of $C$.

Choose two points $w$ and $z$ from $C$ such that $r < w < z < y$. We know that $(a_z, z)=(r, z)$ contains no points of $A$. It follows that $(w, z)$ contains no points of $A$. Since $w$ is a right-sided limit point of $A$, $(w, z)$ would contain points of $A$, a contradiction. Thus the set $A$ must have a two-sided limit point and the lemma is proved. $\blacksquare$