Some Proofs

Table of contents

  1. Theorem 2.3.4
  2. Bolzano-Weierstrass Theorem
  3. Lemma 3.3.1
  4. Min-Max Theorem

Theorem 2.3.4

Let ${x_n}$ be a bounded sequence. Then, there exist subsequences ${x_{n_k}}$ and ${x_{m_k}}$ such that \(\lim_{k\to \infty} x_{n_k} = \limsup_{n\to \infty} x_n\)

and \(\lim_{k\to \infty} x_{m_k} = \limsup_{n\to \infty} x_n.\)

Proof: Let $a_n = \sup{x_k \mid k\geq n}.$ Then, $\exists n_1 \in \mathbb{N}$ such that $a_1 -1 < x_{n_1} \leq a_1$. Now, $\exists n_2>n_1$ such that \(a_{n_1+1}-\frac{1}{2} < x_{n_2} \leq a_{n_1+1}\)

since \(a_{n_1+1} = \sup \{x_k \mid k\geq n_1 +1\}.\)

Similarly, $\exists n_3>n_2$ such that \(a_{n_2+1}-\frac{1}{3} < x_{n_3} \leq a_{n_2+1}.\)

Continuing in this way, we obtain a sequence of integers $n_1< n_2 < n_3< \dots$ such that \(a_{n_k+1} - \frac{1}{k+1} < x_{n_k} \leq a_{n_k+1}.\)

Given $\lim_{k\to \infty} a_{n_k +1} = \limsup_{n\to \infty} x_n$, by the Squeeze Theorem, \(\lim_{k\to \infty} x_{n_k} = \limsup_{n\to \infty} x_n.\)

Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence.

Proof: This follows immediately from the previous theorem, but is so important that it itself is a theorem.

Lemma 3.3.1

If $f:[a,b]\to \mathbb{R}$ is continuous then $f$ is bounded.

Proof: We prove the claim by contrapositive. Suppose $f$ is not bounded. Then for each $n\in \mathbb{N}$, there is an $x_n\in [a,b]$, such that $\vert f(x_n)\vert \ge n$.

The sequence \(\{x_n\}_{n=1}^{\infty}\) is bounded as $a\le x_n\le b$. By the Bolzano-Weierstrass theorem, there is a convergent sequence \(\{x_{n_i}\}_{i=1}^{\infty}\). Let $x:=\lim_{i\to\infty}x_{n_i}$. Since $a\le x_{n_i}\le b$ for all $i$, then $a\le x\le b$. The sequence \(\{f(x_{n_i})\}_{i=1}^{\infty}\) is not bounded as $\vert f(x_{n_i})\vert \ge n_i\ge i$. Thus, $f$ is not continuous at $x$. This contradicts the assumption, so $f$ is bounded.

Note: Find a sequence with a certain property, then use Bolzano-Weierstrass to make such a sequence that also converges.

Min-Max Theorem

Let $f:[a,b]\to \mathbb{R}$. If $f$ is continuous, then $f$ achieves an absolute maximum and absolute minimum.

Proof: We will prove this for the absolute maximum. If $f$ is continuous, then $f$ is bounded by Lemma 3.3.1. Thus, the set \(E = \{f(x) \mid x\in [a,b]\}\)

is bounded above. Let $L = \sup E$. Then:

  1. $L$ is an upper bound for $E$, i.e. $\forall x\in [a,b], f(x)\leq L.$

  2. There exists a sequence ${f(x_n)}_n$ with $x_n\in [a,b]$ such that $f(x_n) \to L$ (however \(\{x_n\}\) is need not converge).

By the Bolzano-Weierstrass theorem, there exists a subsequence \(\{x_{n_k}\}_k\) of \(\{x_n\}\) and $d\in [a,b]$ such that $x_{n_k}\to d$ as $k\to \infty$. Hence,

\[f(d) = \lim_{k\to \infty} f(x_{n_k}) = \lim_{n\to \infty} f(x_n) = L\]

by the continuity of $f$. Thus, $f$ achieves an absolute maximum at $d$.