Showing Closed Sets are Not Necessarily Open in Metric Spaces

[Treadstone 71]

"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 question 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.

 

[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 G_n 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 G_n. Prove (very very easily) that each G_n is open. Figure out what G_n is in terms of F.

 

[Treadstone 71]

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

 

[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, G_n defined with an "inf" is an open ball of radius 2 + 1/n. On the other hand, G_n defined with a "sup" would be empty for all n. Basically, the "inf" definition says that G_n 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 G_n is the set of points Y such that y is close to EVERY x in F.

 

[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.

 

[Treadstone 71]

Nevermind, I thought in terms of boundary points. That is, the set of points in F and those that are less than 1/n away from F. Simple mix up.