On a compact metric space $X$, let $\{f_p\}_{p\in A}$ be a uniformly bounded and equicontinuous family, define $f:X\to \mathbb R$ by
$$f(x)=\sup \{f_p(x):p\in A\}$$ we need to pick out the true statements:
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$\{x:f(x)<t\}$ is an open set $\forall\, t\in\mathbb{R}$.
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$f$ is continuous
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Every sequence $f_{p_n}$ contained in the above family admits a uniformly convergent subsequence.
I think I have to apply Arzelà-Ascoli Theorem, but I just know the statement of it: If $X$ is a compact metric space, then a closed subset of $C(X,\mathbb{R})$ is compact iff it is bounded and equicontinuous.
Please give some hint.
Best Answer
This is for item 2.
Remember that equicontinuity and uniform equicontinuity coincide in a compact space. Fix $\varepsilon>0$. For this $\varepsilon$ there is a $\delta>0$ such that if $d(x,y)<\delta$ then $|f_\alpha(x)-f_\alpha(y)|<\varepsilon$ for all $\alpha$.
Assume that $f(x)\geq f(y)$ then $0\leq f(x)-f(y) \leq f(x)-f_\alpha(y)$ for all $\alpha$. Let $\eta>0$ and pick $\alpha_0=\alpha_0(x,\eta)$ such that $f(x)<f_{\alpha_0}(x) +\eta$. We have $$ f(x)-f(y) \leq f_{\alpha_0}(x)-f_{\alpha_0} +\eta \leq \varepsilon +\eta $$ since $d(x,y)< \delta$. Now make $\eta \to 0$ (which can be done because the same $\delta$ works for every $\alpha_0$), we obtain $0\leq f(x)-f(y)\leq \varepsilon$.
If $f(x)\leq f(y)$ repeat the argument with $x$ replaced with $y$.