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Prove that the limit of a uniformly convergent sequence of bounded functions from $(E, d)$ to $(E', d')$ is bounded.

${f_n(x)}$ is a uniformly convergent sequence of bounded functions. $f_n \to f$ converges uniformly.

Since ${f_n(x)}$ is uniformly convergent, it is Cauchy. So, $\forall \epsilon >0, \exists N \in \mathbb{N}$ such that $\forall n, m > N, d(f_n(x), f_m(x)) < \epsilon $.

So, we can find an $n$ and $m$ such that $\sup_{x \in E}|f_n(x) -f_m(x)| <1 $

I do not know how to proceed in the proof.

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  • $\begingroup$ Are you sure you don’t mean that a uniformly convergent sequence of bounded functions is uniformly bounded? $\endgroup$
    – Boshu
    Nov 19 '19 at 16:58
  • $\begingroup$ No... the proposition I am to prove reads "Prove that the limit of a uniformly convergent sequence of bounded functions (from one metric space to another) is bounded" $\endgroup$
    – kt046172
    Nov 19 '19 at 17:04
  • $\begingroup$ Your initial question never said they were bounded functions. And bounded is a weaker condition than uniformly bounded. Think of bounds for each function and then the maximum of those. $\endgroup$
    – Boshu
    Nov 19 '19 at 19:23
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By uniform convergence there exists $N \in \mathbb{N}$ such that $d'(f(x),f_N(x)) \leqslant 1$ for all $x \in E$.

Since $f_N$ is bounded there exists $y_0 \in E'$ and finite $r_N > 0$ such that $f_n(x) \in B(y_0;r_N)$ for all $x \in E$.

Thus, for all $x \in E$ we have $f(x) \in B(y_0,1+r_N)$ since, by the triangle inequality,

$$d'(f(x),y_0) \leqslant d'(f(x),f_N(x)) + d'(f_N(x),y_0) \leqslant 1 + r_N$$

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