Family of equicontinuous functions on compact set

[mahler1]

Homework Statement .

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.

 

[pasmith]

Homework Statement .

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.

Think about what you need your covering balls to do.

Let x \in X and y \in X be such that d_X(x,y) &lt; \delta. You want to show d_Y(f(x),f(y)) &lt; \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)).

 

[mahler1]

Think about what you need your covering balls to do.

Let x \in X and y \in X be such that d_X(x,y) &lt; \delta. You want to show d_Y(f(x),f(y)) &lt; \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)).

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.

 

[pasmith]

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 $y,z \in B(x,\delta_x) \implies d_Y(f(y),f(z)<\epsilon$.

So far so good.

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.

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
<br /> \delta = \min \{ r(z) : z \in Z \} &gt; 0<br />
and it remains to choose r(z) &gt; 0.

If x \in B(z,r(z)) and d_X(x,y) &lt; \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) &lt; \delta_z.