# Open set and open ball

1. Feb 13, 2014

### sigh1342

The problem statement, all variables and given/known data

Show that S is open if and only if ∀x ∈ S, ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S
And what we have is , let X be normed space, S ⊂X , Then S is close if and only if $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$, x∈ S. A set S is open if and only if X\S is close.

The attempt at a solution
I try to use following approach.
"=>" since S is open, there exist some sequence s.t. its limit is not in S(otherwise X/S is not close), then for any x, we can construct the convergent sequence such that its limit is not in S.(I don't know it is true or not). So by the limit definition. ∃ a open ball B(x; r)(r > 0) such that B(x; r) ⊂ S.

"<=" I try to prove X\S is close, so that X is open. $$∀{x_{n}}⊂X, s.t. x_{n}->x \in X$$ , x should be in X\S, but I have no idea to prove it.um.. Can anyone give me some ideas? :)

2. Feb 13, 2014

### pasmith

I'm not sure this even makes sense.

I would prove this by contrapositive: Assume that there exists an $x \in S$ such that for all $r > 0$, $B(x,r) \not\subset S$, and show that it follows that $X \setminus S$ is not closed.

Your assumption is that for all $x \in S$ there exists $r > 0$ such that $B(x,r) \subset S$. Why does it follow that $x \in S$ cannot be the limit of any sequence in $X \setminus S$?