# Is the space $C[0,1]$ complete?

In order to prove $C[0,1]$ is complete, my functional analysis book says:

"It is only necessary to show that every Cauchy sequence in $C[0,1]$ has a limit".

It goes on by supposing $\{x_n\}$ is a Cauchy sequence in $C[0,1]$. Then for each fixed $t \in [0,1]$ $|x_n(t)-x_m(t)| \le \|x_n - x_m\| \to 0$, so $\{x_n(t)\}$ is a Cauchy sequence of real numbers. I guess this inequality makes sense, since difference of function sequences would be larger than pointwise difference (at a specific $t$) of those functions.. But I did not understand how the author could conclude the sequence $\{x_n(t)\}$ is Cauchy.

Second question: the explanation continues,

"Since the set of real numbers is complete, there is a real number $x(t)$ to which the squence converges; $x_n(t) \to x(t)$".

I know set of real numbers is complete, but how can $\{x_n(t)\}$ represent the entire set of real numbers?

Thanks,

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keep in mind, that $\|x\| := max\{|x(t)|; t\in [0,1]\}$, so if $\{x_n\}$ is Cauchy in $C[0,1]$ and you know, by definition, $|x_n(t)-x_m(t)|\le \|x_n-x_m\|$ then $\{x_n(t)\}$ is Cauchy for every fixed $t$ as the latter is. Therefore you have a Cauchy sequence of real numbers, and by completeness of $\mathbb{R}$ this converges to a real number $r_t:=x(t)$ – math Jan 8 '12 at 12:14
May I suggest the following? Keep this problem aside, and carefully prove -- by yourself, and with all the small details -- that $X \times X$ is complete if $X$ is. Both your questions seem to have little to do with $C[0,1]$. – Srivatsan Jan 8 '12 at 12:14
Just to elaborate on a point made by other people: in your 2nd question, it helps to break down the explanation as follows. (i) Fix $t$ (ii) The sequence $(x_n(t))$ is a Cauchy sequence in ${\mathbb R}$, so by completeness of ${\mathbb R}$ it has a limit, which we call $y_t$. (The subscript denotes the fact that the sequence, and hence the limit, can depend on $t$.) (iii) So for each $t$, we have produced a number $y_t$. (iv) Define $x$ to be the function $t\mapsto y_t$. (v) Show that $x$ is continuous (vi) Show $\Vert x_n -x \Vert_\infty \to 0$. – user16299 Jan 8 '12 at 19:26

Fix $t_0$, $\epsilon > 0$. Since $x_n$ are Cauchy, there exists $N > 0 | \forall n,m > N$ we have $$||x_n - x_m || := \sup_t |x_n(t) - x_m(t)| < \epsilon$$In particular, for $t_0$, and the same $N$, for all $n,m > N$ we have $|x_n(t_0) - x_m(t_0)| < \sup_t |x_n(t) - x_m(t)| < \epsilon$.

Now we know that $x_n(t_0)$, just a sequence of real numbers obtained by evaluating the functions $x_1, x_2, x_3 \ldots$ at $t_0$ is Cauchy. As a Cauchy sequence of real numbers, it must be a convergent sequence.

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This is a particular case of the more general statement that if $K$ is a compact metric space then $C(K)$ with the topology induced by the $\sup$-norm is complete.

To prove this you do the following:

You need to show that the limit of every sequence that is Cauchy with respect to $\| \cdot \|_\infty$ is in $C(K)$. So let $f_n$ be a Cauchy sequence in $C(K)$. First you need to find out what it might converge to. In this case (and in many other cases) the limit function that it converges to in norm is the same as its pointwise limit function. So:

(i) Prove that the pointwise limit function $f$ exists. What does that mean? It means that $f(k)$ is finite for all points $k$ in $K$. So let $k$ be a fixed point in $K$. Then consider $f_n(k)$. Note that this is a sequence of real numbers. Let $\varepsilon > 0$ and now remember that you picked $f_n$ to be a Cauchy sequence with respect to $\| \cdot \|_\infty$. So you have $$| f_n(k) - f_m(k) | \leq \sup_{k \in K} | f_n(k) - f_m(k) | = \| f_n - f_m \|_\infty < \varepsilon$$

Which shows that $f_n(k)$ is a Cauchy sequence in $\mathbb{R}$ with respect to the standard metric. But you know that $\mathbb{R}$ is complete so you know that the limit of $f_n(k)$ is in $\mathbb{R}$. This means that the pointwise limit $f(k) = \lim_{n \to \infty} f_n(k)$ exists.

(ii) Next you go on to showing that $f_n$ also converges to $f$ in norm, that is, $\| f_n - f \|_\infty \xrightarrow{n \to \infty} 0$.

In other words: for every $\varepsilon > 0$ you need to find an $N$ such that for $n > N$ you have $\| f_n - f \|_\infty \leq \varepsilon$.

For this let $\varepsilon > 0$ and then fix an $N$ such that for $n,m \geq N$ you have $\| f_n - f_m \| < \frac{\varepsilon}{2}$. You can do this because $f_n$ is Cauchy by assumption. Then use the triangle inequality to get

$$\| f_n - f \|_\infty \leq \| f_n - f_N \|_\infty + \| f - f_N \|_\infty$$

By how you picked $N$ you have $\| f_n - f_N \|_\infty < \frac{\varepsilon}{2}$.

You also know that for $m \geq N$ you have $\| f_m - f_N \|_\infty < \frac{\varepsilon}{2}$. From this you can conclude that $\lim_{m \to \infty} \| f_m - f_N \|_\infty \leq \frac{\varepsilon}{2}$.

But this means you have shown that $$\| f_n - f \|_\infty \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} =\varepsilon$$ for $n > N$, which means exactly that $\| f_n - f \|_\infty \xrightarrow{n \to \infty} 0$.

(iii) In order to conclude that $C(K)$ is complete, the last thing you need to prove is that $f$ is continuous and therefore actually in $C(K)$. So let $\varepsilon > 0$ and $k$ in $K$. Now you want to find $\delta > 0$ such that for $x$ with $d(k,x) < \delta$ you have $|f(k) - f(x)| < \varepsilon$. Use the triangle inequality to get $$|f(k) - f(x)| \leq |f(k) - f_n(k)| + |f_n(k) - f_n(x)| + |f_n(x) - f(x)|$$

for some $n$. Now you pick $n$ such that $|f(k) - f_n(k)| < \frac{\varepsilon}{3}$ and $|f_n(x) - f(x)| < \frac{\varepsilon}{3}$. Then the $\delta$ you are looking for is now going to be the $\delta$ such that $|f_n(k) - f_n(x)| <\frac{\varepsilon}{3}$ which you know exists because $f_n$ are continuous. Hence $$|f(k) - f(x)| < \varepsilon$$ and therefore $f$ is continuous.

Note that this works for $K$ compact because then the continuous functions on $K$ are bounded and so $\|\cdot\|_\infty$ is actually a norm.

Using a very similar argument you can show that the space of bounded continuous functions on an arbitrary topological space $X$ is complete. I did this in another post here.

Hope this helps.

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I'm a bit late with this answer but I thought I'd add something slightly more general. Seeing as there are already two answers you may ignore me. : ) – Rudy the Reindeer Jan 8 '12 at 13:08
Hardly ;) Thanks for providing another angle. Cheers. – BB_ML Jan 8 '12 at 13:12
@Matt: "is complete", not "is continuous", near the end of your post (just a typo). – gnometorule Jan 8 '12 at 13:55
at the final step of the proof, when you show that the $f$ is continuous using the $\frac{\varepsilon}{3}$ argument, the inequality $$|f(x)-f_n(x)|<\frac{\varepsilon}{3}$$ has to hold $\forall x : d(x,k)<\delta$ so maybe it is needed that the f is the uniform limit of the $(f_n)_{n\in \mathbf{N}}$? – karhas Jan 21 at 19:46
@karhas Yes, that's right: it's definitely needed in the proof that $f$ is the uniform limit of the $f_n$. – Rudy the Reindeer Jan 22 at 1:03

This is simpler to prove if we split the claim up in parts. I'm not giving proofs because others here have already done that.

Theorem 1a. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is pointwise convergent to a bounded function.

Theorem 1b. A uniformly Cauchy sequence of $\mathbb K$-valued bounded functions is uniformly convergent to its pointwise limit.

Corollary 2. $B(X \to \mathbb K)$ is a Banach space when equipped with the uniform norm.

Theorem 3. Let $X$ be a Hausdorff space, $(Y,d)$ a metric space and $f$ and $f_n$ functions from $X$ to $Y$. If $(f_n)$ converges uniformly to $f$ on $X$ (*) and if each $f_n$ is continuous at $x \in X$, then $f$ is continuous at $x$.

Corollary 4. $C_b(X \to \mathbb K)$, the set of bounded continuous functions on a Hausdorff space $X$, is a closed subspace of $B(X \to \mathbb K)$, so it too is a Banach space.

(*): i.e. for all $\varepsilon > 0$ there exists an $N \in \mathbb N$ such that $x \in X \land \mathbb N \owns n \geq N \implies d(f_n(x),f(x)) < \varepsilon$.

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The norm we are working with is $\newcommand{\abs}[1]{|#1|}\newcommand{\norm}[1]{\|#1\|}\norm f= \sup_{x\in[0,1]}\abs{f(x)}$. (You didn't specify the norm in your question, but I guess it's very improbable that you would work with some different norm.)

Then for each fixed $t \in [0,1]$ $|x_n(t)-x_m(t)| \le ||x_n - x_m|| \to 0$, so $\{x_n(t)\}$ is a Cauchy sequence of real numbers. I guess this inequality makes sense, since difference of function sequences would be larger than pointwise difference (at a specific $t$) of those functions.. But I did not understand how the author could conclude the sequence $\{x_n(t)\}$ is Cauchy.

Well, you have just shown that $|x_n(t)-x_m(t)|\to 0$. This is precisely the definition of Cauchy sequence.

The reason why the inequality $\abs{x_n(t)-x_m(t)}\le\norm{x_n-x_m}$ is the definition of $\norm{\cdot}$. For every bounded function $f$ you have $\abs{f(t)} \le \sup_{x\in[0,1]} \abs{f(x)}=\norm f$.

I know set of real numbers is complete, but how can $\{x_n(t)\}$ represent the entire set of real numbers?

You do not need anything like that. Now you have fixed some $t\in[0,1]$ and you're working with one real sequence $\{x_n(t)\}$. This is a sequence of real numbers, by completeness of $\mathbb R$, such sequence converges if it is Cauchy.

After we have shown that the sequence $\{x_n(t)\}$ is convergent, we can denote the limit (for this fixed value of $t$) by $x(t)$.

Perhaps it might be useful for you to have a look at quite detailed proof of a similar results in Matt's answer here.

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I realized the inequality thing soon after I asked.. For Q #2, it seems the author meant the set of real numbers that was created by $\{x_n(t)\}$, not THE set of real numbers, as in $\Re$, or even a closed subset of it. That sequence is complete, obviously, because it is Cauchy and has a limit. – BB_ML Jan 8 '12 at 12:20
I do not think so. Once again: If we fix some $t$ then the sequence $(x_n(t))$ is a sequence of elements of $\mathbb R$. Hence to prove it is convergent we only need to show it is Cauchy. Perhaps the answer posted by uncookedfalcon could help to clarify this. – Martin Sleziak Jan 8 '12 at 12:23
@user6786 It is not a sequence that is complete or incomplete; completeness is a property of the space the sequence resides in. (In fact, considering that the reals are uncountable, it is impossible for any sequence to contain all of the real numbers.) – Srivatsan Jan 8 '12 at 12:25
Correction, yes, property of the space, such as $\Re$. – BB_ML Jan 8 '12 at 12:34
I think I am slowly getting it: author reduces the Cauchy $x_n$ to another Cauchy sequence $\{x_n(t)\}$ in $\Re$, and since he knows the completeness of $\Re$, and the "Cauchyness" of the new sequence, he can conclude $\{x_n(t)\}$ always has a limit. Than he backtraces the steps, takes that limit, back to $x_n$ and states there is a limit for that sequence too, for all $t$ in $x_n(t)$ which is $x_n$. – BB_ML Jan 8 '12 at 12:44