- 15

- 0

## Main Question or Discussion Point

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.

---

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}.

---

So far, my reasoning is this. Supposing x is not isolated, lets 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.

**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.---

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}.

---

So far, my reasoning is this. Supposing x is not isolated, lets 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.
Last edited by a moderator: