Extension: Bolzano-Weierstrass Proof of Extreme Value Theorem

The Bolzano-Weierstrass approach involves a theorem in analysis called the Bolzano-Weierstrass theorem. Using this theorem, we'll prove the Boundedness Theorem, and then use the Boundedness Theorem to prove the Extreme Value Theorem.

Theorem (Bolzano-Weierstrass): Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

We will not prove this theorem (see any real analysis textbook for the proof), but let's look at what this theorem means. Let's say we have a bounded sequence of real numbers $x_1,x_2,x_3\ldots$. Bounded means that there is some $M\in\mathbb{R}$ such that $|x_i|\le M$, for all $i=1,2,3,\ldots$. The Bolzano-Weierstrass theorem then says that there is some subsequence of $(x_n)$ that converges to some $x\in\mathbb{R}$. A subsequence of $(x_n)$ is what is sounds like - you take $x_1,x_2,x_3,\ldots$ and just pick some of those (in order!) to be your subsequence. A simple example is picking all the even terms, or all the odd terms.

Example 1: Consider the sequence $(x_n)$ where $x_n=(-1)^n$. This is the sequence $-1,1,-1,1,\ldots$, which is bounded, with say, $M=1$. This sequence does not converge (try showing this by contradiction, using the definition of a limit of a sequence and the triangle inequality. The key fact is that for every $N$, there exist $n,m>N$ with $x_n=1$ and $x_m=-1$.) However, it does have a subsequence that converges. One such example is all the odd terms, to make the sequence $-1,-1,-1,\ldots$, which clearly converges to $-1$.

Example 2: To see why bounded is necessary in the theorem, consider the sequence $1,2,3,4,5,\ldots$. Here $x_n=n$, and $(x_n)$ diverges to $+\infty$. But every subsequence of $(x_n)$ also diverges to $+\infty$, so there is no subsequence that converges to $x\in\mathbb{R}$.

Now it's time for the Boundedness Theorem.

Theorem (Boundedness Theorem): A continuous function $f:[a,b]\to\mathbb{R}$ is bounded.

Proof: Suppose $f$ is not bounded. Then for every $n\in\mathbb{N}$, there exists some $x_n\in[a,b]$ with $f(x_n)>n$. The sequence $(x_n)$ is bounded since it is inside $[a,b]$, so has a convergent subsequence, call it $(x_{n_k})$ that converges to some $x\in[a,b]$. Now consider $f(x_n)$. Since $f$ is continuous, we have $f(x)=\lim_{n_k\to\infty}f(x_{n_k})$. But since $f(x_{n_k})>n_k$, we have $f(x)=\lim_{n_k\to\infty}f(x_{n_k})=+\infty$, a contradiction. Thus $f$ must be bounded.

Theorem (Extreme Value Theorem): A continuous function $f:[a,b]?\mathbb{R}$ attains its maximum and minimum.

Proof: By the Boundedness Theorem, $f$ is bounded. Consider the supremum (least upper bound) $S$ of $f([a,b])$. We need to show that there is some $x\in[a,b]$ with $f(x)=S$, i.e. $f$ attains this value. Since $S$ is the least upper bound, we can form a sequence $(x_n)\subset[a,b]$ such that $f(x_n)>S-\frac{1}{n}$ (since this is not an upper bound for $f([a,b])$. Moreover, $f(x_n)>S-\frac{1}{n}$ implies that $f(x_n)\to S$. Now consider $(x_n)\subset[a,b]$. By the Bolzano-Weierstrass theorem, this implies that $(x_n)$ has a convergent subsequence $(x_{n_k})\to x\in[a,b]$. By continuity of $f$, then $f(x)=\lim_{n_k\to\infty}f(x_{n_k})=S$, so $f$ attains its maximum $S$.

For the minimum, consider $-f$.

Remark: Why did we need to use the Bolzano-Weierstrass theorem? That is, why doesn't the conclusion $f(x_n)\to S$ finish the proof?

The conclusion $f(x_n)\to S$ doesn't tell us what the $x_n$ actually do. For example, consider $f(x)=x^2$ on $[-1,1]$. Look at the sequence $(x_n)=(-1)^n(1-\frac{1}{n})$. The idea is that $x_n$ oscillates between approaching $1$ and $-1$. Observe that $f(x_n)$ is simply $(1-\frac{1}{n})^2$, which goes to $1$ as $n\to\infty$. However, the sequence $(x_n)$ does not converge to anything. We need the Bolzano-Weierstrass theorem to ensure that we can get a convergent subsequence, and then we use the limit of that convergent subsequence to get the $x$ such that $f(x)=S$. Without having a limit of some sequence, we can't get that $x$.