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.(adsbygoogle = window.adsbygoogle || []).push({});

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)

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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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# X in dA => x isolated or strict limit (Topological space)

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