Continuous functions and Compact Sets

Section 5.4 Continuous functions and Compact Sets

Now we're ready to see how continuous functions interact with compactness.

Proposition 5.4.1.

Let \(f:A\to M\) be a continuous function from a subset of a metric space to a metric space. For any compact \(K\subseteq A\text{,}\) \(f_*(K)\subset M\) is compact.

Theorem 5.4.2. Max-Min Theorem.

Let \(f:A\to \mathbb{R}\) be a continuous function from a subset of a metric space to \(\mathbb{R}\text{.}\) If \(A\) is compact, then there are \(a_+,a_-\in A\) so that \(f(a_+)=\sup\{f(x)\vert x\in A\}\) and \(f(a_-)=\inf\{f(x)\vert x\in A\}\text{.}\)

That is to say, continuous functions achieve maxima and minima on compact domains.

Example 5.4.3.

Here are some functions which lack either a maximum or a minimum on their domains:

\(f:(0,1]\to \mathbb{R}\) given by \(f(x)=\frac{1}{x}\)
\(g:[0,1]\to\mathbb{R}\) given by \(g(x)=\begin{cases}2-x&\text{ if }x\lt \frac{1}{2}\\x+3&\text{ if }x\gt \frac{1}{2}\end{cases}\)
\(h:[1,\infty)\to\mathbb{R}\) given by \(h(x)=\frac{1}{x}\)

Checkpoint 5.4.4.

In \(\mathbb{R}^m\text{,}\) the unit sphere \(S^{m-1}\) (with respect to the 2-norm) is compact. (We showed this in homework.)

For any other \(p\)-norm \(\lVert\cdot\rVert_p\text{,}\) we can see from the formula that \(\lVert\cdot\rVert_p:\mathbb{R}^m\to \mathbb{R}\) is continuous.

Therefore, \(\lVert \cdot\rVert_p\) achieves a maximum and a minimum on \(S^{m-1}\text{.}\)

Use this to explain why every \(p\)-norm on \(\mathbb{R}^m\) is equivalent to \(\lVert\cdot\rVert_2\text{.}\)

Definition 5.4.5.

We call a function \(f:A\to V\) uniformly continuous if

\begin{equation*} \forall \epsilon\gt 0,\exists \delta\gt 0:\forall x,y\in A, d(x,y)\lt \delta\Rightarrow d(f(x),f(y))\lt\epsilon\ \ . \end{equation*}

Checkpoint 5.4.6.

Contrast this definition with Proposition 4.2.10.

Theorem 5.4.7.

If \(f:A\to V\) is continuous and \(A\) is compact, then \(f\) is uniformly continuous.

Checkpoint 5.4.8.

Give a direct proof of Theorem 5.4.7.

Hint.

Given \(\epsilon\gt 0\text{,}\) for each \(x\in A\) there is \(\delta_x\) so that \(d(x,y)\lt \delta_x\Rightarrow d(f(x),f(y))\lt \frac{1}{2}\epsilon\text{.}\) This gives an open cover of \(A\) by balls: \(\mathcal{U}=\left\{B_{\frac{1}{2}\delta_x}(x)\middle\vert x\in A\right\}\text{.}\)