I'm a new user so if my question is inappropriate, non-uniform or ill-shaped; please comment (or edit maybe).

First of all, I'm asking that if my proof is correct or not. Of course, you can prove it nicer and this helps me but my main reason for asking this question is a check for my proof.

Let $a \in V \subseteq X \subseteq \mathbb{R} $. If for an open interval $I$, $I \cap X \subseteq V$, we call $V$ as "$X$-neighbourhood of $a$".

We will prove following statement:

"Let $X \subseteq \mathbb{R}$ and assume that for all $x \in X$ there exist a finite $X$-neighbourhood of $x$. Then, $X$ must be countably infinite."

EDIT: It doesn't have to be infinite, it just must be countable.

My proof: Because of the fact that $X \subseteq \mathbb{R} $, we can just show that $X$ is not uncountably infinite. But$_*$ also, we should show that $X$ cannot be finite.

Let's assume that $X$ is uncountably infinite and let's define a set $A$ as

  • $A = \{Y$: The set $X \setminus Y$ is still uncountably infinite$ \} $.

Now$_{**}$ we will pick a maximal$_{***}$ element $B$ in $A$. Then, if $x \in X\setminus B$ there must be a finite $X$-neighbourhood of $x$. Let's call this neighbourhood as $W$ and call the open interval as $I$ (there might be more than one open interval, just pick one of them) which make $W$ a $X$-neighbourhood of $x$ .Then, obviously $(X \setminus B) \cap I \subseteq W $. But, then $(X \setminus B) \cap I \in A $ and if we take $J = (X \setminus B) \cap I$, $(X \setminus B) \setminus J = X \setminus (B \cup J)$ must be in $A$ and that is a contradiction because we take $B$ as a maximal element.

$_*$: With edit, this part is wrong.

$_{**}$: We need to show that A is well-defined and there exists a maximal element in A.

$_{***}$: If for a $k_1 \in A$, there exists no $k_2 \in A$ such that $k_2 \subset k_1$, we call $k_1$ as the maximal element.

For completion we should prove that $_{**}$ but I think you can easily show it. Also, there are little things, for example, $B$ must be non-empty.

Thanks for any help and please bear in mind that I cannot understand high-level topological explanations.

  • $\begingroup$ Are you sure $X$ must be infinite? What goes wrong if $X$ is a one-point (or finite) set? $\endgroup$ – Ted Shifrin Aug 5 '13 at 20:27
  • $\begingroup$ The Maximal in $A$ is not well-defined. Consider $X = [0, 1]$ and $k_\epsilon = (\epsilon, 1]$ for all $\epsilon > \delta > 0: k_\epsilon \in A$ but $k_{\epsilon - \delta} \supset k_\epsilon$. Thus, the Maximal would have to be $k_0 = (0, 1] \notin A$. $\endgroup$ – AlexR Aug 5 '13 at 20:36
  • $\begingroup$ @AlexR: In this situation, are there a finite $X$-neighbourhood of $x$ for all $x \in X$? $\endgroup$ – aeyalcinoglu Aug 6 '13 at 9:17

The question as posed is false. If $X$ is a point (or any finite set of points for that matter), then any intersection $I\cap X$ is either a point or empty. Then there is clearly a finite $X$ neighborhood for every point in the set (Let $X \subseteq I$).

The easiest way to show $X$ is not uncountably infinite is to look at $X\cap \Bbb Q$, and take the union of finite $X$ neighborhoods for all $x\in X\cap \Bbb Q$ . I feel that this should be enough of a hint to do the problem.

  • $\begingroup$ Okay, I wrote the problem wrongly. It doesn't have to be infinite, It just must be countable. With this edit, is my proof correct? $\endgroup$ – aeyalcinoglu Aug 6 '13 at 9:06
  • $\begingroup$ Your proof does not work as there is no well-defined "maximal" set, and even if there was one it is still not clear you have shown anything. This is not a problem that requires abstract Zorn's Lemma type reasoning to do. When asking why certain subsets are countable and uncountable in $\Bbb R$ your first intuition should be to look at the great advantage of being in $\Bbb R$ (the countably dense subset $\Bbb Q$). $\endgroup$ – PVAL-inactive Aug 6 '13 at 9:42

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