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The problem I tried to solve is the following:

Let $(X,\tau)$ a second countable topological space. Let $A\subset X$ such that $\text{card}(A)>\mathbb{N}$. Then $A$ has accumulation points.

This is what I tried, but I'm not sure it is correct.

We proceed by contradiction. Asume that every $x\in A$ is an isolated point. We set $\mathcal{B}$ be the basis for $\tau$. Let $x\in A$. By definition, $\exists V_x$ a neighbourhood of $x$ such that $V_x\cap A=\{x\}$. By definition of neighbourhood, $\exists O_x \in \tau$ such that $x\in O_x\subset V_x$. By definition of basis, $\exists B_x\in \mathcal{B}$ such that $x\in B_x\subset O_x \subset V_x$, then $B_x\cap A=\{x\}$. If $x\neq y$ are elements in $A$ then $B_x\neq B_y$: If $B_x=B_y$, then $\{x\}=B_x\cap A=B_y\cap A=\{y\}$, a contradiction. Hence, for every $x\in A$ there is at least one element in $\mathcal{B}$ containing $x$ and satisfying all the conditions before. We can conclude $\text{card}(\mathbb{N})\geq\text{card}(\mathcal{B})\geq\text{card}(A)>\text{card}(\mathbb{N})$, a contradiction.

Is this okay?

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  • $\begingroup$ If all a are not accumulation points then there is a disjoint union of balls B(a). But uncountably many disjoint balls cannot possibly be produced from a countable basis. $\endgroup$ – Jacob Wakem Mar 22 '17 at 5:05
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There are some places that could be slightly cleaned up. For example, the first time that you say "by definition" it is not clear which definition you are pointing to (I think, in that case, you are referencing two sentences earlier, i.e., by virtue of $x$ being an isolated point, such a $V_x$ must exist).

It might help to clarify around the final chain of inequalities, too; for example, you twice have "$\text{card}(\mathbb{N})$" in there, so it could be helpful to the reader if you specified precisely whence each of those greater-than signs derives, especially as this leads to the contradiction you ultimately seek.

The overall structure of the argument appears to be right: By assuming there are no limit points, you must have that every point is isolated. But then you could associate each of these uncountably many isolated points with a unique basis element, which would yield uncountably many basis elements: a contradiction, since the basis itself is countable by hypothesis.

For what it's worth, the result here can be strengthened: In such a scenario, not only must the set contain a limit point - it must contain uncountably many limit points.


Edit: To the final point, it is at least worth pointing out why an uncountable set must contain infinitely many limit points. By the exact argument given already, the uncountable set $A$ contains a limit point, $a_1$; but then $A - \{a_1\}$ satisfies all the initial constraints, so it must also contain a limit point, $a_2$, and we can examine $A - \{a_1, a_2\}$. Continuing in this fashion, we prove that there are infinitely many limit points $\{a_k\}_{k \in \mathbb{N}}$ in $A$. For a proof that there are uncountably many limit points, search through already asked questions on MSE and you should find such a demonstration.

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    $\begingroup$ A still more general formulation: if $A$ is any set in a hereditarily Lindelof space, then all but countably many points of $A$ are condensation points of $A.$ (I seem to recall that it's stated this way as an exercise in Kelley's General Topology.) $\endgroup$ – bof Mar 26 '17 at 4:57
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The proof is fine. If I'd write it, I'd go straight from $x \in A$ is isolated in $A$, to $\exists k(x)\in \mathbb{N}: \{x\} = B_{k(x)} \cap A$ and your proof then shows that $x \rightarrow k(x)$ is an injection from $A$ into $\mathbb{N}$, making $A$ at most countable, contradiction.

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