# Homework Help: Open Set

1. Apr 9, 2006

"Let (X,d) be a metrix space, and let F$$\subset$$X be closed. Define G_n to be the set of all those points x in X such that

inf { d(x,y) : y in F } < 1/n

Use these sets G_n to show that a countable intersection of open sets need not be open."

I think the quesiton meant sup { d(x,y) : y in F } < 1/n }, not inf.

Actually I don't think sup or inf should be there at all. If the question meant sup, then Gn are not necessarily open.

Last edited: Apr 9, 2006
2. Apr 9, 2006

### AKG

No, they definitely meant inf. We have:

$$G_n = \{x \in X\, :\, \inf \{d(x,y)\, :\, y \in F\} < 1/n\}$$

A better way to look at Gn is this:

$$G_n = \bigcup _{y \in F} B(y,\, 1/n)$$

where B(y,1/n) is the open ball about y with radius 1/n. Prove that this really is an equivalent way to look at Gn. Prove (very very easily) that each Gn is open. Figure out what Gn is in terms of F.

3. Apr 9, 2006

Wouldn't it say the exact same thing if inf were replaced by sup?

4. Apr 9, 2006

### AKG

No, not even close. Why would you think you could interchange inf and sup without it making a difference? If F is a closed ball of radius 2, then for all natural n, Gn defined with an "inf" is an open ball of radius 2 + 1/n. On the other hand, Gn defined with a "sup" would be empty for all n. Basically, the "inf" definition says that Gn is the set of points y such that there is SOME x in F such that y is close to x. A "sup" definition would say that Gn is the set of points Y such that y is close to EVERY x in F.

5. Apr 9, 2006

### Euclid

Most certainly not. Think about the usual metric in R, with F=[0,1]. Then G_n = (-1/n, 1+1/n), when we use inf. Suppose we were to use sup, we would get G_n is empty (for sufficiently large n), since the sup requirement would mean a point would have to be within 1/n of both 0 and 1.

6. Apr 9, 2006