Infimum calculation bizarre for me

[julypraise]

Homework Statement

Could you please check the following calculation is right?

Let X be a metric space, and A its nonempty subset.

Define \inf_{a \in A} d(x,a) = d(x,A) for any x in X

We have the following facts (don't have to check this)

If a is in the closure of A then d(a,A)=0.

So my calculuation is as follows. Let x be in X and a_0 be in the closure of A

Then

d(x,a_{0}) \leq \inf_{a\in A} \{ d(x,a) + d(a,a_{0}) \} = \inf_{a\in A}d(x,a) + \inf_{a\in A}d(a,a_{0}) = d(x,A)

Homework Equations

The Attempt at a Solution

The calculation looks somehow right (probably because of my bad). I have to show the existence of such a_0. In R^2 dimension, I can draw this and get it, but such a point a_0 is not arbitrary in that case whereas the a_0 in my calculation is arbitrary and works fine which is very stupid...

Probably the equality made in the mid is not right... But if it is not right, how can I show the existence of such a_0?

 

[Ayre]

The last step of your calculation is not correct, and anyway, that approach is not the way you want to proceed.

Instead, you need to use the fact that a_0 lies in the closure of A. Write down as many equivalent definitions you can of a_0 being in the closure of A. One of them will be what you should use.

 

[julypraise]

The last step of your calculation is not correct, and anyway, that approach is not the way you want to proceed.

Instead, you need to use the fact that a_0 lies in the closure of A. Write down as many equivalent definitions you can of a_0 being in the closure of A. One of them will be what you should use.

Thank you very much, I will do that. But if I may aks you, what exactly do you refer to by the last step?

you mean the last equality, I guess?

 

[Ayre]

you mean the last equality, I guess?

Yes. I mean, if you assume the result you're trying to prove, then that equality is correct. But, of course, you're not allowed to assume what you're trying to prove.

 

[julypraise]

Yes. I mean, if you assume the result you're trying to prove, then that equality is correct. But, of course, you're not allowed to assume what you're trying to prove.

Hey, sorry but I'm not sure what you are talking abt.

The thing that if a is in the closure of A then d(a,A)=0 is right.

But I'm trying to prove for arbitrary x in X there exists a_0 in the closure of A

such that d(x,a_0) <= d(x,A). But I'm stuck...

 

[Ayre]

Hey, sorry but I'm not sure what you are talking abt.

The thing that if a is in the closure of A then d(a,A)=0 is right.

But I'm trying to prove for arbitrary x in X there exists a_0 in the closure of A

such that d(x,a_0) <= d(x,A). But I'm stuck...

I'm very sorry. I misread your question, I thought you were trying to show that d(a,A)=0 for a in the closure of A.

But your calculation is still not correct. The formula

\inf_{a\in A} (d(x,a) + d(a,a_0)) = \inf_{a\in A}d(x,a) + \inf_{a\in A} d(a,a_0)\text,

which you are using, does not hold.

Instead, to find an a_0, start by taking a sequence a_n in a such that

d(x, a_n) \rightarrow d(x, A)\text,

and showing that it converges.

Again, apologies for misreading your question. Please ignore everything I said in the earlier posts.

 

[julypraise]

Oh no worries; apology well taken.

And yes, that part is wrong I see.

Okay, I think the sequence is a very good idea.

So if I solve it right now,

letting x be arbitrary in X

observe for every n in N (the set of natural numbers)

there exists a number a(n) in \{ d(x,a) \mid a \in A \} such that

d(x,A) = inf{...} <= a(n) < d(x,A) + 1/n.

Then a(n) = d(x,a'(n)) for some a'(n) in A.

Then it is obvious the sequence (a(n)) = (d(x,a'(n))) converges to d(x,A).

We let a_n := a'(n). By this we construct a sequence (a_n) in A.

Now I'm kinda stuck here again. I know a function a\mapsto d(x,a) defined on the closure of A is continuous, but cannot use it here. I've tried to use this continuity like

\lim_{n \to \infty} d(x,a_{n}) = d(x, \lim_{n \to \infty} a_{n}),

but this holds if we already know a_n converges to some a_0 in the colsure of of A
Another attempt is that

d(x,a_n) converges to a point in the closure of {d(x,a) | a in A}

But is it going to be Cl({d(x,a) | a in A}) = {d(x,a) | a in Cl(A)}??

Need to check..

 

[Ayre]

Let me rephase that. The sequence a_n will not really converge, but a subsequence does. What properties do you know that guarantee that sequences have convergent subsequences?

(I realize now that this solution method will be slightly roundabout. Perhaps you can see how to simplify it later.)

 

[julypraise]

I know Bolzano-Weierstrass thm in R^k. But A is a subspace of a general metric space,
so I don't think I can use it...

It just reminds me now that

actually in a general space it may be possible that such a point a_0 may not exist,

and for such a_0 to exist, I think it is necessary that A is compact.

In this case I define a function f_x

f_x:A \to \mathbb{R}: a \mapsto d(x,a).

Then since d is continuous (and also the restircted d|_A), this function f will be continuous too.

Thus if I use Weierstrass thm,

I get inff(A) is in f(A), which was what I longed to get.