# An uncountable subset of a second countable space has uncountably many of its limit points

Let $X$ have a countable basis , let $A$ be an uncountable subset of $X$. show that uncountably many points of $A$ are limit points of $A$.

This is my attempt:

By way of contradiction, assume that $D$= {$x\in{A}$. $x$ is limit point of $A$} is countable. then the set which contains all points outside $A$ IS UNCOUNTABLE. let $a$ outside $A$ then there exists basic open set $B_a$ contains $a$ such that $B_a\cap A={a}$. we can conclude there exists one-to-one function from all points outside $A$ to elements of countable basis which impossible.

• Typo? Did you mean to say A is uncountable? – DanielWainfleet Jul 26 '16 at 19:06
• Yes, you are right – Gob Jul 26 '16 at 19:10
• Needs some re-wording. Say if $D$ is countable then for $a\in A\backslash D$ let $B_a$ be a member of the countable base $B$ such that $B_a\cap A=\{a\}.$. – DanielWainfleet Jul 26 '16 at 19:15

You essentially seem to be having the right idea, but the proof is very unclear.

Start by picking a countable base $\{B_n: n \in \mathbb{N}\}$ of $X$.

For every $x \in A\setminus A'$ ($A' =$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B_{n(x)}$ such that $B_{n(x)} \cap A = \{x\}$. If $x, y$ in $A \setminus A'$, then if $n(x) = n(y)$, we have that $\{x\} = B_{n(x)} \cap A = B_{n(y)} \cap A = \{y\}$ so $x = y$. Hence $x \rightarrow n(x)$ is 1-1 from $A \setminus A'$ to $\mathbb{N}$, so the former set is countable.

This means that $A \cap A'$ is uncountable, as $A = (A \cap A') \cup (A \setminus A')$ is uncountable, and not a union of two countable sets; it even has the same size as $A$.

• Very nice explanation! +1 for that! – ZFR Dec 2 '18 at 22:44

See Henno Brandsma’s answer for a way to express your idea much more clearly.

A slightly different approach is to let $\mathscr{B}$ be a countable base for $X$, and let

$$\mathscr{B}_0=\{B\in\mathscr{B}:|B\cap A|\text{ is countable}\}\;,$$

the family of basic open sets that contain at most countably many points of $A$. $\mathscr{B}_0$ is a subset of the countable base $\mathscr{B}$, so $\mathscr{B}_0$ is countable. Let

$$A_0=\bigcup_{B\in\mathscr{B}_0}(B\cap A)\subseteq A\;;$$

$A_0$ is the union of countably many countable sets, so $A_0$ is a countable subset of $A$. Let $A_1=A\setminus A_0$; clearly $A=A_0\cup A_1$, so $A_1$ is uncountable (as otherwise $A$, being the union of two countable sets, would be countable). (In fact $|A_1|=|A|$.)

Suppose that $x\in A_1$, and let $U$ be any open nbhd of $x$. $\mathscr{B}$ is a base for the topology, so there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$. Then $x\in B\cap A$, but $x\notin A_0$, so $B\notin\mathscr{B}_0$, and therefore $B\cap A$ is uncountable. In particular, $U\cap(A\setminus\{x\})\supseteq B\cap(A\setminus\{x\})\ne\varnothing$, so we’ve shown that every open nbhd of $x$ contains points of $A\setminus\{x\}$ and hence that $x$ is a limit point of $A$. Since $A_1$ is uncountable, $A$ has uncountably many limit points.

More or less ok, but you should perhaps pay more attention as to the choice of the sets $B_a$? Let $B_n$, $n\geq 1$ be a countable basis. For $a\in A\setminus D$ there is an element $B_{n(a)}$ which does not intersect $A\setminus\{a\}$ and since $a\in B_{n(a)}$ the sets $B_{n(a)}$, $a\in A\setminus D$ must be disjoint. So the map $a\in A\setminus D \mapsto n(a)\in {\mathbb N}$ is injective and $A\setminus D$ is thus countable.

I am by no means an expert, and I would have posted this as a comment. However, I don't have the reputation, so I will write an answer. I seem to be finding one small issue with Henno Brandsma's answer, and I think I also have a fix.

First, I should note that I am following Silverman's translation of Kolmogorov and Fomin which defines a limit point as: A point $x$ in a topological space $T$ is a limit point of a set $M \subset T$ if every open set containing $x$ also contains infinitely many points of $M$''.

Now I think there is a problem here: For every $x\in A \setminus A′$ ($A′=$ the set of limit points of $A$, that you call $D$) we can pick a basic open set $B(x)$ such that $B(x)\cap A=\{x\}$''. What if there only exists a neighbourhood $B(x)$, where $B(x) \cap A= C(x)$, such that $x \in C(x)$, and $C(x)$ is finite. With the provided definition of a limit point, this would be impossible for a metric space (shown in this question) or a $T_1$-space, but I could not find anything regarding a second countable space.

Here is my fix for the proof of showing that $A \setminus A'$ is countable: For every $x\in A \setminus A′$, we can pick a basic open set $B(x)$ such that $B(x)\cap A=C(x)$, such that $x \in C(x)$, and $C(x)$ is finite. Since all $C(x)$ are finite, there exists the finite integer $N=\max_{x\in A} |C(x)|$.

There exists a injection from $A\setminus A'$ to $\mathscr{B} \times N = \{(B(x),i) | x \in A, 1 \leq i \leq N \}$ by mapping each $x$ to $(B(x),i)$ for some $i \leq N$, as every $B(x)$ cannot contain more that $N$ members of $A$. As the set of bases, $\mathscr B$, is countable, so would be $\mathscr B \times N$. Therefore, $A\setminus A'$ should be countable as well.