What condition do I need else to show a uniformly continuous is pointwise bounded and uniformly bounded?

Question

Let a sequence of function f_n uniformly continuous is given. I am wondering what else I need to let it be pointwise bounded and what for uniformly bounded. I am confused with some knowledge about it. So I believe a generalized argument would be helpful.

Answer 1

Suppose $f_n$ is a uniformly continuous sequence of real-valued functions on a metric space $M$ with metric $d$. If $f_n(x_0)$ is bounded for some $x_0 \in M$, and $M$ is connected, then the sequence is pointwise bounded.

Proof: By uniform continuity, there is $\delta > 0$ such that if $d(s, t) < \delta$, then all $|f_n(s) - f_n(t)| < 1$.
Now consider the subset $A$ of $M$ consisting of points $x$ such that there is a finite sequence $x_0, x_1, \ldots, x_m = x$ starting at $x_0$ with $d(x_i,x_{i+1}) < \delta$. $A$ is nonempty because it contains $x_0$. If $p \in A$ and $q \in M \backslash A$, then $d(p,q) \ge \delta$. Thus if $A \ne M$, $A$ and $M\backslash A$ would form a partition of $M$ into disjoint open sets, contradicting the connectedness of $M$.
Therefore $A = M$. Now given such a sequence $x_0, x_1, \ldots, x_m = x$, we have $|f_n(x_i) - f_n(x_{i+1})| < 1$, and therefore $|f_n(x)| \le |f_n(x_0)| + m$. Therefore the sequence is pointwise bounded.

On the other hand, a condition that implies the sequence is uniformly bounded is that $M$ is compact.