# Family of equicontinuous functions on compact set

1. Nov 19, 2013

### mahler1

The problem statement, all variables and given/known data.

Let $X$ be a compact metric space. Prove that if $\mathcal F \subset X$ is a family of equicontinuous functions $f:X \to Y \implies \mathcal F$ is uniformly equicontinuous.

The attempt at a solution.

What I want to prove is that given $\epsilon>0$ there exists $\delta>0$: if $d_X(x,y)<\delta \implies \forall f \in \mathcal F d_Y(f(x),f(y))<\epsilon$. I know that for an arbitrary $x_0 \in X$ and a given $\epsilon$ I can find $\delta$. I also know that $X$ is compact. I think I should write $X$ as a union of open covers involving something with the $\delta$ that works for each point $x_0$, then extract a finite subcover (I would have finite $\delta$'s) and take the minimum of those deltas. I got stuck trying to find the proper union of open covers.

2. Nov 19, 2013

### pasmith

Let $x \in X$ and $y \in X$ be such that $d_X(x,y) < \delta$. You want to show $d_Y(f(x),f(y)) < \epsilon$. You have a finite subcover by open balls so there must exist some $z \in X$ such that $x \in B(z,r(z))$ for some ball $B(z,r(z))$ in the subcover, and the triangle inequality will then give you a bound for $d_X(y,z)$. You can then, by appropriate choice of $\delta$ and $r(z)$, ensure that you have bounds on $d_Y(f(x),f(z))$ and $d_Y(f(y),f(z))$, and a further application of the triangle inequality will give a bound on $d_Y(f(x),f(y))$.

3. Nov 19, 2013

### mahler1

Let $\epsilon>0$, I know that for each $x \in X$, there exists $\delta_x$: if $y \in B_X(x,\delta_x) \implies \forall f \in \mathcal F d_Y(f(x),f(y))<\frac {\epsilon} {2}$. If $y,z \in B(x,\delta_x) \implies d_Y(f(y),f(z)<\epsilon$. I can write $X$ as $X=\bigcup_{x \in X} B_X(x,\delta_x)$, which is an open cover of $X$. By hypothesis, there exists $\delta>0$ a Lebesgue number for this open cover. This means that for any $y,z$ such that $d(y,z)<\delta \implies y,z \in B_X(x,\delta_x)$ for some $x \in X \implies d_Y(f(y),f(z))<\epsilon$, this proves $\mathcal F$ is uniformly equicontinuous.

I know you've suggested me to get a finite subcover but I've tried to to that before and I didn't know how to get the appropiate $\delta$, I am still thinking how could I solve it without using the Lebesgue number.

4. Nov 20, 2013

### pasmith

So far so good.

In my earlier post, I suggested that you look at a finite subcover obtained from the open cover $\{ B(x, r(x)) : x \in X\}$. Thus there exists a finite $Z \subset X$ such that $\{ B(z, r(z)) : z \in Z \}$ is an open cover of $X$.

The entire point of the construction is that we obtain a finite set of radii of covering balls, which allows us to take
$$\delta = \min \{ r(z) : z \in Z \} > 0$$
and it remains to choose $r(z) > 0$.

If $x \in B(z,r(z))$ and $d_X(x,y) < \delta \leq r(z)$ then the triangle inequality will give a bound on $d_X(y,z)$ in terms of $r(z)$. Your aim is to ensure that $d_X(y,z) < \delta_z$.