Mistakes in wikipedia ? Pointwise bounded set of $\mathcal C(X,\mathbb R)$.

Question

In wikipedia they say that $F\subset \mathcal C(X,\mathbb R)$ is pointwise bounded if for every $x\in X$, $$\sup\{|f(x)|\mid f\in F,x\in X\}<\infty .$$

Question 1 Is there a mistakes ? I would say either

1) for all $x\in X$ $$\sup\{|f(x)|\mid f\in F\}<\infty ,$$

2) or $$\sup\{|f(x)|\mid f\in F, x\in X\}<\infty .$$

but as written it looks strange. So, what is the good definition ?

Question 2 So, at the end, what is needed to use Arzela-Ascoli theorem ? $$\sup\{f(x)|\mid f\in F\}<\infty ,$$ for all $x\in X$ ? Or $$\sup\{|f(x)|\mid x\in X, f\in F\}<\infty \ \ ?$$

Question 3 I guess that if $X$ is compact, both definition is equivalent, but if $X$ not compact, the second definition looks stronger. What do you think ?

Answer 1

Which condition do you need to apply A-A? If $X$ is not compact I'm not aware of a version of the theorem that applies in the first place. (Ok, if $X$ is locally compact there's a version for $C_0(X)$. But that's essentially equivalent to the theorem for compact spaces, considering the one-point compactification...) If you do find a version for non-compact $X$ you can determine what sort of boundedness is needed by looking at the hypotheses...

You say " I guess that if $X$ is compact, both definition is equivalent". Yes if $F$ is equicontinuous, not in general:

Suppose $X$ is a compact Hausdorff space and $F\subset C(X)$ is equicontinuous. If $F$ is pointwise bounded then $F$ is uniformly bounded.

Proof: Say a small set is an open set $V\subset X$ such that $|f(x)-f(y)|<1$ for every $f\in F$ and $x,y\in V$. Since $F$ is equicontinuous every point of $X$ is in some small set. Since $X$ is compact it is covered by finitely many small sets $V_1,\dots, V_n$.

Say $x_j\in V_j$ and $|f(x_j)|\le M_j$ for every $f\in F$. Then$$|f(x)|\le 1+\max(M_1,\dots, M_n)\quad(x\in X, f\in F).$$

But not so if we don't assume $F$ is equuicontinuous.

Choose $f_n\in C[0,1]$ such that $f_n(0)=0$, $0\le f_n(t)\le 1/t$ for $0<t\le 1$, and $f_n(1/n)=n$. Then $F=(f_n)$ is pointwise bounded but not uniformly bounded.