# Thinking about the Space of Irrationals Topologically

Let $\mathbb{R}$ denote the real number line and $\mathbb{P}$ denote the set of all irrational numbers. The irrational numbers and the set $\mathbb{P}$ occupy an important place in mathematics. The set $\mathbb{P}$ with the Euclidean topology inherited from the full real line is a topological space in its own right. Thus the space $\mathbb{P}$ has some of the same properties inherited from the Euclidean real line, e.g., just to name a few, it is hereditarily separable, hereditarily Lindelof, paracompact and metrizable. The space of the irrational numbers $\mathbb{P}$ has many properties apart from the full real line (e.g. $\mathbb{P}$ is totally disconnected). In this post, we look at a topological description of the space $\mathbb{P}$.

Let $\omega$ be the set of all nonnegative integers. Then the space $\mathbb{P}$ of irrational numbers is topologically equivalent (i.e. homeomorphic to) the product space $\prod \limits_{i=0}^\infty X_i$ where each $X_i=\omega$ has the discrete topology. We will also denote the product space $\prod \limits_{i=0}^\infty X_i$ by $\omega^\omega$. We have the following theorem.

Theorem
The space $\mathbb{P}$ of irrational numbers is homeomorphic to the product space $\omega^\omega$.

Because of this theorem, we can look at irrational numbers as sequences of nonnegative integers. Specifically each irrational number can be identified by a unique sequence of nonnegative integers. We can think of each such unique sequence as an address to locate an irrational number. In the remainder of the post, we describe at a high level how to define the unique addresses, which will also give us a homeomorphic map between the space $\mathbb{P}$ and the product space $\omega^\omega$.

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Step 0
Put the rational numbers in a sequence $r_0,r_1,r_2,r_3,\cdots$ such that $r_0=0$. We divide the real line into countably many non-overlapping intervals. Specifically, let $A_0=[0,1]$, $A_1=[-1,0]$, $A_2=[1,2]$, etc (see the following figure).

To facilitate the remaining construction, we denote the left endpoint of the interval $A_i$ by $L_i$ and denote the right endpoint by $R_i$.

Step 1
In each of the interval $A_i$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_i$ are $L_i$ and $R_i$, respectively. We choose a sequence $x_{i,0}, x_{i,1}, x_{i,2},\cdots$ of rational numbers converging to the right endpoint $R_i$. Then let $A_{i,0}=[L_i,x_{i,0}]$, $A_{i,1}=[x_{i,0},x_{i,1}]$, $A_{i,2}=[x_{i,1},x_{i,2}]$, etc (see the following figure).

Two important points to consider in Step $1$. One is that we make sure the rational number $r_1$ is chosen as an endpoint of some interval in Step 1. The second is that the length of each $A_{i,j}$ is less than $\displaystyle \frac{1}{2^1}$.

Step 2
In each of the interval $A_{i,j}$, we further divide it into non-overlapping intervals. The left endpoint and the right endpoint of the interval $A_{i,j}$ are $L_{i,j}$ and $R_{i,j}$, respectively. We choose a sequence $x_{i,j,0}, x_{i,j,1}, x_{i,j,2},\cdots$ of rational numbers converging to the left endpoint $L_{i,j}$. Then let $A_{i,j,0}=[x_{i,j,0},R_{i,j}]$, $A_{i,j,1}=[x_{i,j,1},x_{i,j,0}]$, $A_{i,j,2}=[x_{i,j,2},x_{i,j,1}]$, etc (see the following figure).

As in the previous step, two important points to consider in Step $2$. One is that we make sure the rational number $r_2$ is chosen as an endpoint of some interval in Step 2. The second is that the length of each $A_{i,j,k}$ is less than $\displaystyle \frac{1}{2^2}$.

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Remark

In the process described above, the endpoints of the intervals $A_{f(0),\cdots,f(n)}$ are rational numbers and we make sure that all the rational numbers are used as endpoints. We also make sure that the intervals from the successive steps are nested closed intervals with lengths $\rightarrow 0$. The consequence of this point is that the nested decreasing closed intervals will collapse to one single point (since the real line is a complete metric space) and this single point must be an irrational number (since all the rational numbers are used up as endpoints of the nested closed intervals).

In Step $i$ where $i$ is an odd integer, we make the endpoints of the new intervals converge to the right. In Step $i$ where $i>1$ is an even integer, we make the endpoints of the new intervals converge to the left. This manipulation is to ensure that the nested closed intervals will never share the same endpoint from one step all the way to the end of the process.

Thus if we have $f \in \omega^\omega$, then $\bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)} \ne \varnothing$ and in fact has only one point that is an irrational number.

On the other hand, for each point $x \in \mathbb{P}$, we can locate inductively a sequence of intervals, $A_{f(0)}, A_{f(0),f(1)}, A_{f(0),f(1),f(2)}, \cdots$, containing the point $x$. This sequence of nested closed intervals must collapse to a single point and the single point must be the irrational number $x$.

The process described above gives us a one-to-one mapping from $\mathbb{P}$ onto the product space $\omega^\omega$. This mapping is also continuous in both directions, making it a homeomorphism. the nested intervals defined above form a base for the Euclidean topology on $\mathbb{P}$. These basic open sets have a natural correspondance with basic open sets in the product space $\omega^\omega$.

For example, for $f \in \omega^\omega$, $\left\{ A_{f(0),\cdots,f(n)} \cap \mathbb{P}: n \in \omega \right\}$ is a local base at the point $x \in \bigcap \limits_{n=0}^\infty A_{f(0),\cdots,f(n)}$. One the other hand, each $A_{f(0),\cdots,f(n)} \cap \mathbb{P}$ has a natural counterpart in a basic open set in the product space, namely the following set:

$\displaystyle . \ \ \ \ \left\{g \in \omega^\omega: g \restriction n = f \restriction n \right\}$

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The above process establishes that the countable product of the integers, $\omega^\omega$, is equivalent topologically to the Euclidean space $\mathbb{P}$.

# An elementary example of a productively Lindelof space

A topological space is productively Lindelof if its product with every Lindelof space is Lindelof. It is well known that the product of a compact space with any Lindelof space is Lindelof. As a corollary, the product of a $\sigma$-compact space with any Lindelof space is Lindelof. Another way to state this basic topological fact is that $\sigma$-compact spaces are productively Lindelof. We present an example of a productively Lindelof space that is not $\sigma$-compact, demonstrating that these two notions are not equivalent. This example is an elementary one. No heavy machinery is required to define the example. References for productively Lindelof spaces include [1] and [3].

The fact that the product of any $\sigma$-compact space with any Lindelof space is Lindelof is due to the Tube Lemma.

We now define a productively Lindelof space that is not $\sigma$-compact. Let $X= \left\{p\right\} \cup D$ where $D$ is any uncountable set and $p \notin D$. The set $D$ is discrete in $X$ and open neighborhoods at $p$ have the form $\left\{p\right\} \cup (D-A)$ where $A \subset D$ is countable. The only compact subsets of this space are finite sets. Thus $X$ is not $\sigma$-compact.

To see that $X$ is productively Lindelof, let $Y$ be any Lindelof space. Let $\mathcal{U}$ be any open cover of $X \times Y$. Assume that $\mathcal{U}$ consists of open sets of the form $G \times H$ where $G$ is open in $X$ and $H$ is open in $Y$.

There exists a countable $\mathcal{V} \subset \mathcal{U}$ such that $\mathcal{V}$ covers $\left\{p\right\} \times Y$. Suppose that $\mathcal{V}=\left\{G_1 \times H_1,G_2 \times H_2,G_3 \times H_3,\cdots \right\}$. Also assume that for each $i$, $G_i=\left\{p\right\} \cup (D-A_i)$ where $A_i$ is countable.

Note that each $A_i \times Y$ is a Lindelof space since it is the product of a countable space (thus $\sigma$-compact) with a Lindelof space. It is also clear that each point $(x,y) \in X \times Y$ either belongs to a set in $\mathcal{V}$ or to $A_i \times Y$ for some $i$.

For each $i$, choose countable $\mathcal{W}_i \subset \mathcal{U}$ such that $\mathcal{W}_i$ covers $A_i \times Y$. Then $\mathcal{V} \cup \mathcal{W}_1 \cup \mathcal{W}_2 \cup \cdots$ is a countable subcover of $\mathcal{U}$.

Reference

1. Alster, K., On spaces whose product with every Lindelof space is Lindelof, Colloq. Math. 54 (1987), 171–178.
2. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
3. Tall, F., Productively Lindelof spaces may all be D, Canad. Math. Bull. to apear.
4. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.

# An observation about hereditarily separable function spaces

For any completely regular space $X$, by $C_p(X)$ we mean the space of all real-valued continuous functions on $X$ endowed with the pointwise convergent topology. It is known that $C_p(X)$ is hereditarily separable if and only if $X^n$ is hereditarily Lindelof for all positive integer $n$ if and only if $X^\omega$ is hereditarily Lindelof where $\omega$ is the first infinite ordinal (a result that follows from a theorem of Zenor in [4]). This result points to a duality between hereditary separability of the function space $C_p(X)$ and the hereditary Lindelof property of the domain space $X$ and is restated below.

Theorem
Let $X$ be a completely regular space. Then the following conditions are equivalent:

1. $C_p(X)$ is hereditarily separable.
2. $X^n$ is hereditarily Lindelof for all positive integer $n$.
3. $X^\omega$ is hereditarily Lindelof.

As an introduction to this theorem, we present the proof to one direction of this theorem for $n=1$.

Observation
Let $X$ be any completely regular space. We have the following obervation:

If $C_p(X)$ is hereditarily separable, then $X$ is hereditarily Lindelof.

Suppose $X$ is not hereditarily Lindelof. We aim to show that $C_p(X)$ is not hereditarily separable by producing a non-separable subspace $\mathcal{F}$ of $C_p(X)$.

Let $Y \subset X$ be a subspace that is not Lindelof. Let $\mathcal{U}$ be a collection of open subsets of $X$ such that $\mathcal{U}$ covers $Y$ and no countable subcollection of $\mathcal{U}$ covers $Y$.

For each $y \in Y$, choose $U_y \in \mathcal{U}$ such that $y \in U_y$. By the completely regularity of $X$, choose a continuous $f_y: X \rightarrow \mathbb{R}$ such that $f_y$ maps $X-U_y$ to $0$ and $f_y(y)=1$. Let $\mathcal{F}=\left\{f_y:y \in X \right\}$. It can be shown that $\mathcal{F}$ is a non-separable subspace of $C_p(X)$. That is, no countable subset of $\mathcal{F}$ can be dense in $\mathcal{F}$. $\blacksquare$

For any completely regular space $X$, it is also known (see [2]) that $C_p(X)$ is separable if and only if $X$ has a weaker separable metrizable topology (i.e. $X$ has a weaker topology such that $X$ with this weaker topology is a separable metrizable space). The result in [2] combined with the observation presented here provides a way to obtain sepearable $C_p(X)$ that is not hereditarily separable. Look for any $X$ that is not hereditarily Lindelof but has a weaker separable metrizable topology. One such example is the Michael Line.

The observation we make here is a rather weak result. The double arrow space $Z$ is hereditarily Lindelof. Yet $C_p(Z)$ is not even separable since $Z$ is compact space that is not metrizable. Note that $Z^2$ is not hereditarily Lindelof since it contains a copy of the Sorgenfrey plane (see the previous post on double arrow space).

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Noble, N., The density character of function spaces, Proc. Amer. Math. Soc. 42:1 (1974) 228-233.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.
4. Zenor, P., Hereditarily m-separability and the hereditarily m-lindelof property in product spaces and function spaces, Fund. Math. 106 (1980), 175-180

# Closed uncountable subsets of the real line

This is post #12 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the previous post Perfect sets and Cantor sets, II, we show that every subset of the real line that is a perfect set contains a Cantor set and thus has cardinality continuum. We would like to add an observation that every uncountable closed subset of the real line contains a perfect set. Thus every uncountable closed subset of the real line contains a Cantor set and thus has the same cardinality as the real line itself.

Recall that a subset $A$ of the real line is a perfect set if it is closed and every point of $A$ is a limit point of $A$. In the previous post Perfect sets and Cantor sets, II, we demonstrate a procedure for constructing a Cantor set out of any nonempty perfect set. In another post The Lindelof property of the real line, we show that every uncountable subset $A$ of the real line contains at least one of its limit points. Thus all but countably many points of $A$ are limit points of $A$. Consequently, for any uncountable closed subset $A$ of the real line, all but countably many points of $A$ are limit points of $A$. By removing the countably many non-limit points (isolated points), we have a perfect set.

# Perfect sets and Cantor sets, II

This is post #11 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the previous post Perfect sets and Cantor sets, I, we show that every nonempty perfect set is uncountable. We now show that any perfect contains a Cantor set. Hence the cardinality of any perfect set is continuum.

Given a perfect set $W$, we construct a Cantor set within $W$. Consider the following cases:

• Case 1. $W$ contains some bounded closed interval $[a,b]$.
• Case 2. $W$ does not any bounded closed interval.

If case 1 holds, then we can apply the middle third process on $[a,b]$ and produce a Cantor set. So in the remaining dicussion of this post, we assume case 2 holds. This means that for each closed interval $[a,b]$, there is some $x \in [a,b]$ such that $x \notin W$.

Background Discussion
Let $A \subset \mathbb{R}$ and $p \in A$. The point $p$ is a right-sided limit point of $A$ if for each open interval $(a,b)$ containing $p$, the open interval $(p,b)$ contains a point of $A$. The point $p$ is a left-sided limit point of $A$ if for each open interval $(a,b)$ containing $p$, the open interval $(a,p)$ contains a point of $A$. The point $p$ is a two-sided limit point of $A$ if it is both a right-sided limit point and a left-sided limit point of $A$. For the proof of the following lemma, see the post labeled #10 listed below.

In Lemma 2 below, we apply the least upper bound property and the greatest lower bound property. See the post labeled #4 listed below.

Key Lemmas for Construction

Lemma 1
Suppose that $X \subset \mathbb{R}$ is an uncountable set. Then $X$ contains a two-sided limit point.

As a corollary to the lemma 1, for the perfect set $W$ in question, all but countably many points of $W$ are two-sided limit points of $W$.

Lemma 2
Suppose $E \subset \mathbb{R}$ is a nonempty perfect set that satisfies Case 2 indicated above. Suppose that for the closed interval $[a,b]$, we have:

• $(a,b) \cap E \ne \phi$,
• the left endpoint $a$ is a right-sided limit point of $E$,
• the right endpoint $b$ is a left-sided limit point of $E$.

Then we have $a such that:

• there are no points of $E$ in the open interval $(a^*,b^*)$,
• the point $a^*$ is a left-sided limit point of $E$,
• the point $b^*$ is a right-sided limit point of $E$.

Proof. Since Case 2 holds, for the closed interval $[a,b]$ in question, there is some $x \in (a,b)$ such that $x \notin E$. Then we can find an open interval $(c,d)$ such that $x \in (c,d)$ and $a and $(c,d) \cap E = \phi$.

Any point in $(c,d)$ is an upper bound of $W_1=[a,c) \cap E$. By the least upper bound property, $W_1$ has a least upper bound $a^*$. Any point in $(c,d)$ is an lower bound of $W_2=(d,b] \cap E$. By the greatest lower bound property, $W_2$ has a greatest lower bound $b^*$. Then $a^*$ and $b^*$ satisfy the conclusion of the lemma. $\blacksquare$

Lemma 3
Suppose $E \subset \mathbb{R}$ is a nonempty perfect set. Suppose we have a closed interval $[s,t]$ such that the left endpoint $s$ is a right-sided limit point of $E$ and the right endpoint $t$ is a left-sided limit points of $E$. Then we have $s such that:

• the open interval $(s_*,t_*)$ contains points of $E$,
• both endpoints $s_*$ and $t_*$ are two-sided limit points of $E$,
• $t_*-s_*<0.5(t-s)$.

Proof. Suppose we have a closed interval $[s,t]$ as described in the lemma. Then $E_1=[s,t] \cap E$ is a nonempty perfect set. Thus $E_1$ is uncountable. So pick $p \in (s,t)$ such that $p$ is a two-sided limit point of $E_1$.

Choose open interval $(c,d)$ such that $s and $d-c<0.5(t-s)$. Since $p$ is a two-sided limit point of $E_1$, choose $s^*$ and $t^*$ such that $c, $p and both $s^*$ and $t^*$ are two-sided limit points of $E_1$. It follows that $s^*$ and $t^*$ satisfy the conclusion of the lemma. $\blacksquare$

Lemma 4
Suppose $E \subset \mathbb{R}$ is a nonempty perfect set that satisfies Case 2 indicated above. Suppose we have a closed interval $[a,b]$ such that the endpoints are two-sided limit points of $E$. Then we have disjoint closed intervals $K_0=[p_0,q_0]$ and $K_1=[p_1,q_1]$ such that

• $K_0 \subset [a,b]$ and $K_1 \subset [a,b]$,
• the lengths of both $K_0$ and $K_1$ are less then $0.5(b-a)$,
• for each of $K_0$ and $K_1$, the endpoints are two-sided limit points of $E$.

Proof. This is the crux of the construction of a Cantor set from a perfect set and is the result of applying Lemma 2 and Lemma 3.

Applying Lemma 2 on $[a,b]$ and obtain the open interval $(a^*,b^*)$. We remove the open interval $(a^*,b^*)$ from $[a,b]$ and obtain two disjoint closed intervals $[a,a^*]$ and $[b^*,b]$. Each of these two subintervals contains points of the perfect set $W$ since the endpoints are limit points of $W$ in the correct direction.

Now apply Lemm 3 to shrink $[a,a^*]$ to obtain a smaller subinterval $K_0=[p_0,q_0]$ such that the length of $K_0$ is less than $0.5(a^*-a)$ and is thus less than $0.5(b-a)$. Likewise, apply Lemma 3 on $[b^*,b]$ to obtain $K_1=[p_1,q_1]$ such that the length of $K_1$ is less than $0.5(b-b^*)$ and is thus less than $0.5(b-a)$. Note that both $K_0$ and $K_1$ constain points of $E$, making both $K_0 \cap E$ and $K_1 \cap E$ perfect sets and compact sets. $\blacksquare$

Construction
Suppose $W \subset \mathbb{R}$ is a nonempty perfect set that satisfies Case 2. Pick two two-sided limit $a_0$ and $b_0$ of $W$. Obtain $B_0=K_0$ and $B_1=K_1$ as a result of applying Lemma 4. Let $A_1=B_0 \cup B_1$.

Then we apply Lemma 4 on the closed interval $B_0$ and obtain closed intervals $B_{00}$, $B_{01}$. Likewise we apply Lemma 4 on the closed interval $B_1$ and obtain closed intervals $B_{10}$, $B_{11}$. Let $A_2=B_{00} \cup B_{01} \cup B_{10} \cup B_{11}$.

Continue this induction process. Let $C=\bigcap \limits_{n=1}^{\infty} A_n$. The set $C$ is a Cantor set and has all the properties discussed in the posts labled #6 and #7 lised at the end of this post.

We claim that $C \subset W$. For any countably infinite sequence $g$ of zeros and ones, let $g_n$ be the first $n$ terms in $g$. Let $y \in C$. Then $\left\{y\right\}=\bigcap \limits_{n=1}^{\infty} B_{g_n}$ for some countably infinite sequence $g$ of zeros and ones (see post #6 listed below). Then every open interval $(a,b)$ containing $y$ would contain some closed interval $B_{g_n}$. Thus $y$ is a limit point of $W$. Hence $y \in W$.

# The Lindelof property of the real line

This is post #10 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

In the real line, the set $\mathbb{Z}$ of all integers (positive, negative and zero) is a closed set as well as a discrete set. It is closed because the complement is an open set. It is discrete since no point of $\mathbb{Z}$ is a limit point of $\mathbb{Z}$. It turns out that in the real line, only countable sets can be both closed and discrete. Equivalently, the real line satisfies the following property:

$* \$ Every uncountable subset has a limit point.

A set $A \subset \mathbb{R}$ is discrete if no point of $A$ is a limit point of $A$. The set $H=\left\{\frac{1}{n}: n=1,2,3,\cdots\right\}$ is a discrete set, as is the set $\mathbb{Z}$. However, unlike $\mathbb{Z}$, the set $H$ is not a closed set. It turns out that in the real line, only countable sets can be discrete. Equivalently, the real line satisfies the following property:

$** \$ Every uncountable subset contains one of its limit points.

It follows that for any uncountable subset $A$ of the real line, all but countably many points of $A$ are limit points of $A$.

We show that both the statements $*$ and $**$ are intimately tied to the Lindelof property. Specifically, both $*$ and $**$ hold for the real line because the real line itself and every subset of the real line has the Lindelof property.

We also prove that for every uncountable subset $A$ of the real line, there is at least one $p \in A$ such that $p$ is a two-side limit point of $A$. It follows that for every uncountable subset $A$ of the real line, all but countably many points of $A$ are two-sided limit points of $A$.

Definitions
Let $X$ be a topological space. A collection $\mathcal{U}$ of subsets of $X$ is said to be a cover of $X$ if every point of $X$ belongs to some set in $\mathcal{U}$. The collection $\mathcal{U}$ is said to be an open cover if it is a cover of $X$ and it consists entirely of open subsets of $X$. If $\mathcal{U}$ is an open cover of $X$ and if $\mathcal{V} \subset \mathcal{U}$ also covers $X$, then $\mathcal{V}$ is said to be a subcover of $\mathcal{U}$.

The space $X$ is said to be a Lindelof space (or has the Lindelof property) if every open cover of $X$ has a countable subcover. The space $X$ is said to be hereditarily Lindelof if every subspace of $X$ is a Lindelof space.

Theorem 1
Let $X$ be a topological space.

1. If $X$ is Lindelof, then $X$ satisfies the property $*$.
2. If $X$ is hereditarily Lindelof, then $X$ satisfies the property $**$.

Proof. We prove the contrapositive, that is, if $X$ has an uncountable subset that has no limit point, then $X$ is not Lindelof. Let $A \subset X$ be uncountable such that $A$ has no limit point. Then $A$ is a closed set. For each $x \in A$, there is an open set $O_x$ such that $x \in O_x$ and $O_x$ contains no point of $A$ other than $x$. Then the following is an open cover of $X$:

$\mathcal{U}=\displaystyle \left\{O_x: x \in A\right\} \cup \left\{X-A\right\}$

Note that only countably many $O_x$ cannot cover $A$. Thus $\mathcal{U}$ is an open cover of $X$ that has no countable subcover, showing that $X$ is not Lindelof.

To show 2, we also prove the contrapositive. Let $A \subset X$ be some uncountable set such that no point of $A$ is a limit point of $A$. Then for each $x \in A$, $\left\{x\right\}$ is an open set relative to $A$. Thus the subspace $A$ is not Lindelof, showing that $X$ is not hereditarily Lindelof. $\blacksquare$

Theorem 2
Every subset of the real line $\mathbb{R}$ has the Lindelof property, i.e. $\mathbb{R}$ is hereditarily Lindelof.

Remark. This follows from Lemma 4 in the post labeled #4 listed below. This theorem follows from the fact that the Euclidean topology of the real line is generated by a countable base (specifically, the open intervals with rational numbers as endpoints).

Corollary 3
The real line satisfies properties $*$ and $**$. Thus we have:

1. Every uncountable subset of the real line has a limit point.
2. Every uncountable subset of the real line contains one of its limit points.

Remark
This follows from Theorem 1 and Theorem 2. It also follows that if $A$ is an uncountable subset of the real line, all but countably many points of $A$ are limit points of $A$. $\blacksquare$

Let $A \subset \mathbb{R}$ and $p \in A$. The point $p$ is said to be right-sided limit point of $A$ if for each open interval $(a,b)$ such that $p \in (a,b)$, we have $(p,b) \cap A \ne \phi$. The point $p$ is said to be left-sided limit point of $A$ if for each open interval $(a,b)$ such that $p \in (a,b)$, we have $(a,p) \cap A \ne \phi$. The point $p$ is said to be a two-sided limit point of $A$ if it is both a right-sided and left-sided limit point.

Based on the discussion in the previous post on completness (the post labeled #3 below), we can assume the least upper property whenever we work with the real line. This property states that for every subset $A$ of the real line, if $A$ is bounded above, then $A$ has a least upper bound (the upper bound that is the smallest among all the upper bounds). This assumption simplies the proof of the next theorem. The least upper bound property is equivalent to the greatest lower bound property, which states that for every subset $A$ of the real line, if $A$ is bounded below, then $A$ has a greatest lower bound (the lower bound that is the greatest among all the lower bounds).

Theorem 4
Every uncountable subset of the real line has at least one point that is a two-sided limit point.

Proof. Suppose that $A \subset \mathbb{R}$ is an uncountable set such that none of its points is a two-sided limit point. By corollary 3, all but countably many points of $A$ is a limit point of $A$. So by removing these countably many points if necessary, assume that every point of $A$ is a limit point of $A$.

For each $x \in A$, $x$ is either a right-sided limit point of $A$ or a left-sided limit point of $A$ but not both. So either $(1)$ there are uncountably many points of $A$ that are left-sided limit points of $A$ or $(2)$ there are uncountably many points of $A$ that are right-sided limit points of $A$. Assume $(1)$ and use the least upper bound property. The proof assuming $(2)$ is similar and uses the greatest lower bound property.

Let $B$ be the set of all points of $A$ that are left-sided limit points of $A$. By assumption $(1)$, $B$ is uncountable. For each $x \in B$, choose a rational number $b_x$ such that $(x,b_x)$ contains no point of $A$. Then there must be some rational number $r$ such that $r=b_x$ for uncountably many $x \in B$. Thus the following set $C$ is uncountable:

$\displaystyle C=\left\{x \in B: b_x=r\right\}$

Note that the rational number $r$ is an upper bound of $C$. By the least upper bound property, let $u$ be the least upper bound of $C$. Choose some $y \in C$ such that $y. This is possible since $x \le u$ for all $x \in C$ and $C$ is uncountable. There must be some $z \in C$ such that $y. Otherwise, $y$ would be an upper bound. Since $z$ is a left-sided limit point of $A$, $(y,z)$ contains infinitely many points of $A$. But $(y, b_y)=(y,r)$ is supposed to contain no points of $A$, a contradiction.

Under assumption $(2)$, the proof is similar and uses the greatest lower bound property. Thus we establish that in each uncountable subset of the real line, one of its points must be a two-sided limit point. As a corollary to this fact, all but countably many points of an uncountable set of real numbers must be two-sided limit points of the set. $\blacksquare$

# Perfect sets and Cantor sets, I

This is post #9 of the series on the Euclidean topology of the real line. See the links at the bottom for other posts in the series.

Recall that a subset $A$ of the real line is a perfect set if $A$ is closed in Euclidean topology of the real line and that every point of $A$ is a limit point of $A$. Any closed and bounded interval $[a,b]$ is a perfect set. The Cantor sets (the middle third version and other variations) are perfect sets (see the links #7 and #8 below). It turns out that any nonempty perfect set contains a Cantor set. In this series of posts on Euclidean topology of the real line, by Cantor sets we mean any set that can be constructed by a binary process of splitting closed intervals into two halves at each stage (see links for #6 and #8 below). We demonstrate the algorithm of constructing a Cantor set from any perfect set. This post (part I) shows that any nonempty perfect set is uncountable. Knowing that a perfect is uncountable will simplify the construction process (next post).

Suppose $W \subset \mathbb{R}$ is a nonempty perfect subset. We show that $W$ is uncountable. Since $W$ has at least one point and every point is a limit point, $W$ is infinite. The key to showing $W$ is uncountable is that every nested decreasing sequence of compact subsets of the real line (actually in any topological space) has nonempty intersection. If $W$ happens to be countable, we can define a nested sequence of compact subsets of $W$ with empty intersection. Thus $W$ cannot be countable.

The following lemma is a corollary to Theorem 3 in the post # 4 listed below. The lemma applies to any abstract spaces where compactness can be defined. We state the lemma in terms of the real line since this is our focus.

Lemma
Suppose $C_1,C_2,C_3, \cdots$ are compact subsets of the real line such that

$\displaystyle C_1 \supset C_2 \supset C_3 \supset \cdots$.

Then $\bigcap \limits_{n=1}^{\infty} C_n \ne \phi$.

To make the argument that $W$ is uncountable more precise, suppose that $W$ is countable. Then we can enumerate $W$ in a sequence indexed by the positive integers. We have:

$\displaystyle W=\left\{w_1,w_2,w_3,\cdots\right\}$

Pick a bounded open interval $O_1$ such that $w_1 \in O_1$. Next, pick an open interval $O_2$ such that $\overline{O_2} \subset O_1$ and $w_2 \notin \overline{O_2}$ and $O_2 \cap W \ne \phi$.

In the $n^{th}$ stage where $n \ge 2$, pick an open interval $O_n$ such that $\overline{O_n} \subset O_{n-1}$ and $w_n \notin \overline{O_n}$ and $O_n \cap W \ne \phi$. Since $W$ is a perfect set, the induction step can continue at every stage.

Now, let $C_n=\overline{O_n} \cap W$. Note that $C_n$ is a compact set since $\overline{O_n}$ is compact. By the lemma, the intersection of the $C_n$ must be nonempty. By the induction steps, no point of $W$ belongs to all the sets $\overline{O_n}$, implying the intersection of $C_n$ is empty, a contradiction. Thus $W$ must be uncountable.

Remark
As a corollary, the real line and the unit intervals are uncountable. A more interesting corollary is that any nonempty perfect set has a two-sided limit point. In fact all but countably many points of a nonempty perfect set are two sided limit points. See the post The Lindelof property of the real line for a proof that any uncountable subset of the real line has a two sided limit point. This fact will simplify the construction of a Cantor set from a perfect set.

Reference

1. Rudin, W., Principles of Mathematical Analysis, Third Edition, 1976, McGraw-Hill, Inc, New York.