Equicontinuous set of functions. This question comes up after going over Arzela-Ascoli theorems.
For a set of continuous functions $\mathbb F$ from $\mathbb R$ to $\mathbb R$  that is equicontinuous. How do I show that if sup{$|f(0)|:f \in F$} $< \infty$ , then $\mathbb F$ is pointwise bounded?
I know I need to show that since sup{$|f(0)|:f \in F$} = $M_0 < \infty$ then for each $x \in \mathbb R$ sup{$|f(x)|:f \in F$} = $M_x < \infty$.
What I have gotten so far is that I should fix $\epsilon$ and use the fact that the domain is $\mathbb R$.
I think I just need some help using equicontinuity and sup together.
 A: Assume that the set
$
Z=\{x: M_x<\infty
\}
$is non empty.
Then let $u = \inf Z$. 
As the familly is equicontinuous, consider $\delta$ corresponding to the definition of equicontinuity with $\epsilon=1$.
Consider $u-\frac\delta2<x<u$, $y<u+\frac\delta 2$ such as
$y\in Z$.
For $A= M_x + 1$, you then have for a certain
$f\in F$: 
$$
|f(y)|>A
$$
but you also have
$$
|f(y)| \le |f(x)| + |f(y)-f(x)|\le M_x + 1 = A$$
This is impossible.
A: As an alternative to mookid's answer (which looks good), consider this one, which uses compactness (actually, mookid's answer reminds me of the proof of the compactness of $[0, 1]$, so maybe I'm just doing this at a less fundamental level).
Anyway, consider any $x \in \mathbb{R}$, and consider the set $[0, x]$ (we suppose WLOG that $x > 0$; if $x < 0$, we just examine $[-x, 0]$ and the rest of the proof is the same).
For $\varepsilon = 1$, for each $y \in [0, x]$, choose $\delta_y > 0$ so that $z \in (y - \delta_y, y + \delta_y) \Rightarrow |f(z)-f(y)| < 1$ for each $f \in F$; $\delta_y$'s existence is guaranteed by the equicontinuity of $F$.
Define $A_y = (y - \delta_y, y + \delta_y)$. Then $\{A_y\}_{y \in [0,x]}$ is an open cover of $[0, x]$, a compact set, and thus has a finite subcover, $\{A_{y_i}\}_{i=1}^n$.
Now, we know $M_0 < \infty$. If we order the $y_i$'s so that $y_1 < y_2 < \ldots < y_n$, then by the definition of equicontinuity of $F$, we know that $y \in A_{y_1} \Rightarrow |f(y)| \leq M_0 + 1$ for all $f \in F$, and more generally, $y \in A_{y_i} \Rightarrow |f(y)| \leq M_0 + i$ for all $f \in F$ (this might require a more formal proof--perhaps one using induction--but it shouldn't be too difficult). So, since $x \in A_{y_i}$ for some $i \in \{1, \ldots, n\}$, we know that $|f(x)| \leq M_0 + n$ for all $f \in F$, and therefore that $M_x \leq M_0 + n < \infty$.
I think explicitly using compactness makes the intuition a little more clear: because there is a bound for every function in $F$ at $0$, and because there is a bound on how much every function can change in a neighborhood of each point of $\mathbb{R}$, it can never be the case that some subset of $F$ increases locally at such a rate that it "breaks away" from the rest of the group toward infinity.
