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I'm practicing an exam question and would appreciate having my proof verified.

Let $(x_n)$ be a Cauchy sequence in $X$. Suppose $f: X \rightarrow \mathbb{R}$ is a continuous (and therefore bounded) function.

Then, $(f(x_n))$ is a bounded sequence in $\mathbb{R}$. By Bolzano-Weierstrass, there exists a subsequence $(f_{n_k})$ that converges to some $y \in \mathbb{R}$. Therefore, there exists a ball $B$ of radius $\epsilon$ centered at $f(x_{n_k})$ that contains $y$ and all $f(x_{n_j})$ for $j \geq k$.

Let $B' = f^{-1}(B)$ denote the preimage of $B$ in $X$. By the continuity of $f$, $B'$ is an open ball of in $X$ which contains all $x_{n_j}$ for $j \geq k$ and $f^{-1}(y) =: x$. Thus, the subsequence $x_{n_k}$ converges to $x \in X$.

Lastly, this implies that $x_n$ converges to $x$ since $$d(x_n, x) \leq d(x_n, x_{n_k}) + d(x_{n_k}, x) < \epsilon$$ by the triangle inequality.

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    $\begingroup$ Some comments: (1) $B’$ is an open set in $X$ but needs not to be an open ball. (2) You say that $x:=f^{-1}(y)$, but what if $f$ is not injective? (3) the hypothesis is that any continuous function from $X$ to $\Bbb{R}$ is bounded, however in your proof you only use that there is one continuous bounded function. $\endgroup$ Aug 26 '19 at 22:23
  • $\begingroup$ Stronger conclusion: $X$ is compact: if $X$ is not compact the there is a sequence $\{x_n\}$ with no limit point. Define $f(x_n)=n$. Then $f$ is continuous on $\{x_n: n \geq 1\}$ and it can be extended to a continuous function on $X$ by Tietze's Theorem. $\endgroup$ Aug 26 '19 at 23:39
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$B'$ is not an open ball. It's open, yes, but it doesn't have to be a ball. Also, $f^{-1}(y)$ can have several values at once, and you don't really know that any of them are the limit of the $x_n$.

You have only used that there exists a continuous and bounded function. That's not the whole strength of the hypothesis, and it's not enough. I would suggest you construct a specific function in a clever way, and then use the fact that if it is continuous, then it is bounded. And then use the cleverness with which it was constructed to show that the space is complete. Or at least that some given Cauchy sequence converges.

Here is one approach, showing the contrapositive instead (I always consider the contrapositive; some times the proof for one variation comes much easier than the proof for the other). Which is to say, if $X$ is not complete, then there is a continuous, unbounded function $X\to\Bbb R$.

Assume $X$ is not complete. Let $x_n$ be a Cauchy sequence in $X$ which is not convergent. Then define the function $f:X\to\Bbb R$ as follows: $$ f(x)=\frac1{\lim_{k\to\infty}d(x,x_k)} $$ It is well-defined and continuous (this must be shown), but $f(x_n)$ goes to $\infty$ as $n$ increases, so it isn't bounded (this must also be shown).

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  • $\begingroup$ Actually $X$ has to be compact and the proof is even simpler. $\endgroup$ Aug 26 '19 at 23:40

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