Not equicontinuous

I am trying to show that if a family of continuous functions $\mathcal{F}$ in $C(X)$ is uniformly bounded and is not equicontinuous, then there exist a sequence $(x_n) \in X$ convergent to $x_0$ and a sequence $(f_n) \in \mathcal{F}$ such that $(f_n(x_0)) \rightarrow a$ and $(f_n(x_n)) \rightarrow b \ne a$.

I started from the epsilon-delta definition. $\mathcal{F}$ is not equicontinuous implies theres an $x_0$ such that $\mathcal{F}$ is not equicontinuous at $x_0$. This means $\exists \epsilon$ $\forall \delta$ $\exists_{f,x}$ such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|\ge \epsilon$.

In particular, swapping $\delta=n$, the above implies there exists a convergent sequence $(x_n) \rightarrow x_0$ and $|f_n(x_n)-f_n(x_0)| \ge \epsilon$.

We also have $f_n$ continuous functions, and so $(f_m(x_n)) \rightarrow f_m(x_0)$. Here I deliberately put $m$ as the index for $f$ because I think the index for functions is different from the index for the x's. This is where I am stuck at, do we need extra information to prove $(f_m(x_0))$ is a convergent sequence? Or is it there but I cant see it?

When I draw a little diagram tring to visualise these convergent sequences, I feel like $(f_n(x_n))$ is a "diagonal sequence" but I still don't quite know how to show its convergent to a limit different from $(f_m(x_0))$.

Any help is deeply appreciated!

• You are right, I have added the uniformly bounded condition to the question. I'm still unclear about the second part. Why are you able to choose such subsequences? – Floquet May 13 '14 at 10:15

This is where I am stuck at, do we need extra information to prove $(f_m(x_0))$ is a convergent sequence?
There is no reason to expect that sequence to converge. However, by the uniform boundedness assumption, it is a bounded sequence (in $\mathbb{C}$ or $\mathbb{R}$), and therefore has convergent subsequences. Thus after passing to a subsequence, we can assume that the sequence converges.
After extracting the first subsequence, we have a sequence $f_k$ in $\mathcal{F}$ and a sequence $(x_k)$ in $X$ with $x_k \to x_k$, such that $f_k(x_0) \to a$, and, for all $k$, $\lvert f_k(x_k) - f_k(x_0)\rvert \geqslant \varepsilon$. Then $(f_k(x_k))$ is a bounded sequence, and we can extract a further subsequence such that $f_{k_r}(x_{k_r})$ converges to some $b$. Since $\lvert f_{k_r}(x_{k_r}) - f_{k_r}(x_0)\rvert \geqslant \varepsilon$ for all $r$, we can deduce that $\lvert b-a\rvert \geqslant \varepsilon$.