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Theorem: Let $(Y,d)$ be a complete metric space. Then the metric space $(\mathcal{B}([a,b], Y), d^{\sup})$ is complete, where $B(C,D)$ is the set of all bounded functions form $C$ to $D$ and $d^{\sup}(f,g) := \sup_{x \in [a,b]} d(f(x), g(x))$ for $f,g: [a,b] \to Y$ is the supremum-metric induced by $d$.

I now want to find a simple counterexample showing this implication doesn't hold the other way around but haven't been able to get any help from this related question.

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The converse also holds:

For any nonempty set $X$, we have an isometric embedding $c:Y\hookrightarrow \mathcal B(X,Y)$, taking the constant maps.
Now, if $(y_n)$ is a Cauchy sequence, then so is $c(y_n)$, so it converges to some $f\in\mathcal B(X,Y)$.
Then, by definition of $d^\sup$, evaluating at any $x\in X$, we get $y_n\to f(x)$.

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  • $\begingroup$ Could you please elaborate? I am not very familiar with isometric embeddings. $(y_n)_{n \in \mathbb{N}}$ is a sequence in $Y$? $\endgroup$ – Viktor Glombik Mar 26 at 21:38
  • $\begingroup$ Or: in other words: it looks like your proof of the opposite direction can be applied to a more general case. How would one proof this direction in this specific setting? $\endgroup$ – Viktor Glombik Mar 27 at 8:41
  • $\begingroup$ The point is that the constant functions $c(y)$ behave exactly the same way as the points $y$, meaning that $d^\sup(c(y), c(y')) =d(y, y')$. For your special case, just take $X:=[a,b]$. $\endgroup$ – Berci Mar 27 at 8:56
  • $\begingroup$ ok, but can you please explain the isometric embedding? $\endgroup$ – Viktor Glombik Mar 27 at 17:15

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