X in Boundary => x Isolated or Strict Limit (topological space)

[NickAlger]

Although the intuition makes sense, I am having trouble determining why the following proposition is correct. The document leaves this as an exercise for the reader... great.

Proposition: Suppose A is a set in a topological space, and dA is the boundary of A. If x is in dA, then x is either an isolated point, or a strict limit point of both A and A closure.

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The definitions I am working from are:
1) dA is (A closure)∩((A complement) closure)

2) x in A is isolated if there is an open set E containing x such that E\{x} does not intersect A. x is sooo lonely, surrounded by strangers... ;)

3) x is a strict limit point of A if there is a sequence {x_n} contained in A\{x} such that x_n -> x.

4) x_n -> x if every open set containing x also contains the tail of the sequence {x_n}.

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So far, my reasoning is this. Supposing x is not isolated, let's show that it is a strict limit point of A by constructing a sequence {x_n} in A that converges to x. After this, showing it for A closure will just requiring twiddling a few words.

If the number of open sets containing x were countable, {E1, E2, E3, ...}, then I would put x1 in E1\{x}, x2 in (E1 intersect E2)\{x}, x3 in (E1 intersect E2 intersect E3)\{x}, and so forth.

We know these intersections can't be empty - if Ei and Ej are open and contain x, then (Ei intersect Ej) is also an open set containing x. Since x is not isolated, (Ei intersect Ej) must contain some y in A (distinct from x). Thus for a countable number of open sets, we can construct the proper convergent sequence.

However, this logic will not work if there are an uncountable number of open sets. I might be able to extend my argument if only the basis is countable, but we don't know that. For example, in the real numbers centered at 0, you might start making up your sequence based on a particular countable collection of open sets like {Ball(.51), Ball(.501), Ball(.5001), etc} and never get to any of the "smaller" open sets.

 

[fresh_42]

$x$ not isolated means: every open set $U$ containing $x$ has a second element $x\neq a \in A$.
We further know that $x \in \partial A = A^{-}\cap (X\backslash A)^{-}$.
You didn't give us your definition of closure, so I assume we can say, that the closure of a set is the set itself plus all its limit points, strict or not. Since $x\in A$ we have $x \notin X\backslash A$. Hence it is a strict limit point of $X\backslash A$, i.e. there is a sequence $y_n \longrightarrow x$ in $X\backslash A$. This means any open set containing $x$, contains the tail of this sequence, too.

We want to show that $x$ is a strict limit point of $A$, i.e. that there is a sequence $x_n \longrightarrow x$ in $A\backslash \{\,x\,\}$, i.e. any open set which contains $x$ must contain the tail of such a sequence.

Let $x\in U$ be any open set. Now we build a sequence of open sets $U_n$ defined by $x\in U_n$ and $y_k \in U\cap U_n$ for all $k \geq n$. Every $U\cap U_n$ contains all elements $x\neq a_k \in A$ with $k\ge n$, hence $x$ is a strict limit point of $A$.