Is a Uniformly Convergent Sequence of Bounded Functions Also Bounded?

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Here is this week's POTW:

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Consider a sequence $f_n : M\to M'$ between two metric spaces $M$ and $M'$, where $n = 1,2,3,\ldots$. Prove that if each $f_n$ is bounded and $f_n$ converges uniformly to $f$, then $f$ is bounded.
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This week's problem was solved correctly by Janssens. You can read his solution below.

Spoiler

By one common definition, a subset $A$ of a metric space $(X, \rho)$ is bounded if there exist $x \in X$ and $r > 0$ such that $A \subseteq B(x,r)$, where the latter set denotes the open $\rho$-ball at $x$ with radius $r$. By the triangle inequality it is immediate that $A \subseteq X$ is bounded iff for each $x \in X$ there exists $r > 0$ such that $A \subseteq B(x,r)$.

Now, let $d'$ be the metric on $M'$ and fix a point $y \in M'$. Since each $f_n$ is bounded, for every $n \in \mathbb{N}$ there exists $r_n > 0$ such that $f_n(M) \subseteq B(y, r_n)$. So, by the triangle inequality,
\begin{align*}
d'(f(x), y) &\le d'(f(x), f_n(x)) + d'(f_n(x), y)\\
&< d'(f(x), f_n(x)) + r_n,
\end{align*}
for all $x \in M$ and all $n \in \mathbb{N}$. By the given uniform convergence, there exists $m \in \mathbb{N}$ such that $d'(f(x), f_m(x)) < 1$ for all $x \in M$. This implies that $d'(f(x),y) < 1 + r_m$, so $f(M) \subseteq B(y, 1 + r_m)$, showing that $f$ is bounded.