# Lecture 24

## Reviews

Let $f: X\to Y$. Consider the following statement:

"$f$ is continuous $\iff$ for every open set $V\in Y$, $f^{-1}(V)$ is open in $X$."

1. To give a direct proof of the $\implies$ direction, what must be the first few steps be?

2. To give a direct proof of the $\impliedby$ direction, what must be the first few steps be?
3. Try to complete the proofs of both directions.

> A function $f:X\to Y$ is continuous if $\forall p\in X$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))$. (_For every point in a ball of $B_\delta(p)$, there is a ball of $B_\epsilon(f(p))$ that contains the image of the point._)
> 
> A set $V\subset Y$ is open if $\forall q\in V$, $\exists r>0$ such that $B_r(q)\subset V$.

## New materials

### Continuity and open sets

#### Theorem 4.8

A function $f:X\to Y$ is continuous if and only if for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$.

Proof:

$\implies$: Suppose $f$ is continuous. Let $V\subset Y$ be open. Let $p\in f^{-1}(V)$. Since $f(p)\in V$, $\exists \epsilon > 0$ such that $B_\epsilon(f(p))\subset V$.

Since $f$ is continuous, $\exists \delta > 0$ such that $f(B_\delta(p))\subset B_\epsilon(f(p))\subset V$. Therefore, $B_\delta(p)\subset f^{-1}(V)$. This shows that $f^{-1}(V)$ is open.

$\impliedby$: Suppose for every open set $V\subset Y$, $f^{-1}(V)$ is open in $X$. Let $p\in X$ and $\epsilon > 0$. Let $B_\epsilon(f(p))\in V$. Then $f^{-1}(B_\epsilon(f(p)))$ is open in $X$. 

Since $p\in f^{-1}(B_\epsilon(f(p)))$ and $f^{-1}(B_\epsilon(f(p)))$ is open, $\exists \delta > 0$ such that $B_\delta(p)\subset f^{-1}(B_\epsilon(f(p)))$. Therefore, $f(B_\delta(p))\subset B_\epsilon(f(p))$. This shows that $f$ is continuous.

QED

#### Corollary 4.8

$f$ is continuous if and only if for every closed set $C\subset Y$, $f^{-1}(C)$ is closed in $X$.

Ideas of proof:

- $C$ closed in $Y\iff Y\backslash C$ open in $Y$
- $f^{-1}(C)$ closed in $X\iff f^{-1}(Y\backslash C)$ open in $X$
- $f^{-1}(Y\backslash C) = X\backslash f^{-1}(C)$

Continue this proof by yourself.

#### Theorem 4.7

Composition of continuous functions is continuous.

Suppose $X,Y,Z$ are metric spaces, $E\subset X$, $f:E\to Y$ is continuous, and $g:Y\to Z$ is continuous. Then $g\circ f:E\to Z$ is continuous.

Ideas of proof:

- Let $B_\epsilon(g(f(p)))\subset Z$
- $g(f(B_\delta(p)))\subset B_\epsilon(g(f(p)))$
- $f(B_\delta(p))$ is open in $Y$
- $g^{-1}(B_\epsilon(g(f(p)))$ is open in $Y$
- $(g\circ f)^{-1}(B_\epsilon(g(f(p)))) = f^{-1}(g^{-1}(B_\epsilon(g(f(p)))))$
- $f^{-1}(g^{-1}(B_\epsilon(g(f(p))))) = (g\circ f)^{-1}(B_\epsilon(g(f(p))))$

Apply Theorem 4.8 to complete the proof.

#### Theorem 4.9

For $f:X\to \mathbb{C},g:X\to \mathbb{C}$ are continuous, then, $f+g,f/g$ are continuous.

Ideas of proof:

We can reduce this theorem to Theorem about limits and apply what you learned in chapter 3.

#### Examples of continuous functions 4.11

> $\forall p\in \mathbb{R}$, $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in \mathbb{R}$, $|x-p|<\delta\implies |f(x)-f(p)|<\epsilon$.

(a). $f(x) = \mathbb{R}\to \mathbb{R},f(x) = x$ is continuous. boring.

Proof:

Let $p\in \mathbb{R}$ and $\epsilon > 0$. Let $\delta = \epsilon$. Then, $\forall x\in \mathbb{R}$, if $|x-p|<\delta$, then $|f(x)-f(p)| = |x-p| < \delta = \epsilon$.

QED

Therefore, by **Theorem 4.9**, $f(x) = x^2$ is continuous. $f(x) = x^3$ is continuous... So all polynomials are continuous.

(b). $f:\mathbb{R}^k\to \mathbb{R},f(x)=|x|$ is continuous.

Ideas of proof:

- $|f(x)-f(p)| = ||x|-|p||\leq |x-p|$ By reverse triangle inequality.
- Let $\epsilon > 0$. Let $\delta = \epsilon$.

### Continuity and compactness

#### Theorem 4.13

A mapping of $f$ of a set $E$ into a metric space $Y$ is said to be **bounded** if there is a real number $M$ such that $|f(x)|\leq M$ for all $x\in E$.

#### Theorem 4.14

$f:X\to Y$ is continuous. If $X$ is compact, then $f(X)$ is compact.

Proof strategy:

For every open cover $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$, there exists a corresponding open cover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$.

Since $X$ is compact, there exists a finite subcover $\{f^{-1}(V_\alpha)\}_{\alpha\in A}$ of $X$. Let the finite subcover be $\{f^{-1}(V_\alpha)\}_{i=1}^n$.

Then, $\{V_\alpha\}_{i=1}^n$ is a finite subcover of $\{V_\alpha\}_{\alpha\in A}$ of $f(X)$.

See the detailed proof in the textbook.

#### Theorem 4.16 (Extreme Value Theorem)

Suppose $X$ is a compact metric space and $f:X\to \mathbb{R}$ is continuous. Then $f$ has a maximum and a minimum on $X$.

i.e.

$$
\exists p_0,q_0\in X\text{ such that }f(p_0) = \sup f(X)\text{ and }f(q_0) = \inf f(X).
$$

Proof:

By Theorem 4.14, $f(X)$ is compact.

By Theorem 2.41, $f(X)$ is closed and bounded.

By Theorem 2.28, $\sup f(X)$ and $\inf f(X)$ exist and are in $f(X)$. Let $p_0\in X$ such that $f(p_0) = \sup f(X)$. Let $q_0\in X$ such that $f(q_0) = \inf f(X)$.

QED

---

Supplemental materials:

_I found this section is not covered in the lecture but is used in later chapters._

#### Definition 4.18

Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. $f$ is **uniformly continuous** on $X$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x, y\in X$, $|x-y| < \delta \implies |f(x)-f(y)| < \epsilon$.

#### Theorem 4.19

If $f$ is a continuous mapping of a compact metric space $X$ into a metric space $Y$, then $f$ is uniformly continuous on $X$.

Proof:

See the textbook.

QED

---

### Continuity and connectedness

> **Definition 2.45**: Let $X$ be a metric space. $A,B\subset X$ are **separated** if $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.
> 
> $E\subset X$ is **disconnected** if there exist two separated sets $A$ and $B$ such that $E = A\cup B$.
> 
> $E\subset X$ is **connected** if $E$ is not disconnected.

#### Theorem 4.22

$f:X\to Y$ is continuous, $E\subset X$. If $E$ is connected, then $f(E)$ is connected.

Proof:

We will prove the contrapositive statement: if $f(E)$ is disconnected, then $E$ is disconnected.

Suppose $f(E)$ is disconnected. Then there exist two separated sets $A$ and $B\in Y$ such that $f(E) = A\cup B$.

Let $G = f^{-1}(A)\cap E$ and $H = f^{-1}(B)\cap E$.

We have:

$f(E)=A\cup B\implies E = G\cup H$

Since $A$ and $B$ are nonempty, $A,B\subset f(E)$, this implies that $G$ and $H$ are nonempty.

To complete the proof, we need to show $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.

We have $G\subset f^{-1}(A)\cap E\subset f^{-1}(A)\subset f^{-1}(\overline{A})$ Since $\overline{A}$ is closed, $f^{-1}(\overline{A})$ is closed. This implies that $\overline{G}\subset f^{-1}(\overline{A})$.

So $\overline{G}\subset f^{-1}(\overline{A})$ and $\overline{H}\subset f^{-1}(\overline{B})$.

Since $A$ and $B$ are separated, $\overline{A}\cap B = \phi$ and $\overline{B}\cap A = \phi$.

Therefore, $\overline{G}\cap H = \phi$ and $\overline{H}\cap G = \phi$.

QED

#### Theorem 4.23 (Intermediate Value Theorem)

Let $f:[a,b]\to \mathbb{R}$ be continuous. If $c$ is a real number between $f(a)$ and $f(b)$, then there exists a point $x\in [a,b]$ such that $f(x) = c$.

Ideas of proof:

Use Theorem 2.47. A subset $E$ of $\mathbb{R}$ is connected if and only if it has the following property: if $x,y\in E$ and $x<z<y$, then $z\in E$.

Since $[a,b]$ is connected, by **Theorem 4.22**, $f([a,b])$ is connected.

$f(a)$ and $f(b)$ are real numbers in $f([a,b])$, and $c$ is a real number between $f(a)$ and $f(b)$.

By **Theorem 2.47**, $c\in f([a,b])$.

QED
