Show that the sets of isolated point of $E$ is a countable set - Axiome of choice

Question

Let $E \subset \mathbb{C}$. Show that the sets of isolated point of $E$ is a countable set.

That question is related to this question. However, my question somewhat different.

Define $Acc(E)^c$ the set of isolated points of $E$. Then there exists an open set in $V \subset \mathbb{C}$ for each point $z \in Acc(E)^c$. It is therefore reasonable to define the set, noted $O = \{V \subset \mathbb{C}\}$, of all the open set. By the axiome of choice, we can take an element in each of this open set, and formed a countable set.

In conclusion, I decided to take each isolated point of each open set to form this countable set. Hence, this new set is only $Acc(E)^c$.

Do I have the right to take the axiom of choice to show the question?

Your question is exactly the $n=2$ case of the question you link to because of the isomorphism between the topologies of $\Bbb R^2$ and $\Bbb C$ — Ross Millikan, Feb 07 '16 at 16:32
I know this is the same question, but it is proved differently. I want to know if this kind of trial proof is well managed or not. — Taj Mohamed Bandalandabad, Feb 07 '16 at 16:36

score 1 · Accepted Answer · answered Feb 07 '16 at 16:37

No. Recall that $\Bbb C$ has a countable basis, which we can enumerate. Now given a set of isolated points, they can be separated by open sets, each containing a unique point in the set. So they can be separated by basic open sets. We can choose the least basic open set in our enumeration which works.

So there is an injection from the set today isolated points into the countable basis, so it has to be countable.

(Note we haven't used any assumption here other than the space being second countable.)

Show that the sets of isolated point of $E$ is a countable set - Axiome of choice

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