What is the difference between pointwise boundedness and boundedness? In my textbook, the Arzela-Ascoli Theorem states that for a compact set $A$ in a metric space $M$ and $B\subset\mathcal{C}\left(A, N\right)$, where $N$ is another metric space, then $B$ is compact if and only if $B$ is closed, equicontinuous, and pointwise compact.
I also see two corollaries in my book that state "for a compact set $A\subset M$, $N=\mathbb{R^{n}}$, and an equicontinuous and pointwise bounded set $B\subset\mathcal{C}\left(A, \mathbb{R^{n}}\right)$, then every sequence in $B$ has a uniformly convergent subsequence" and "If $A\subset M$ is compact and $N=\mathbb{R^{n}}$ then $B\subset\mathcal{C}\left(A, \mathbb{R^{n}}\right)$ is compact if and only if it is bounded, closed, and equicontinuous."
I have a couple of questions:
$\left(1\right)$ What is pointwise compactness?
$\left(2\right)$ What is the difference between boundedness, pointwise boundedness, and uniform boundedness?
$\left(3\right)$ If my function is bounded for all $x$, does this imply that it is pointwise bounded at all $x$?
$\left(4\right)$ Does pointwise convergence imply pointwise boundedness?
For $\left(3\right)$, I think that if it the set $B$ is bounded, then for all $x$ it is bounded, so for each fixed $x$, it is pointwise bounded. But I am not sure if $x$ should really be $f$ in this case or not since it is a set of functions.
For $\left(4\right)$, I want to say that this is true because if it is pointwise convergent, then it must be bounded at that point.
The rest I am unsure of.
 A: 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.
