# Let $Y$ be a complete metric space. Then $C^0 (X,Y)$ is complete under the uniform convergence metric.

I'm trying to show that the set of continuous functions $$f: X \to Y$$ is complete under the uniform convergence metric if $$Y$$ is complete.

Just to be clear, the metric is:

$$d(f,g) = \text{sup}\, \{d(f(x), g(x))\, ; \,x\in X \}$$

So far what I got is:

$$C^0(X,Y)$$ being complete under the above metric means that every Cauchy sequence of functions is uniformly convergent and that is converges to a continuous function, right?

Lemma: Let $$f_n \in C^0(X,Y)$$, $$f:X \to Y$$, $$\underset{n \to \infty} \lim d(f_n, f) = 0$$, then $$f$$ is continuous.

Proof: Take $$x_0 \in X$$ and $$\epsilon > 0$$. There exists $$n_0 \in \mathbb{N}\,; \, n > n_0 \implies d(f_n, f) < \frac{\epsilon}{5}$$. We know $$f_{n_0}$$ is continuous. Then there exists $$\delta > 0$$ such that the euclidian distance between $$f_{n_0}(x_0)$$ and $$f_{n_0}(x)$$ is smaller than $$\frac{\epsilon}{5}$$. If the euclidian distance between $$x$$ and $$x_0$$ is smaller than $$\delta$$, then we have:

$$d(f(x), f(x_0)) \leq d(f(x), f_{n_0}(x)) + d(f_{n_0}(x), f_{n_0}(x_0)) + d(f_{n_0}(x_0), f(x_0)) \leq \frac{3}{5}\epsilon < \epsilon$$

$$\blacksquare$$

We have then that any Cauchy sequence in $$C^0(X,Y)$$ converges - because the limit of a sequence of continuous functions with domain $$X$$ and range $$Y$$ is itself a continuous function, which makes $$C^0(X,Y)$$ complete.

And I think this solves it, but I feel like I'm missing something. I haven't used the fact that $$Y$$ is complete, for instance. $$Y$$ being complete guarantees that any function $$f \in C^0(X,Y)$$ maps a Cauchy sequence to a convergent one and this feels like an useful observation, but I can't see where it fits in a proof.

• Feel free to ask questions on my answer Jan 20 '19 at 17:46
• For mathematicians the question in very inaccurate. For instance, there still are several problems with the lemma. First, if $d(f_n,f)$ is defined then $f$ is already in $C(X,Y)$ and there is nothing to prove. By the way, you didn’t define $C^0(X,Y)$. Next, you introduced $\delta$, but missed the condition that $x$ and $x_0$ are $\delta$-close. Moreover, you even not defined what is $X$ (A metric space? A topological space? A set?). Also, a metric of a general metric space is not necessarily Euclidean distance induced from $\Bbb R^n$. Jan 25 '19 at 8:25
• I did not downvote you, but here are people who like to downote and close questions with smaller lacks. You were lucky this time, but, anyway this is mathematics and we work with exact and explicit definitions. In fact, we need such definitions to work with. Jan 25 '19 at 8:25

The problem is not well posed. Are $$X,Y$$ general metric spaces? If so, what does the euclidean metric $$e$$ on $$Y$$ mean? If not, please give the precise assumptions on $$X,Y.$$

We have another problem: Suppose $$X=Y=\mathbb R$$ with the euclidean metric. Then both $$f(x)=0,g(x)=x$$ belong to $$C(X,Y).$$ But what is $$d(f,g)$$ supposed to be? You have it as

$$d(f,g)= \sup_{x\in\mathbb R}|f(x)- g(x)| = \sup_{x\in\mathbb R}|0-x| = \infty.$$

There is also a problem with the lemma. You have $$f_n \in C(X,Y)$$ but there is no condition on $$f.$$ So why is $$d(f_n,f)$$ even defined? (Note $$d_n$$ should be $$d$$ there.)

Finally, any proof that doesn't use the completeness of $$Y$$ is doomed. For if $$Y$$ is not complete, there is a Cauchy sequence $$y_n$$ in $$Y$$ that fails to converge to a point of $$Y.$$ Define $$f_n(x)\equiv y_n$$ for all $$n.$$ Then $$f_n$$ is Cauchy in $$C(X,Y)$$ but fails to converge to any $$f\in C(X,Y).$$

Right now my answer is in the form of comments/questions, I know. I need your answers to these questions to give an answer.

Added later: Here's a way to fix things: Since you repeatedly mention the euclidean metric, let's stay in that setting and suppose $$X,Y$$ are both subsets of some $$\mathbb R^m.$$ We assume that $$Y$$ is complete in the usual $$\mathbb R^m$$ metric. Define $$B =B(X,Y)$$ to be the set of all bounded functions from $$X$$ to $$Y.$$ For $$f,g\in B,$$ define

$$d(f,g)= \sup_{x\in X} |f(x)-g(x)|.$$

Now we have something that is well defined. Verify that $$d$$ is a metric on $$B.$$ Note that $$f_n\to f$$ in $$B$$ iff $$f_n\to f$$ uniformly on $$X.$$

Now define $$C_B= C_B(X,Y)$$ to be the set of functions in $$B$$ that are continuous on $$X.$$ The result we want to prove is

Thm: $$C_B$$ is a complete metric space in the $$d$$ metric.

Your lemma can be stated as

Lemma: If $$f_n\in C_B,$$ $$f\in B$$ and $$d(f_n,f)\to 0,$$ then $$f\in C_B.$$

Your proof then goes through.

To prove the theorem, suppose $$(f_n)$$ is a Cauchy sequence in $$C_B.$$ Then from the definition of the $$d$$-metric, for each $$x\in X,$$ $$f_n(x)$$ is a Cauchy sequence of points in $$Y.$$ And $$Y$$ is complete!! Thus for each $$x\in X$$ the limit $$\lim_{n\to \infty}f_n(x)$$ exists as a point in $$Y.$$ We can therefore define $$f:X\to Y$$ to be this limit at each $$x\in X.$$

If we can show $$f\in B$$ and $$f_n\to f$$ uniformly on $$X,$$ we'll be done by the lemma. This should be familiar territory. I'll leave the proof here for now, but ask if you have any questions.

• I've edited what I can, but there are points you raised which were simply not presented to me. This was given to me by a professor not as homework, but as a curious problem which I could attempt to solve for fun and these are all the specifics of it I have. Jan 24 '19 at 4:59

You are right to be skeptical. One tip off is you haven't used the completeness of $$Y$$ anywhere.

You have shown that if a sequence $$f_n\to f$$ with this metric, then $$f$$ is continuous. This is not the same as proving that if $$f_n$$ is a Cauchy sequence, then it converges and the limit point happens to be continuous as well.

To do this, use completeness of $$Y$$ and the uniform estimate $$\sup_{x\in X}|f_{n}(x)-f_m(x)|<\epsilon$$ for $$n,m\geq N$$ sufficiently large, to extract a pointwise limit. Then, the final step is to prove that this candidate is actually a uniform limit.

Let’s fix it.

We assume that $$X$$ is a topological space, $$(Y,d)$$ is a complete metric space, and $$C^0(X,Y)$$ is a subset of a set $$C(X,Y)$$ of continuous functions from $$X$$ to $$Y$$. A problem (already noted in zhw’s answer) is that given $$f,g\in C(X,Y)$$,

$$\bar d(f,g) = \text{sup}\, \{d(f(x), g(x))\, ; \,x\in X \}$$ (I renamed $$d$$ to $$\bar d$$, because $$d$$ is already used as the metric on $$Y$$), is not always finite, and in this case $$(C(X,Y),\bar d)$$ is not a metric space. There are several ways to fix it.

• Put $$C^0(X,Y)=C(X,Y)$$ and $$d’(f,g)=\min\{\bar d(f,g),1\}$$ for all $$f,g\in C(X,Y)$$.

• If we want to keep $$\bar d$$ as metric and make a natural assumption that the set $$C$$ of all constant functions from $$X$$ to $$Y$$ belong to $$C^0(X,Y)$$. Then we can put

$$C^0(X,Y)=\{f\in C(X,Y):\forall g\in C (\bar d(f,g)<\infty)\}= \{f\in C(X,Y):\exists g\in C (\bar d(f,g)<\infty)\}.$$

• We can put $$C^0(X,Y)=C(X,Y)$$ and allow to $$\bar d$$ have infinite values, and work with this generalized metric space.

But in any of the above cases, a sequence $$\{f_n\}$$ of functions of $$C^0(X,Y)$$ is Cauchy provides that $$\bar d(f_n,f_m)$$ is finite for sufficiently big $$n$$ and $$m$$.

Now, since $$(Y,d)$$ is a complete metric space, for each $$x\in X$$ there exists $$f(x)=\lim_{n\to\infty} f_n(x)$$. Fix any $$\varepsilon>0$$. There exists $$N$$ such that $$\bar d(f_n,f_m)\le\varepsilon$$ for all $$n,m\ge N$$. Then for each $$x\in X$$ we have

$$d(f_N(x),f(x))=\lim_{m\to\infty} d(f_N(x),f_m(x))\le\varepsilon.$$

Thus the sequence $$\{f_n\}$$ converges to $$f$$ uniformly. Remark, that it is yet not the same that $$\lim_{n\to\infty}\bar d(f_n,f)=0$$, because we have not proved yet that $$f\in C^0(X,Y)$$, so $$\bar d(f_n,f)$$ may be undefined. We finish the proof as follows.

Since the function $$f_N$$ is continuous, there exists a neighborhood $$U$$ of the point $$x$$ such that $$d(f_N(x),f_N(x’))<\varepsilon$$ for each point $$x’\in U$$. Then

$$d(f(x),(x’))\le d(f(x),f_N(x))+d(f_N(x),f_N(x’))+ d(f_N(x’),f(x’))< \varepsilon+\varepsilon+\varepsilon,$$

so the function $$f$$ is continuous.