# 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. $\blacksquare$