# Compactness of $(M, d)$ in terms of continuous functions $f:M\to \mathbb R$

Suppose $(M,d)$ a metric space. I want to show that

If every continuous real-valued function on $M$ attains a maximum, then $(M, d)$ is compact.

I was trying to do this by assuming $M$ non-compact and showing explicitly a function which does not attain a maximum. If $M$ is not bounded it's enough to take the distance from a fixed point as we can always find a sequence which "exits" every compact and pushes the distance at infinity. But what if $M$ is bounded?

If $M$ is not compact then there is a sequence $(x_n)$ without convergent subsequences. The sequence is a closed discrete subspace so the map $f: \{x_n:n\in\mathbb N\}\to \mathbb R$ given by $x_n\to n$ is continuous. By the Tietze extension theorem it has a continuous extension $F: M\to\mathbb R$. Clearly, $F$ is unbounded.

• Small point: you need to make sure your sequence has no duplicate entries for that function to be well-defined. Feb 27, 2012 at 22:09
• You don't really need to appeal to Tietze. Define $f(x) = \sum_n n \max(1 - d(x, x_n)/\epsilon_n, 0)$ where $\epsilon_n > 0$ is small enough that there are no $x_j \ne x_n$ with $d(x_j, x_n) < 4 \epsilon_n$. Feb 27, 2012 at 23:19
• Thanks a lot to you all! All the answers really helped me a lot! Thanks! Feb 28, 2012 at 10:38

This is not meant to replace azarel’s answer, but rather to show in more detail just how unbounded real-valued functions arise in non-compact metric spaces.

Recall that if a metric space $$\langle X,d\rangle$$ is not compact, then either it’s not complete, or it’s not totally bounded.

If it’s not complete, it has a Cauchy sequence $$\langle x_n:n\in\omega\rangle$$ that does not converge; define $$f:X\to\mathbb{R}:x\mapsto\lim_{n\to\infty}d(x,x_n)\;.$$

Intuitively, $$f(x)$$ is the distance from $$x$$ to the non-existent point to which the Cauchy sequence wants to converge, so you’d expect $$f$$ to exist and be continuous, and indeed this isn’t hard to prove. Moreover, $$f(x)$$ is never $$0$$, because the Cauchy sequence has no limit in $$X$$. Thus, the function $$g:X\to\mathbb{R}:x\mapsto\frac1{f(x)}$$ is continuous, and it’s also clearly unbounded, since $$\lim_{n\to\infty}g(x_n)=\infty\;.$$ In effect we’ve pushed that non-existent limit of the Cauchy sequence out to be a point at infinity.

If $$X$$ is not totally bounded, there is an $$r>0$$ such that no finite collection of open $$r$$-balls covers $$X$$. Thus, we may recursively construct a sequence $$\langle x_n:n\in\omega\rangle$$ such that for each $$n\in\omega$$, $$x_n\in X\setminus\bigcup_{k where $$B(x_k,r)$$ is the open ball of radius $$r$$ centred at $$x_k$$. Note that $$d(x_m,x_n)\ge r$$ whenever $$m, so $$\{B(x_n,r/3):n\in\omega\}$$ is an infinite family of pairwise disjoint open balls of radius $$r/3$$. Now let

\begin{align*} g&:X\to\mathbb{R}:\\ &x\mapsto\begin{cases} 0,&\text{if }x\in X\setminus\bigcup_{n\in\omega}B\left(x_n,\frac{r}3\right)\\\\ \frac{3n}r\left(\frac{r}3-d(x,x_n)\right),&\text{if }x\in B\left(x_n,\frac{r}3\right)\;. \end{cases}\end{align*}

Clearly $$g(x_n)=n$$ for each $$n\in\omega$$, so $$g$$ is unbounded, and once again it isn’t hard to show that $$g$$ is continuous.

This time the intuitive idea is to replace the continuous function $$B(x,s)\to\mathbb{R}:y\mapsto d(x,y)$$ that measures distance from the centre of a ball with the function $$B(x,s)\to\mathbb{R}:y\mapsto s-d(x,y)\tag{1}$$ that in a sense measures distance from the boundary of the ball, and then ‘blow up’ the function in $$(1)$$, making the ball’s radius look bigger than $$s$$. Since we have infinitely many balls to play with, and they don’t interfere with one another, we can get arbitrarily large ‘blow-up’.