HELP real analysis question: continuity and compactness

[amanda_ou812]

HELP! real analysis question: continuity and compactness

Homework Statement

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).

Homework Equations

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...

 

[Fredrik]

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)?

 

[canis89]

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?

 

[Fredrik]

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.

 

[amanda_ou812]

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

 

[Fredrik]

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.

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.

Fredrik - f(K) is compact

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

 

[amanda_ou812]

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?

 

[Fredrik]

I used x-0 to mean x-nought as an arbitrary point in X.

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

[itex]x_0[/itex]

to get x₀ or $x_0$ respectively.

If I use x and y's doesn't that prove its uniformly continuous? (of course, that would mean continuous as well)

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.

f(K) compact implies f(K) is a closed and bounded subset of R which implies f(K) has a sup and inf.

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

When p-nought is contained in R, the d(p-nought, f(K)) = inf {d d(p-nought, q-nought)) : q-nought is contained in f(K)}

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.

[remember f(K) = d(p, K)] ...

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

 

[amanda_ou812]

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 k_m such that for any k contained in K, f(k_m) = inf{f(K) : k in K} and f(k_m) <= f(k). But, f(k_m) = 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.

 

[Fredrik]

there exists a t contained in R so that d(t, f(K)) = inf { |t-s|: s in contained in f(K)}

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.

there is a number k_m such that for any k contained in K, f(k_m) = inf{f(K) : k in K} and f(k_m) <= f(k).

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.

But, f(k_m) = d(t, f(K))...

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.