Under what metric spaces are pointwise and uniform convergence equivalent?

Question

In exercise 1 of Charles Chapman Pugh's Real Mathematical Analysis, is a question of when pointwise and uniform convergence are 'equivalent' for metric spaces $M$ and $N$ and a sequence of functions $f_n: M \to N.$ Under my understanding, this would mean that any sequence of functions $f_n \to f$ would immediately imply $f_n \rightrightarrows f$ for a particular combination of $M$ and $N$, since the converse is already true.

What I have found:

If $M$ is finite, then the convergences are equivalent. In other words, $$\forall \epsilon>0,\forall x \in M, \exists L(\epsilon,x) \in \mathbb{N} \hspace{3mm} \text{s.t.} \hspace{3mm} d_M(f_n(x), f(x)) < \epsilon \hspace{3mm} \text{if} \hspace{3mm} n \ge L(\epsilon,x)$$ implies uniform convergence since $L$ and fixed $\epsilon > 0$ is a function on a finite space so $\max_{x}{L(\epsilon,x)} = L(\epsilon)$ exists. Using this maximum, we swap the $x$ and $L$ quantifiers giving us the definition for uniform convergence for $f_n$.

In trying to find more examples or a totality of every combination is where I am stuck. One lead is the fact that if $M$ is compact and $f_n$ is pointwise equicontinuous and pointwise bounded, then $f_n$ being pointwise convergent would imply uniform convergence by generalized Arezelà-Ascoli theorem. Yet, I can't think of any properties to attach to a metric space which would force every sequence of pointwise convergent functions to be pointwise equicontinuous and pointwise bounded.

Answer 1

If $M$ is finite they are equivalent.

If $M$ is not finite, they are not equivalent, unless $N$ is empty or contains one element. To see why they are not equivalent in the nontrivial case, take points $a,b \in N$, and points $x_1,x_2,x_3,\dots \in M$. Define $f$ by $f(x) = a$ for all $x \in M$. Define $f_n$ by $f_n(x) = a$ for all $x \in M\setminus\{x_{n+1},x_{n+2},x_{n+3},\dots\}$ and $f_n(x_{n+j}) = b$ for $j \ge 1$. Then $f_n \to f$ but not uniformly.