1. ## Topology Problem

i think i'm just having trouble visualizing this problem...

they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.

how would i go about solving this problem?

2. Originally Posted by squarerootof2
i think i'm just having trouble visualizing this problem...

they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.

how would i go about solving this problem?
Can you explain what you mean by "disjoint union"?

Maybe the following will help: The closure of S is the union of S and its limit points.

3. i think disjoint union just means a collection. i just looked at the solutions in the back and it seems to suggest a proof by contradiction, i.e. assume that x is not a limit point of S and to mess around with an epsilon ball containing only finitely many pts of S. i was just wondering what the best way to approach this problem was. thanks.

4. Originally Posted by squarerootof2
i think i'm just having trouble visualizing this problem...

they define x in S as an isolated point of S if there exists r>0 such that B(x;r)∩S={x}. the problem asks to show that the closure of subset S of X is the disjoint union of limit pts of S and isolated pts of S.
It should be clear from the definition that no isolated point is a limit point of S.
It follows that every non-isolated point is a limit point of S.
Now let $S'$ denote the set of all limits points of S.
Then $S\backslash S'$ would be the set of all isolated points of S.
Observing that the closure of S consists of all points of S union all limits points of S, the proof follows.