# HELP real analysis question: continuity and compactness

1. Oct 27, 2011

### amanda_ou812

HELP!!! real analysis question: continuity and compactness

1. The problem statement, all variables and given/known data
Let (X,d) be a metric space, fix p ∈ X and define f : X → R by f (x) = d(p, x). Prove that f is continuous. Use this fact to give another proof of Proposition 1.126.

Proposition 1.126. Let (X, d) be a metric space, let K ⊆ X and let p ∈ X. If K is compact, then there is a point q0 ∈ K, such that dist(p, K) = d(p, q0).

2. Relevant equations

3. The attempt at a solution
I think I know how to do the first part of the question but I don't know how to do the second part.

part 1
given ε>0 let delta = ε, then when d(p, x)< delta we have abs( f(p) - f(x)) = abs( d(p, p) - d(p, x)) = abs (0 - d(p, x)) = d(p, x) < delta = ε.

part 2
I am not sure how to do this. Suppose K is compact and that f: K-> R is defined by f(q0) = d(p, q0). Well f continuous implies....I am not sure where to go from here....

2. Oct 28, 2011

### Fredrik

Staff Emeritus
Re: HELP!!! real analysis question: continuity and compactness

Define f:X→ℝ by f(x)=d(x,p) for all x. f is continuous and K is compact. What does that tell you about the range f(K)?

3. Oct 28, 2011

### canis89

Re: HELP!!! real analysis question: continuity and compactness

Hi amanda, about the first part, you only show that f is continuous on p. If you want to show that f is continuous on X, then you have to show that for any $x_0\in X,$ for every $\epsilon>0,$ there exists $\delta>0$ such that $d(x,x_0)<\delta$ implies $|f(x)-f(x_0)|<\epsilon$. Or maybe you only want to show that f is continuous on p?

4. Oct 28, 2011

### Fredrik

Staff Emeritus
Re: HELP!!! real analysis question: continuity and compactness

Good point canis89. I didn't even notice that. We need f to be continuous everywhere (or at least on all of K) for my approach to part 2 to work.

5. Oct 28, 2011

### amanda_ou812

Re: HELP!!! real analysis question: continuity and compactness

Ah, I see your point. Given e>0 let delta = e the for x-0 contained in X we have d(x, x-0)< delta implies abs (f(x) - f(x-0)) = abs ( d(p, x) - d(p, x-0)) <= abs (d(x, x-0)) [reverse triangle inequality] < delta = e.

Fredrik - f(K) is compact

6. Oct 28, 2011

### Fredrik

Staff Emeritus
Re: HELP!!! real analysis question: continuity and compactness

You seem to have the right idea, but I don't know what you mean by x-0 (or - or 0). The proof should go like this:

We're going to prove that f is continuous. This is equivalent to proving that f is continuous at x for all x in X. Let x be an arbitrary member of X. Let ε>0 be arbitrary. Choose δ=ε. For all y in X such that d(x,y)<δ, we have |f(y)-f(x)|= ... <ε.

That's what I had in mind. Now that you just need to figure out how to use that.

7. Oct 28, 2011

### amanda_ou812

Re: HELP!!! real analysis question: continuity and compactness

I used x-0 to mean x-nought as an arbitrary point in X. If I use x and y's doesn't that prove its uniformly continuous? (of course, that would mean continuous as well)

f(K) compact implies f(K) is a closed and bounded subset of R which implies f(K) has a sup and inf. When p-nought is contained in R, the abs((p-nought - f(K)) = inf { abs(p-nought - q-nought) : q-nought is contained in f(K)} [remember f(K) = d(p, K)] so there exists a q-nought contained in f(K) and p-nought contained in R s.t. dist(p-nought, f(K)) = dist (p-nought, q-nought). That seems to be what 1.126 says except its for the images of K but would it be ok to take f(K) = A where A is an arbitrary subset of a metric space (X, d)? Then would that statement prove 1.126?

Last edited: Oct 28, 2011
8. Oct 28, 2011

### Fredrik

Staff Emeritus
Re: HELP!!! real analysis question: continuity and compactness

I see. I recommend that you use either vBulletin's sub tags, or the latex code x_0. In other words, you type either [noparse]x0[/noparse] or
Code (Text):
$x_0$
to get x0 or $x_0$ respectively.

What symbols we use certainly doesn't matter, but the fact that the delta you found doesn't depend on the point under consideration means that this f is uniformly continuous.

Every bounded subset of ℝ has a supremum and an infimum.

You seem to be saying that for all t in ℝ, d(t,f(K))=inf{|t-s| : s is in f(K)}? This correct, but we're not interested in d(t,f(K)). We only care about d(x,K). You also shouldn't have used the variable q0, since it was already being used for something else.

f(K) is a set of non-negative real numbers. d(p,K) is a non-negative real number.

9. Oct 28, 2011

### amanda_ou812

Re: HELP!!! real analysis question: continuity and compactness

So, since there exists a t contained in R so that d(t, f(K)) = inf { |t-s|: s in contained in f(K)} there is a number km such that for any k contained in K, f(km) = inf{f(K) : k in K} and f(km) <= f(k). But, f(km) = d(t, f(K))...

I don't think I am understanding the problem. So, I defined a function f with the based on part 1 and I know that K is compact. I am not understanding what I need to show to prove that there is a point q0 ∈ K, such that dist(p, K) = d(p, q0). I mean, 1,126 just seems intuitive to me. The questions for this class are always so difficult for me to do.

Last edited: Oct 28, 2011
10. Oct 28, 2011

### Fredrik

Staff Emeritus
Re: HELP!!! real analysis question: continuity and compactness

Hm, I suppose you could say that, since every real number t has that property (by definition of d(t,f(K))). But it's a weird thing to say, like saying that there's an integer that's also a real number.

This statement is flawed in several ways. You're saying that there's a $k_m\in\mathbb R$ such that for all $k\in K$, we have $f(k_m)=\inf\{f(K):k\in K\}$ and $f(k_m)\leq f(k)$. What is $\{f(K):k\in K\}$ supposed to mean? It should be "the set of all f(K) such that k is a member of K". That would be the set {f(K)}, i.e. the set that has f(K) as its only member. Maybe you meant $\{f(k')|k'\in K\}$. But this is just $f(K)$. Either way, the statement is of the form "there's a $k_m\in\mathbb R$ such that for all $k\in K$, we have $f(k_m)=$<something that that has nothing to do with K>". And I haven't even mentioned the biggest problem with that statement yet: $k_m$ isn't a member of the domain of f.

All you have said about t so far is that it's a real number that has a property that all real numbers have. Since f(K)≠ℝ, it wouldn't be hard to find a t that's not at distance $f(k_m)$ from f(K)?

What you need to do is to figure out how to use the fact that f(K) is closed and bounded.

Last edited: Oct 28, 2011