Uniform convergence and pointwise convergence Having the standard definitions for pointwise resp. uniform convergence of sequences of functions in a general metric space ($X,d$). What special conditions should X fulfill such that pointwise convergence <=> uniform convergence?
Since uniform convergence => pointwise convergence we just need to find the conditions on $X$ such that pointwise => uniform.
If X is finite could be enough?
 A: Hint for finite $X$: Suppose $X = \{x_1,\dots,x_n\}$ and that $f_n$ converges pointwise to some $f(x)$ on $X$.  Fix $\epsilon>0$.  By the definition of convergence, state the following:
We may select an $N_i \in \mathbb{N}$ such that $|f_n(x_i) - f(x_i)|< \epsilon$ whenever $n>N_i$.
In order for $\{f_n\}$ to satisfy uniform convergence, we'd have to say that there is some $N \in \mathbb{N}$ for which $|f_n(x) - f(x)| < \epsilon$ for all $x \in X$ whenever $n>N$.  Using the above, can you find such an $N$?
A: Let's analize this question for arbitrary sequences of functions and for sequences of continuous functions:

Let $(X,d)$ be a metric space. Then the following are equivalent:
  (a) $X$ is finite;
  (b) Every sequence of functions $f:X\rightarrow\mathbb{R}$ that converges pointwise also converges uniformly;
  (c) Every sequence of continuous functions $f:X\rightarrow\mathbb{R}$ that converges pointwise also converges uniformly;

Proof
Obviously, $(a)\Rightarrow(b)\Rightarrow(c)$, so it only remains to show that $(c)\Rightarrow(a)$, or, equivalently, its counter-positive $\lnot(a)\Rightarrow\lnot(c)$.
Suppose $X$ is infinite. We have two cases:
(i) $X$ is discrete, that is, all points of $X$ are isolated.
In this case, let $x_1,x_2,\ldots$ be distinct points of $X$. For every $n$, define the function $f_n:X\rightarrow\mathbb{R}$ by $f_n(x_n)=1$ and $f_n(y)=0$ for $y\neq x_n$. Then $\left\{f_n\right\}$ is a sequence of continuous functions such that converges pointwise to $0$ but not uniformly.
(ii) $X$ contains a point $x_0$ that is not isolated.
For every $n\in\mathbb{N}$, define $f_n:X\rightarrow\mathbb{R}$ by
$$f_n(y)=\begin{cases}1-n\cdot d(y,x_0)&\text{, if }d(y,x_0)\leq 1/n\\0&\text{, otherwise} \end{cases}$$
Then $\left\{f_n\right\}$ is a sequence of continuous functions that converges pointwise to a non-continuous function (specifically, the characteristic of $\left\{x_0\right\}$), hence does not converges uniformly. Q.E.D.
(this proof can be adapted to Hausdorff locally compact spaces.)
