Distance between Compact Subspaces: Proving Existence of Minimum Distance

[fireisland27]

Homework Statement

For a metric space (M,d) and two compact subspaces A and B define the distance d(A,B) between these sets as inf{d(x,y): x in A and y in B}. Prove that there exists an x in A and a y in B such that d(x,y)=d(A,B).

Homework Equations

The Attempt at a Solution

I want to say that d can be thought of as a function from AxB to R and because A and B are compact AxB is compact. And so if I can show that d is continuous then it achieves it's minimum by the extreme value theorem. I'm not sure how to show the continuiuty however, or if this is even the proper way to go about this.

 

[arkajad]

Well, you want to show that if x_n\rightarrow x, y_n\rightarrow y then d(x_n,y_n)\rightarrow d(x,y), right?

Do you know already that d(x,y) is continuous in each variable separately?

 

[fireisland27]

I believe that if y is fixed then d_y(x) = d(x,y) is continuous by a very simple epsilon delta proof. Is this what you mean? If so d(x_n,y) converges to d(x,y). Does this automatically give us that d(x_n,y_n) converges to d(x,y)?

 

[Fredrik]

Start with |d(x_n,y_n)-d(x,y)|, add and subtract something clever, and use the triangle inequality.

 

[fireisland27]

Ok, I think it all comes together now. The only part I want to be sure of is, given a sequence {(x,y)_n} in AxB which converges to (x,y), then is it as obvious as it seems that x_n converges to x in A and y_n converges to y in B? Does this require a proof?
If not than I think I can now show that for a sequence {(x,y)_n} converging to (x,y), {d(x,y)_n} converges to d(x,y) making d continuous and thus attaining its minimum on AxB which would be the distance between A and B. Does this all seem correct?

 

[Fredrik]

I think you should prove it. The general approach looks correct to me (but I didn't think very hard about it).