If every sequence of pointwise equicontinuous functions $M \rightarrow \mathbb{R}$ is uniformly equicontinuous, does this imply that $M$ is compact?
Real Analysis – Does Pointwise Equicontinuous and Uniformly Equicontinuous Imply Compactness?
compactnessequicontinuityreal-analysis
Related Solutions
Heuristically, if you have a family of equicontinuous functions, then you could essentially treat them as a "single continuous function" because for any $\epsilon>0$ you could always find a single $\delta>0$ that will work for your entire family. Moreover, we know a continuous function on a compact metric space attains its maximum, i.e. bounded. Hence together you can see why the family should be uniformly bounded.
Edit:
Here's the proof.
Let $(X, d)$ denote the compact metric space and $\mathcal{F}$ is our equicontinuous family of functions.
Fix $\epsilon>0$. Then for every fixed $x \in X$, by the equicontinuity property of $\mathcal{F}$, we can find a $\delta_x>0$ (i.e. $\delta$ depending on $x$) such that \begin{align} |f_n(x)-f_n(y)|<\epsilon \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} In particular, it follows \begin{align} |f_n(y)| < \epsilon +|f_n(x)| \leq \epsilon +M_x \ \ \text{ whenever } \ \ |x-y|<\delta_x. \end{align} Note that we have used pointwise boundedness of $\mathcal{F}$. Next, observe $X \subset \bigcup_{x \in X} B(x, \delta_x)$ is an open cover of our compact metric space. Hence by compactness there exists a finite subcover say $\bigcup^N_{i=1} B(x_i, \delta_i)$. Let $M = \sup_{1\leq i \leq N} M_{x_i}<\infty$, then we see that \begin{align} |f_n(y)| <\epsilon+M \end{align} for every $y \in X$ and $n\in \mathbb{N}$.
Answering your questions in order:
1) Since $B$ is a set I think pointwise compactness here is limit point compactness, which means that every infinite set contains a limit point. In metric spaces this is equivalent to sequential compactness, which is that every infinite sequence possesses a convergent subsequence (cf. Bolzano-Weierstrass).
2) A sequence of functions $f_n(x)$ is pointwise bounded if for each $x$ there is a finite constant $M(x)$ depending on $x$ such that $|f_n(x)| \leq M(x)$. It is uniformly bounded if there is a single finite constant $M$ that does not depend on $x$ such that $|f_n(x)| \leq M \ \forall x$ and for all $n\in \mathbb N$.
When you only have one function $n=1$ and these two definitions reduce to the same thing. For example: $f(x) = \frac{1}{1+x^2}$ is (uniformly) bounded on the whole of $\mathbb R$; it's easy to see that $|f(x)|\leq 1 \ \forall x\in \mathbb R$. See this ( A sequence of functions $\{f_n(x)\}_{n=1}^{\infty} \subseteq C[0,1]$ that is pointwise bounded but not uniformly bounded. ) question for examples of sequences of functions that are pointwise but not uniformly bounded.
3) Yes, if you have a set (family) of functions that are bounded then each function is pointwise bounded. Note that a function defined on a bounded set of points need not be bounded (consider $f(x)=\frac{1}{x}$ defined on $[0,1]$).
4) Yes.
Best Answer
Consider the subspace $M=\mathbb{Z}$ of the reals, an infinite discrete space, certainly not compact. But every sequence of functions is uniformly equicontinuos in that space.