How to find an unusual proof that continuous functions on the closed interval [0,1] are bounded.

Question

Timothy Gowers tells us how to find an unusual proof that continuous functions on the closed interval $[0,1]$ are bounded.

He gives us a brief outline of the usual proofs:

"One proof starts with the Heine-Borel theorem: however you write $[0,1]$ as a union of a collection of open sets, you can also write it as the union of a finite subcollection. To deduce the statement, for each $x$ let $U_x$ be the set of all $y$ such that $|f(y)-f(x)| < 1$. The sets $U_x$ are open and their union is obviously all of $[0,1]$. By the Heine-Borel theorem there are $x_1,...,x_n$ such that the union of the sets $U_{x_i}$ is all of [0,1]. But then f is bounded above by 1+max f(xi).

Another proof starts with the Bolzano-Weierstrass theorem: every sequence in [0,1] has a convergent subsequence. Then, if the theorem is false, we can find an infinite sequence (xn) such that f(xn) > n for every n. Pick a subsequence converging to x, and check that $f$ cannot possibly be continuous at $x$."

He then says that there are existing proofs of this that are "unusual" for e.g. when you prove it without knowing anything about compactness.

Does anyone else know any other ways of proving the above that are deemed "unusual" and isn't the normal proof?

Answer 1

I don't think that Spivak's proof uses compactness.

Let $A$ be the set of all values $a\in [0,1]$ of $a$ such that $f[0,a]$ is bounded.

Notice that $0\in A$, and $1$ is an upper bound for $A$.

Therefore $A$ has a least upper bound $\alpha$.

We first prove $\alpha\in A$, if $\alpha=0$ it is clear, otherwise:

Taking $\epsilon=1$ there is a $\delta$ such that $f(\alpha-\delta,\alpha]$ is bounded by continuity, we also have that there is $a\in A$ with $a>\alpha-\delta$ with $a\in A$ (because $\alpha$ is a least upper bound). Therefore $f([0,a])$ and $f((\alpha-\delta,\alpha])$ is bounded. So $f([0,\alpha]$ is bounded, and $\alpha\in A$.

We now prove $\alpha=1$:

Notice that if $\alpha\neq 1$ then taking $\epsilon=1$ there is a $\delta$ such that $f[\alpha,\alpha+\delta)$ is bounded, since $f([0,\alpha]$ is also bounded we conclude that $f([0,\alpha+\frac{\delta}{2}])$ is bounded, a contradiction, since $\alpha+\delta\in A$.

Answer 2

First show that continuous function on interval is uniformly continuous. (This answer addresses multiple ways to do it without compactness)

Then use the fact that any uniformly continuous image of a bounded set is bounded.

Answer 3

I'm not sure it is unusual, but you don't necessarily have to use compactness directly.

For example let's consider the set: $S = \{y\in[0,1], \exists M \ge 0, \forall x\in[0,y], |f(x)| \le M\}$. Because $S$ is bounded and nonempty, $m=\sup S$ is well-defined.

If $m<1$, by continuity in $m$ there exists $\delta>0$ s.t. $[m-\delta,m+\delta]\subset[0,1]$ and $|f(x)-f(m)| \le 1$ for all $x \in [m-\delta,m+\delta]$. However $m-\delta \in S$, thus for all $x\in[m-\delta,m+\delta]$:

$$|f(x)| \le |f(x)-f(m)| + |f(m)-f(m-\delta)|+|f(m-\delta)| \le M+2$$

This is also true for $0 \le x \le m-\delta$, hence $m+\delta \in S$ and the expected contradiction.

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UPD: Thus $m=1$.We now have to show $m \in S$ (hence $S=[0,1]$). This is done exactly like before.