# For a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$, if $F$ is compact on compact subsets of $X$, then $F$ is compact on $X$

The problem as stated in the title isn't quite correct. Let $X$ be a topological space. What I have is a family of functions $F\subset C(X)$ in the metric space $(C(X),d)$ which on compact subsets of $X$ is closed, pointwise bounded, and equicontinuous. Furthermore there exists an ascending chain of compact subsets of $X$, $K_1\subset K_2\subset\dots$, such that $\bigcup K_n=X$. The metric $d$ is given by:

$$d(f,g):=\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{||f-g||_{K_n}}{1+||f-g||_{K_n}},$$ where $||f||_{K_n} := \sup_{x\in K_n}|f(x)|$.

I'm not sure the metric is important though.

$(*)$ We also know that $f_n\rightarrow f$ on compact subsets of $X$ iff $d(f_n,f)\rightarrow 0$. (proved in a previous exercise)

Prove that $F$ is compact on all of $X$.

By Theorem 3 on page 209 of Royden's Real Analysis, I'm able to conclude from the assumptions that $F$ is compact when the domain of its functions is restricted to a compact subset of $X$. Thus my strategy has been to take an arbitrary sequence $\{f_n\}$ in $F$ defined on all of $X$, restrict its domain to some $K_n\subset X$, and then find a convergent subsequence $\{f_{n_k}\}$. If I could prove that this subsequence is the same convergent subsequence in all the $K_n$, then I could use $(*)$ to prove that $F$ is sequentially compact and thus compact, on all of $X$.

Unfortunately compactness only guarantees the existence of some convergent subsequence, and thus for each successive $K_n$ I may have to consider a new convergent subsequence of $\{f_n\}$ ad infinitum, and thus I can't use $(*)$ to complete my proof.

Is my method wrong? Do I need a new approach?

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I don't think this is true without some restrictions on $X$: Take $X=[0,1)$ and $f_n(x)=x^n$. This sequence converges in compact subsets to $0$, but it doesn't converge uniformly on the whole interval. –  Jose27 Mar 24 '13 at 5:46
@Jose27, there seems to be two things going on here. First, I don't think a lack of uniform convergence poses a problem, as the $\frac{1}{2^n}$ in my metric allows one to find a fixed $N$ independent of the point chosen from $X$. However your counterexample does work to disprove the problem as stated in the title for a different reason: $f=0$ is not in your family, and thus your family isn't closed (read: not compact). However in the problem as stated in the body of the post, $F$ is required to be closed, so in that case your counterexample doesn't satisfy the assumptions. –  Thoth Mar 24 '13 at 6:43
In $C([0,1))$ the set $(f_n)$ in my previous comment is closed (since any accumulation point must be zero by the pointwise convergence). On the other hand your metric is used precisely because it doesn't see things like these: Things at the "boundary" or at "infinity" are ignored. –  Jose27 Mar 24 '13 at 6:48
@Jose27, oh you're completely right, so $(f_n)$ doesn't converge at all with the $d$ metric. –  Thoth Mar 24 '13 at 6:53
@Jose27, but I don't think the family you gave is equicontinuous –  Thoth Mar 24 '13 at 18:15