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## Main Question or Discussion Point

I am trying to show the following proposition, but I can only do so in the special case where the basis for the topology is countable. I posted this in the calc & analysis forum and didn't get any responses - I think that this is a more appropriate forum judging by the other posts.

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If the number of open sets containing x were countable, [tex]\left\{ E_{1}, E_{2}, E_{3}, ... \right\}[/tex], then I would put [tex]x_{1} \in \left(A \cap E_{1}\right) \backslash\left\{x\right\}[/tex], [tex]x_{2} \in \left(A \cap E_{1}\cap E_{2}\right) \backslash\left\{x\right\}[/tex], [tex]x_{3} \in \left(A \cap E_{1}\cap E_{2}\cap E_{3}\right)\backslash\left\{x\right\}[/tex], and so forth.

We know these intersections can't be empty - if [tex]E_{i}[/tex] and [tex]E_{j}[/tex] are open and contain x, then [tex]E_{i}\cap E_{j}[/tex] is also an open set containing x. Since x is not isolated, [tex]E_{i}\cap E_{j}[/tex] must contain some [tex]y \in A[/tex] distinct from x. Thus for a countable number of open sets containing x, we can construct the proper convergent sequence to prove the result.

**Proposition**. If x is in [tex]\partial A[/tex], then x is either an isolated point, or a strict limit point of both [tex]A[/tex] and [tex]A ^{c}[/tex]. ([tex]A[/tex] is a set in a topological space, [tex]\partial A[/tex] is the boundary)---

**So far, my reasoning is this**. Suppose x is not isolated - then lets show that it is a strict limit point of [tex]A[/tex] by constructing a sequence [tex]\left\{ x_{n}\right\}[/tex] in [tex]A[/tex] that converges to x.If the number of open sets containing x were countable, [tex]\left\{ E_{1}, E_{2}, E_{3}, ... \right\}[/tex], then I would put [tex]x_{1} \in \left(A \cap E_{1}\right) \backslash\left\{x\right\}[/tex], [tex]x_{2} \in \left(A \cap E_{1}\cap E_{2}\right) \backslash\left\{x\right\}[/tex], [tex]x_{3} \in \left(A \cap E_{1}\cap E_{2}\cap E_{3}\right)\backslash\left\{x\right\}[/tex], and so forth.

We know these intersections can't be empty - if [tex]E_{i}[/tex] and [tex]E_{j}[/tex] are open and contain x, then [tex]E_{i}\cap E_{j}[/tex] is also an open set containing x. Since x is not isolated, [tex]E_{i}\cap E_{j}[/tex] must contain some [tex]y \in A[/tex] distinct from x. Thus for a countable number of open sets containing x, we can construct the proper convergent sequence to prove the result.

*However, this logic will not work if there are an uncountable number of open sets.*I could extend my argument if only the basis is countable, but not to a topology without a countable basis.
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