1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.