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.

  • $\begingroup$ Thanks a lot for pating attention to my post! So you mean that i can argue even without contradiction right? $\endgroup$ – ZFR Oct 19 '19 at 20:00
  • $\begingroup$ @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 $f(x)=f(x') \to x = x'$. $\endgroup$ – Henno Brandsma Oct 19 '19 at 20:01
  • $\begingroup$ 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 $\endgroup$ – ZFR Oct 19 '19 at 20:48
  • 1
    $\begingroup$ @ZFR your proof works but is more indirect. I wanted to show that was not needed. $\endgroup$ – Henno Brandsma Oct 20 '19 at 12:28

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