NoteNextra-origin/pages/Math4111/Math4111_L24.md

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.

EOP

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.

EOP

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).

EOP

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.

EOP

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]).

EOP