# 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. $\blacksquare$

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