# Limit points of second countable topological space

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 $$A$$.

Remark: Probably this question have appeared couple of times in this forum but I solved it by myself and I would be very grateful if you can check my details.

Proof: Suppose not then only countably many points of $$A$$ are limit points of $$A$$. Since $$A$$ is uncountable then it has uncountable subset $$C$$ whose elements are not limit points of $$A$$.

Hence for any $$c\in C$$ $$\exists U_c$$ - neighborhood of $$c$$ such that $$U_c\cap A=\{c\}$$. Since $$X$$ has countable basis $$\mathcal{B}=\{B_n\}_{n\geq 1}$$ then $$\exists n:=n(c)$$ such that $$B_n\cap A=\{c\}$$.

Let's consider the following: for any $$c\in C$$ consider $$I_c=\{n\in \mathbb{Z}_+: B_n\cap A=\{c\}\}$$. Since $$I_c\subset \mathbb{Z}_+$$ and $$I_c\neq \varnothing$$ then by well-ordering $$I_c$$ has smallest element, say $$n(c)$$.

Define the function $$f:C\to \mathcal{B}$$ by equation: $$f(c)=B_{n(c)}.$$

Let's show that this function is injective: let $$c_1,c_2\in C$$ such that $$c_1\neq c_2$$ but $$f(c_1)=f(c_2)$$. Then $$B_{n(c_1)}=B_{n(c_2)}$$ then by definition of $$n(c_1)$$ and $$n(c_2)$$ we will get: $$B_{n(c_1)}\cap A=B_{n(c_1)}\cap A$$ $$\Rightarrow$$ $$\{c_1\}=\{c_2\}$$ $$\Rightarrow$$ $$c_1=c_2$$ which is absurd. Hence $$f(c_1)\neq f(c_2)$$.

So we have an injective function $$f:C\hookrightarrow \mathcal{B}$$ but since $$\mathcal{B}$$ is countable set then it implies that $$C$$ is countable which is contradiction because by construction $$C$$ was uncountable.

Is the proof correct in all details?

The idea is OK, but you don't have to use a contradiction at all:

Let $$C$$ be the subset of $$A$$ that are not limit points. Then for each $$x \in C$$ we let $$f(x)$$ be the minimal $$n \in \Bbb Z^+$$ such that $$B_n \cap A=\{x\}$$. (No need to go all formal about it, it's clearly well-defined because the $$B_n$$ form a base and $$x$$ is not a limit point)

This defines an injection from $$C$$ into $$\mathbb{Z}^+$$ (because $$f(c)=f(c')$$ implies

$$\{c\}= B_{f(c)} \cap A = B_{f(c')} \cap A = \{ c' \}$$

so that $$c=c'$$) and so $$C$$ is countable and so $$A\setminus C$$ (the points of $$A$$ that are limit points of $$A$$) is uncountable.

• Thanks a lot for pating attention to my post! So you mean that i can argue even without contradiction right? – ZFR Oct 19 '19 at 20:00
• @ZFR Yes, the idea of the 1-1 map using the base is fine, but no contradiction is needed. If a set maps injectively into a countable set, it's countable. The 1-1 proof also needs no contradiction: injective means $f(x)=f(x') \to x = x'$. – Henno Brandsma Oct 19 '19 at 20:01
• Yes indeed you are right. But i guess my reasoning is also correct, right? However i did not think in the same way as you did – ZFR Oct 19 '19 at 20:48
• @ZFR your proof works but is more indirect. I wanted to show that was not needed. – Henno Brandsma Oct 20 '19 at 12:28