# Closed set: definition

1. Sep 21, 2008

### dirk_mec1

1. The problem statement, all variables and given/known data
http://img527.imageshack.us/img527/6049/48193240ao5.png [Broken]

I don't understand this proof. The first two lines are clear to me: the sequence xn is in F and F is closed so its complement is open so there is a ball with radius r around x in Fc.

But I don't understand the last two lines. Of course there y larger then the radius od the ball but what's the relation with the converging sequence?

Last edited by a moderator: May 3, 2017
2. Sep 21, 2008

### morphism

If x_n -> x then the sequence x_n is eventually in O.

3. Sep 21, 2008

### dirk_mec1

But then you do not need the line that there are y which are greater than the radius and inside F, right?

4. Sep 21, 2008

### HallsofIvy

Staff Emeritus
Yes, you do. That's the whole point. Since {xn} converges to x, given any r> 0, there exist N such that if n> N, d(x, xn)< r. Taking r such that Br(x) is in O, it follows that xn, for n> N s in Br(x) so in O. IF, as in the hypothesis, all xn are in F, then some members of O (they "y" they mention) are in F, a contradiction.