What is the relationship between a closed set and a converging sequence?

[dirk_mec1]

Homework Statement

http://img527.imageshack.us/img527/6049/48193240ao5.png

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

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?

 

[morphism]

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

 

[dirk_mec1]

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

 

[HallsofIvy]

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