# Counterexample for $(\mathcal{B}([a,b], Y), d^{\sup})$ complete $\implies Y$ complete

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.

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 $$y_n\to f(x)$$.
• Could you please elaborate? I am not very familiar with isometric embeddings. $(y_n)_{n \in \mathbb{N}}$ is a sequence in $Y$? – Viktor Glombik Mar 26 at 21:38
• 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 $X:=[a,b]$. – Berci Mar 27 at 8:56