Complex-valued functions on a compact Hausdorff space.

[Fredrik]

A book I'm reading says (on page 5) that if X is compact and Hausdorff, every continuous function from X into ℂ is bounded. Why is that? I have only been able to prove it for metric spaces:

Suppose that f:X→ℂ is continuous but not bounded. Choose y_n such that |f(y_n)|≥n. Since X is compact, the sequence (y_n) has a convergent subsequence (x_n). Since X is Hausdorff, the limit is unique. So there's a unique x in X such that x_n→x. The assumption that f is continuous now implies that f(x_n)→f(x), but this is impossible because of how we chose the y_n in the first place.

As far as I know, it's not true in general that every sequence in a compact topological space has a convergent subsequence. That's where my proof fails for arbitrary topological spaces.

If the link doesn't work, try replacing the .se in the URL with .com or the code for your own country. (Alternatively, don't bother clicking on it. I included all the relevant information in this post).

 

[Office_Shredder]

It should be easier than that.

The continuous image of a compact space is always compact. So f(X) is a compact subset of $$\mathbb{C}$$. If it's not bounded, then the open cover B'_n of open balls at the origin of radius n is a cover with no finite subcover, which is a contradiction.

The Hausdorff part doesn't even play a role here.

 

[Fredrik]

The continuous image of a compact space is always compact.

D'oh. I should have thought of that. Thanks.

 

[klackity]

It can actually be proven directly from the definitions! (Without needing to know that the continuous image of a compact set is compact)

Consider the open cover {B_n} of open balls of radius n about the origin in $$\mathbb{C}$$. By continuity, f^-1(B_n) is open for each n, and the sets {f^-1(B_n)} cover X. As X is compact, there is a finite subcover f^-1(B_1) ... f^-1(B_N). Thus f is bounded by N.

The moral: always keep in mind the topological definitions of continuity and compactness!

(It's always useful so make clear in your mind which properties are metric properties, and which are topological properties. Continuity, compactness, openness, and convergence are purely topological notions. Bounded and Cauchy only make sense in a metric. A lot of proofs simply involve translating information about a metric or euclidean space into purely topological terms. In this case, we just needed to know that "Bounded in $$\mathbb{C}$$" really just means "Is contained in some open set in this particular collection {B_n}". Once you have the collection {B_n}, you don't ever have to think about the metric again. And that's good!)

(This is actually what Differential Geometry is based upon: we say things about abstract spaces by peering at them through spaces we know a lot about (euclidean space).)

 

[Fredrik]

Consider the open cover {B_n} of open balls of radius n about the origin in $$\mathbb{C}$$. By continuity, f^-1(B_n) is open for each n, and the sets {f^-1(B_n)} cover X. As X is compact, there is a finite subcover f^-1(B_1) ... f^-1(B_N). Thus f is bounded by N.

This is a really nice proof. Thanks.

 

[Fredrik]

A book I'm reading says (on page 5) that if X is compact and Hausdorff, every continuous function from X into ℂ is bounded.

The author also says (page 62, just above 3.1.1) that the set C(X) of continuous functions from X into ℂ, with the norm defined by

$$\|f\|=\sup_{x\in X}|f(x)|$$

is a Banach algebra. Any thoughts on that? Do I need to use that X is Hausdorff for this. (I thought I had proved that C(X) is a Banach algebra without using the Hausdorff condition, but I found a flaw in the proof while typing this, so I need to keep thinking).

 

[klackity]

The author also says (page 62, just above 3.1.1) that the set C(X) of continuous functions from X into ℂ, with the norm defined by

$$\|f\|=\sup_{x\in X}|f(x)|$$

is a Banach algebra. Any thoughts on that? Do I need to use that X is Hausdorff for this. (I thought I had proved that C(X) is a Banach algebra without using the Hausdorff condition, but I found a flaw in the proof while typing this, so I need to keep thinking).

Once we prove C(X) is a Banach space, then proving it is a Banach algebra is not too hard:

Let f, g be continuous functions on X, bounded by M1 and M2 respectively (as X is compact).
Fix c in X. Notice for each open ball about f(c)g(c),... We can find an open neighborhood U1 of c in X such that |f(x) - f(c)| < $$\epsilon$$/2M2 for all x \in U1. We can also find an open neighborhood U2 of c in X such that |g(x) - g(c)| < $$\epsilon$$/2M1 for all x \in U2. Let U = U1 $$\cap$$ U2 (also an open set in X). Thus in U, |f(x)g(x) - f(c)g(c)| < |f(x)g(x) - f(x)g(c)| + |f(x)g(c) - f(c)g(c)| < $$\epsilon$$/2 + $$\epsilon$$/2 < $$\epsilon$$.

Thus C(X) is closed under multiplication.

Then we have: $$\|fg\|=\sup_{x\in X}|f(x)g(x)|$$ < $$\sup_{x\in X}|f(x)| \sup_{x\in X}|g(x)| = \|f\|\|g\|$$

To prove C(X) is in fact a banach space, you need to prove the above for pointwise addition of functions instead of pointwise multiplication, and you need to prove C(X) is complete.

I'm not sure where Hausdorff comes in... perhaps in proving completeness?

 

[Fredrik]

It was the completeness proof that was causing me difficulties. Perhaps I should have been more clear about that.

This is the first half of my failed attempt to prove completeness: (The half I think is OK).

Let (f_n) be an arbitrary Cauchy sequence in C(X), and let ε>0 be arbitrary. There exists an integer N such that

$$n,m\geq N\Rightarrow\varepsilon>\|f_n-f_m\|=\sup_{x\in X}|f_n(x)-f_m(x)|\geq|f_n(y)-f_m(y)|$$

for all y in X. So for each x in X, (f_n(x)) is a Cauchy sequence in ℂ. ℂ is a complete metric space, so these sequences have limits, which I'll write as f(x).Now I want to show that this implies that f_n→f, but I haven't been able to do so. I'll try again tomorrow.

 

[Fredrik]

I still don't get it. I would appreciate if someone could give me a hand here. I need to show that if X is compact and Hausdorff, the set C(X)={f:X→ℂ|f continuous} with the norm

$$\|f\|=\sup_{x\in X}|f(x)|$$

is a Banach algebra. I think I understand why it's a normed algebra, but not why it must be complete.

 

[disregardthat]

I still don't get it. I would appreciate if someone could give me a hand here. I need to show that if X is compact and Hausdorff, the set C(X)={f:X→ℂ|f continuous} with the norm

$$\|f\|=\sup_{x\in X}|f(x)|$$

is a Banach algebra. I think I understand why it's a normed algebra, but not why it must be complete.

If I have understood you correctly, you want to show that the set of continuous functions from X to the complex numbers C(X) is complete under the uniform metric? I think a proof might go as follows:

$$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$ for any integer n. Choose N large enough so that $$|f_n(x)-f(x)|,|f_n(y)-f(y)|< \frac{\epsilon}{3}$$ for $$n \geq N$$, and by continuity of $$f_N$$ choose a $$\delta$$ such that $$|x-y|<\delta \Rightarrow |f_N(x)-f_N(y)|<\frac{\epsilon}{3}$$. Then $$|x-y| \leq \delta \Rightarrow |f(x)-f(y)| \leq |f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|<\epsilon$$, and f is continuous.

Now $$|f_n-f|$$ is continuous on a compact set X to R and hence bounded, so it has a maximum, say $$|f_n(x_n)-f(x_n)|$$. Then $$||f_n-f|| = |f_n(x_n)-f(x_n)| \leq |f_n(x_n)-f_m(x_n)|+|f_m(x_n)-f(x)|$$. Choose N large enough so that $$||f_n-f_m|| < \frac{\epsilon}{2}$$ when $$n,m \geq N$$. Choose T such that $$|f_m(x_N)-f(x_N)| < \frac{\epsilon}{2}$$ when $$m \geq T$$, and let M be the largest of N and T. Then $$||f_N-f|| \leq |f_N(x_N)-f_M(x_N)|+|f_M(x_N)-f(x_N)|<\frac{\epsilon}{2}+ \frac{\epsilon}{2} = \epsilon$$.

Hence f_n converges to f in C(X) in the uniform norm.

Tricky stuff, might have made a mistake or two. Are you sure you haven't used the Hausdorff condition already? I don't think it is required for completeness.

 

[Fredrik]

Thanks, I think that helps. But I think I see a mistake in the first part of the proof. In the first few lines, x and y are arbitrary members of X. You choose N such that n≥N implies |f_n(x)-f(x)|<ε/3 and |f_n(y)-f(y)|<ε/3. This N may depend on x and y, so things get weird when you choose x close to y. Also, when you do that, δ may depend on N.

This is still better than what I did, because I completely overlooked the fact that I have to prove that the f I found in #8 is continuous.

I haven't used the Hausdorff condition anywhere, at least not deliberately. The proofs are pretty straightforward, so it seems unlikely that I used it without realizing it. My book makes the following claims on page 5:

If X is a non-empty open set, B(X)={f:X→ℂ|f bounded} with the uniform norm, is a normed vector space. If X is compact and Hausdorff, C(X)={f:X→ℂ|f continuous} with the same norm, is a subspace of the former. If X is locally compact and Hausdorff, C₀(X)={f:X→ℂ|For each ε>0, there's a compact K such that |f|<ε on ℂ-K}, with the same norm, then C₀(X) is also a subspace of B(X).

It seems like the assumptions he mentions should be important. He's talking about Banach spaces here instead of Banach algebras because the definition of a Banach algebra doesn't appear until later in the book, but once he has defined "Banach algebra", he says that this C(X) is a Banach algebra.

Edit: I thought I understood your completeness proof, but now I think I don't. You showed that for arbitrary n,m, we have

$$\|f_n-f\|=|f_n(x_n)-f(x_n)|\leq |f_n(x_n)-f_m(x_n)|+|f_m(x_n)-f(x_n)|\leq \|f_n-f_m\|+|f_m(x_n)-f(x_n)|$$

where x_n is the point where |f_n-f| has its maximum. You chose N such that

$$n,m\geq N\Rightarrow \|f_n-f_m\|<\frac{\varepsilon}{2}$$

This is fine since (f_n) is a Cauchy sequence. Then you chose T_n (one for each n≥N or just one for n=N?) such that

$$m\geq T_n\Rightarrow |f_m(x_n)-f(x_n)|<\frac{\varepsilon}{2}$$

This is also fine, since f_m(x_n)→f(x_n). You defined M=max(N,T_N), and you chose m=M in the inequality that holds for arbitrary n,m. This turns it into

$$\|f_n-f\|\leq \|f_n-f_M\|+|f_M(x_n)-f(x_n)|$$

Now for all n≥M, the first term is <ε/2, but the second is only guaranteed to be <ε/2 specifically when n=N, not for all n≥M.

 

[disregardthat]

Thanks, I think that helps. But I think I see a mistake in the first part of the proof. In the first few lines, x and y are arbitrary members of X. You choose N such that n≥N implies |f_n(x)-f(x)|<ε/3 and |f_n(y)-f(y)|<ε/3. This N may depend on x and y, so things get weird when you choose x close to y. Also, when you do that, δ may depend on N.

You are right. I will see if I can make it work somehow.

If X is a non-empty open set, B(X)={f:X→ℂ|f bounded} with the uniform norm, is a normed vector space. If X is compact and Hausdorff, C(X)={f:X→ℂ|f continuous} with the same norm, is a subspace of the former. If X is locally compact and Hausdorff, C0(X)={f:X→ℂ|For each ε>0, there's a compact K such that |f|<ε on ℂ-K}, with the same norm, then C0(X) is also a subspace of B(X).

Ok, it seems to me that he is suggesting that X must be hausdorff in order for continuous functions to be bounded. However that is not required.

 

[disregardthat]

I can't figure it out for now. I don't think we can use that f is continuous before showing that f_n converges uniformly towards f. Showing continuity from that however is easy.

So I guess we will need to show that ||f_n-f|| --> 0 without the continuity condition on f.

 

[Fredrik]

You made a couple of posts while I was adding stuff to #11. You might want to take a look at those comments too. I also noticed that exercise 1.5.2 (b) (on page 13 in the book, and page 17 in this pdf) asks us to prove something similar to what we're trying to do.

The link is to the author's web page, so it's a legal download.

 

[disregardthat]

This is also fine, since f_m(x_n)→f(x_n). You defined M=max(N,T_N), and you chose m=M in the inequality that holds for arbitrary n,m. This turns it into

$$\|f_n-f\|\leq \|f_n-f_M\|+|f_M(x_n)-f(x_n)|$$

Now for all n≥M, the first term is <ε/2, but the second is only guaranteed to be <ε/2 specifically when n=N, not for all n≥M.

An error on my part there, but I think it is minor. M can be chosen for each n here, I will rewrite the whole argument for clarity:

Now $$|f_n-f|$$ is continuous on a compact set X to R and hence bounded, so it has a maximum, say $$|f_n(x_n)-f(x_n)|$$. Then $$||f_n-f|| = |f_n(x_n)-f(x_n)| \leq |f_n(x_n)-f_m(x_n)|+|f_m(x_n)-f(x_n)|$$ for all m. Choose N large enough so that $$||f_n-f_m|| < \frac{\epsilon}{2}$$ when $$n,m \geq N$$. Now, for each $$m,n \geq N, |f_n-f| <\frac{\epsilon}{2} + |f_m(x_n)-f(x_n)|$$. Now, for each n, choose M_n larger than N such that $$|f_m(x_n)-f(x_n)|<\frac{\epsilon}{2}$$ for all $$m \geq M_n$$. Then, for each $$n \geq N$$, we have $$|f_n-f| <\frac{\epsilon}{2} + |f_{M_n}(x_n)-f(x_n)|<\epsilon$$.

However this is under the assumption that f is continuous, which we have not proved.

 

[disregardthat]

All right... I have some supplement here, I think this will work:

We know that (f_n) is a cauchy-sequence under the sup-norm from a compact space to a metric space. Thus f_n are all bounded. We define f(x) as the limit of f_n(x).

$$|f(x)-f_n(x)| \leq |f(x)-f_m(x)|+|f_m(x)-f_n(x)| \leq |f(x)-f_m(x)|+||f_m-f_n|| \leq |f(x)-f_m(x)| +\sup_{p,q \geq M}||f_p-f_q|| \to \sup_{p,q \geq M}||f_p-f_q||$$

as $$m \to \infty$$ for all $$n \geq M$$ for any M. This applies for all x, so $$||f-f_n|| \leq \sup_{p,q \geq M}||f_p-f_q||$$ Choose N such that $$||f_q-f_p|| < \epsilon$$ whenever $$p,q \geq N$$. Then $$||f-f_n|| \leq \epsilon$$ when $$n \geq N$$, so $$||f-f_n|| \to 0$$.

Now continuity of f can be shown:

$$|f(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f(y)| \leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$ for any integer n. Choose N large enough so that $$||f_n-f|| < \frac{\epsilon}{3}$$ for all $$n \geq N$$. Then $$|f_N(x)-f(x)|,|f_N(y)-f(y)|< \frac{\epsilon}{3}$$ for $$n \geq N$$ for all x,y in X, and by continuity of $$f_N$$ choose a $$\delta$$ such that $$|x-y|<\delta \Rightarrow |f_N(x)-f_N(y)|<\frac{\epsilon}{3}$$. Then $$|x-y| < \delta \Rightarrow |f(x)-f(y)| \leq |f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y)-f(y)|<\epsilon$$, and f is continuous.

 

[Fredrik]

Thank you very much. I understand the proof in #15 now. I have no objections against this version of it. I need to go out for a while, but I'll take a look at #16 when I come back.

 

[Fredrik]

Your continuity proof has the same kind of problem as your original completeness proof. Your δ will depend on x and y. I've been trying to find a way around it, but so far without success. This is my version of what you did in #16:

For each x, choose N_x so that

$$n\geq N_x\Rightarrow |f_n(x)-f(x)|<\frac{\varepsilon}{3}$$

(This is possible because of the pointwise convergence). For each pair (x,y), define N_xy=max{N_x,N_y}.

For each n, choose δ_n so that

$$|x-y|<\delta_n\Rightarrow |f_n(x)-f_n(y)|<\frac{\varepsilon}{3}$$

(This is possible because each f_n is continuous). Then we have

$$|x-y|<\delta_{N_{xy}}\Rightarrow|f(x)-f(y)|\leq|f(x)-f_{N_{xy}}(x)|+|f_{N_{xy}}(x)-f_{N_{xy}}(y)|+|f_{N_{xy}}(y)-f(y)|<\varepsilon$$



Another problem is that the original assumptions don't say that X is a normed vector space, and |x-y| is undefined when it isn't.

Edit: I have an idea that looks promising. One of the equivalent definitions of continuity says that f is continuous at x if x_n→x implies f(x_n)→f(x). So suppose that x_n→x. We have

$$|f(x_n)-f(x)|=|\lim_k f_k(x_n)-\lim_k f_k(x)|=\lim_k |f_k(x_n)-f_k(x)|$$

and also

$$|f_k(x_n)-f_k(x)|\rightarrow 0$$

when n→∞.

 

[disregardthat]

Your continuity proof has the same kind of problem as your original completeness proof. Your δ will depend on x and y. I've been trying to find a way around it, but so far without success. This is my version of what

Are you sure? delta can and will always depend on x however, but by the continuity of f_N (only one of the functions), such a delta will exist, and I don't see how it depends on y. The proof shows continuity of f in x. N does not depend on x or y since N is a number extracted from the sup-norm limit, not the pointwise one.

 

[disregardthat]

Edit: I have an idea that looks promising. One of the equivalent definitions of continuity says that f is continuous at x if x_n→x implies f(x_n)→f(x). So suppose that x_n→x. We have

$$|f(x_n)-f(x)|=|\lim_k f_k(x_n)-\lim_k f_k(x)|=\lim_k |f_k(x_n)-f_k(x)|$$

and also

$$|f_k(x_n)-f_k(x)|\rightarrow 0$$

when n→∞.

Hm, interesting, but are you sure you can interchange limits like that? I don't know the criterion, but it is not allowed if the elements can be negative. In this case they are all positive, so it might work.

 

[disregardthat]

$$|x-y|<\delta_{N_{xy}}\Rightarrow|f(x)-f(y)|\leq|f(x)-f_{N_{xy}}(x)|+|f_{N_{xy}}(x)-f_{N_{xy}}(y)|+|f_{N_{xy}}(y)-f(y)|<\varepsilon$$

delta cannot depend on y as it does here, it can only depend on epsilon and x. I believe it is essential that you establish uniform convergence before showing continuity, because there are many examples of pointwise converging continuous functions not having continuous limits.

Another problem is that the original assumptions don't say that X is a normed vector space, and |x-y| is undefined when it isn't.

Wow, I have completely overlooked that. Maybe the epsilon-delta proof can be turned into a proof of continuity in some way. In my proof (which I think is correct, if the possible delta-issue is not an issue at all), instead of choosing delta at that point in the argument:
- We choose a neighbourhood U of x such that if y is an element of U, then |f_N(x)-f_N(y)|< epsilon. This is the continuity condition on f_N (since epsilon balls form a basis in the range space). Keeping U as the restriction on y (ignoring the delta part), the rest of the proof goes as it stands.

 

[Fredrik]

N does not depend on x or y since N is a number extracted from the sup-norm limit, not the pointwise one.

Ah, you chose your N in a smarter way than I did. Let me try again then...

Let (f_n) be a Cauchy sequence in C(X). We have already shown that f_n→f. This means that there's an integer N such that

$$n\geq N\Rightarrow \|f_n-f\|<\frac{\varepsilon}{3}$$

Let x be arbitrary. We're going to prove that f is continuous at x. For each n, choose δ_n so that

$$|x-y|<\delta_n\Rightarrow |f_n(x)-f_n(y)|<\frac{\varepsilon}{3}$$

(This is possible because each f_n is continuous). Then we have

$$|x-y|<\delta_{N}\Rightarrow|f(x)-f(y)|\leq|f(x)-f_{N}(x)|+|f_{N}(x)-f_{N}(y)|+|f_{N}(y)-f(y)|$$

$$\leq \|f-f_N\|+\frac{\varepsilon}{3}+\|f-f_N\|<\varepsilon$$

OK, this looks good. (δ_N depends on x, but it's supposed to do that, since it's the radius of the appropriate open ball around x. What's important is that it doesn't also depend on y).

Hm, interesting, but are you sure you can interchange limits like that?

No, I'm not sure. I still have details to work out. I won't be doing that tonight, but I'll try again tomorrow.

 

[Fredrik]

I think I can do the whole thing now. Suppose that (f_n) is a Cauchy sequence in C(X), and let ε>0 be arbitrary.

Pointwise convergence:

There's an N such that

$$n,m\geq N\Rightarrow \varepsilon>\|f_n-f_m\|=\sup_{x\in X}|f_n(x)-f_m(x)|\geq|f_n(y)-f_m(y)|$$

for each y in X. So for each x in X (f_n(x)) is a Cauchy sequence in ℂ, which is a complete metric space, so these sequences are convergent. Denote the limits by f(x). This defines a function f:X→ℂ.

Uniform convergence/Completeness (without assuming continuity):

I'm going to show that f_n→f. Choose N such that

$$n\geq N\Rightarrow\|f_n-f_m\|<\frac{\varepsilon}{3}$$

For each x, choose N_x such that

$$n\geq N_x\Rightarrow |f_n(x)-f(x)|<\frac{\varepsilon}{3}$$

For each n, let x_n be such that

$$\|f_n-f\|<|f_n(x_n)-f(x_n)|+\frac{\varepsilon}{3}$$

Such an x_n must exist, because otherwise there's an upper bound that's smaller than the supremum. For all n≥N, we have

$$\|f_n-f\|<|f_n(x_n)-f(x_n)|+\frac{\varepsilon}{3}\leq|f_n(x_n)-f_{N_{x_n}}(x_n)|+|f_{N_{x_n}}(x_n)-f(x_n)|+\frac{\varepsilon}{3}$$

$$<\|f_n-f_{N_{x_n}}\|+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}<\varepsilon$$

Continuity:

I'm going to show that f is continuous, by showing that it's continuous at each x. Let x be arbitrary. Choose N such that

$$n\geq N\Rightarrow\|f_n-f\|<\frac{\varepsilon}{3}$$

Let E be an arbitrary open subset of ℂ that contains f(x). Since E is open, there's an open ball B with radius r<ε/2 around f(x) such that

$$f(x)\subset B\subset E$$

Let (x_i) be a net such that x_i→x. f_N^-1(B) is open since f_N is continuous, so there's an i₀ such that

$$i\geq i_0\Rightarrow x_i\in f_N^{-1}(B)$$

$$x_i\in f_N^{-1}(B)\Leftrightarrow f_N(x_i)\in B\Leftrightarrow |f_N(x_i)-f(x)|<r<\frac{\varepsilon}{2}$$

For all i≥i₀, we have

$$|f(x_i)-f(x)|\leq|f(x_i)-f_N(x_i)|+|f_N(x_i)-f(x)|<\|f_N-f\|+\frac{\varepsilon}{2}<\varepsilon$$

So f(x_i)→f(x), and therefore f is continuous at x. But x was arbitrary so f is continuous on all of X.

I was also able to find an alternative proof of continuity, that doesn't use nets. But it's getting late, so I won't type it up now. Maybe tomorrow.

 

[Fredrik]

Here's a continuity proof that doesn't involve nets. f is continuous at x, if f^-1(E) contains an open set that contains x, for each open set E that contains f(x).

Let x be arbitrary, and let E be an arbitrary open set that contains f(x). Since E is an open set in a metric space, there's an open ball B around f(x) with radius r, such that

$$f(x)\in B\subset E$$

Choose N such that

$$n\geq N\Rightarrow \|f_n-f\|<\frac{r}{2}$$

and let B' be the open ball around f(x) with radius r/2. Then

$$|f_N(x)-f(x)|\leq\|f_N-f\|<\frac{r}{2}\quad\Rightarrow f_N(x)\in B'\quad\Rightarrow x\in f_N^{-1}(B')$$

This set is open, because f_N is continuous. So if we can prove that it's a subset of f^-1(E), we're done. We will do this by showing that it's a subset of f^-1(B), which clearly is a subset of f^-1(E). Let y be an arbitrary member of f_N^-1(B'). We have

$$y\in f_N^{-1}(B')\Rightarrow f_N(y)\in B'\Rightarrow |f_N(y)-f(x)|<\frac{r}{2}$$

and therefore

$$|f(y)-f(x)|\leq |f(y)-f_N(y)|+|f_N(y)-f(x)|<\|f-f_N\|+\frac{r}{2}=r$$

which implies that

$$y\in f^{-1}(B)$$

Thanks again to everyone who helped, especially Jarle.

 

[Landau]

Sorry for digging up an old thread.

Hm, interesting, but are you sure you can interchange limits like that? I don't know the criterion, but it is not allowed if the elements can be negative. In this case they are all positive, so it might work.

I don't understand you remark about positive/negative. Isn't this just the 'sum law' of limits? i.e. you know lim f_n and lim g_n exist, then lim (f_n-g_n) = lim f_n - lim g_n. As the absolute value function is continuous, you can also interchange limit and absolute value.

As for the rest of this thread: I would do it like this (in short):

If (f_n) is a Cauchy sequence in C(X), then it is a pointwise Cauchy sequence in C. Hence we can form the pointwise limit f(x)=lim f_n(x). The Cauchy assumption means that for every e>0 there is N such that if n,m>N then $\|f_m-f_n\|<e$. So for fixed x, we have

$$|f(x)-f_n(x)|\leq|f(x)-f_m(x)|+\|f_m-f_n\|<|f(x)-f_m(x)|+e.$$

if n,m>N. Letting $m\to\infty$ we get that, for n>N, $|f(x)-f_n(x)|\leq e$. But x was arbitrary, so in fact, for n>N, $\|f-f_n\|\leq e$. I.e. f_n converges to f in the sup-norm, as was to be shown.
It is then a basic fact from elementary analysis that the uniform limit of continuous functions is continuous. Indeed, we have the estimate

$$|f(x)-f(y)|\leq |f(x)-f_n(x)|+|f_n(x)-f_n(y)|+|f_n(y)-f(y)|$$
$$\leq 2\|f-f_n\|+|f_n(x)-f_n(y)|.$$

Now given e>0, the first term is <e if n is large enough, say larger than N. Subsequently, for a fixed n>N, the second term is <e if we take |x-y| smaller than some delta (by continuity of all the f_n, so in particular the f_n for our fixed n>N). Hence we have found delta such that the total is <2e if |x-y|<delta, rendering f continuous.