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Theorem : Continuous Maps Preserve Compactness

$(X, d_X), (Y, d_Y). f: X \to Y$ continuous.

$K \subseteq X$, $K$ compact.

Then the image $f(K) := \set{ f(x) : x \in K }$ of $K$ is also compact.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

If $\set{U_\alpha}$ is some open cover of $f(K)$,

then exists a finite subcover.

Note

Since f is continuous, we have all the set $\inv{ f }(U_ \alpha) $ are open.

Thus $\set{ \inv{ f } (U_\alpha)} $ is an open cover of $K$.

By compactness, exists some finite subcover.

$\inv{ f } (U_1), \inv{ f }(U_2) …, \inv{ f }(U_n)$

But then (pushing forward) we got $U_1, U_2, …, U_n $ is a finite subcover.

Note

Q: Why is $\set{ \inv{ f } (U_ \alpha)}$ a cover for $K$.

Note $f(k) \subset \bigcup_{ \alpha } U_ \alpha$

Apply $\inv{ f } $ we get

$K \subset \inv{ f } (f (K)) \subset \inv{ f } (\bigcup_ \alpha) = \bigcup_{ \alpha } \inv{ f } (U_ \alpha)$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Fix $\epsilon > 0$.

Since $f$ is continuous at every $x_0 \in X $,

there exists $\delta_{ x_0 } > 0 $, such that $d(x, x_0) < \delta _ {x_0} \implies d(f(x), f(x_0)) < \frac{ \epsilon }{ 2 }$.

Notice $X \subseteq \bigcup_{ x_0 \in X } B(x_0, \delta_{x_0}) $

$X \subseteq \bigcup_{ x_0 \in X } B(x_0, \frac{\delta_{x_0}}{ 2 }) $

$\delta = \min{ \frac{ \delta_{x_1}}{ 2 }, \frac{ \delta_{x_2}}{ 2 }, …, \frac{ \delta_{x_n}}{ 2 }}$

If $x, y \in X $, $d(x, y) < \delta $, $(\delta \leq \frac{ \delta_{x_i}}{ 2 } \forall i)$

$x \in B(x_i, \delta _{ \frac{ x_i }{ 2 }})$ for some $i$.

$d(x, x_0) < \frac{ \delta_{x_i}}{ 2 } $ and $d(y, x_i) < \delta_{x_i} $

$x, y \in B(x_i, \delta_{x_i}) $

$d(f(x), f(y)) < \epsilon$

Proposition : Maximum Principle

Let $(X, d) $ be a compact metric space, and $f: X \to \R$ continuous.
Then $f$ is bounded.

Furthermore,
$f $ attains its maximum at some point $x_{ \text{max}} \in X$,
and attains its minimum at some point $x_{ \text{min}} \in X$.

Definition: Uniform Continuity

$(X, d_X) $, $(Y, d_Y) $, $f: X \to Y $.

We say that $f $ is uniformly continuous iff

$$\forall \epsilon > 0, \exists \delta > 0\text{ s.t. }\forall x, x' \in X, d_X(x, x') < \delta\text{ implies }d_Y(f(x), f(x')) < \epsilon $$

Remarks

Every uniformly continuous function is continuous, but not conversely. (Why?)

Theorem

$(X, d_X), (Y, d_Y)$. Suppose that $(X, d_X) $ is compact. $f: X \to Y$.

$$f\text{ is continuous }\iff f \text{ is uniformly continuous}$$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

(necessity)

$f$ is uniformly continuous, then $f$ is continuous.

(sufficiency)

$f$ is continuous.

$\forall x_0 \in X$, $f$ is continuous at $x_0$.

Fix $\epsilon > 0$. There exists a $\delta(x_0) > 0 $, depending on $x_0 $,
such that $d_Y(f(x), f(x_0)) < \epsilon / 2$ whenever $d_X(x, x_0) < \delta(x_0)$.

By the triangle inequality, this implies that $d_Y(f(x), f(x')) < \epsilon$ whenever $x \in B_{(X, d_X)}(x_0, \delta(x_0) / 2)$ and $d_X(x', x) < \delta(x_0) / 2$.

Now consider the (possibly infinite) collection of balls
$$\set{ B_{(X, d_X)}(x_0, \delta(x_0) / 2): x_0 \in X } $$ Each ball in this collection is of course open, and the union of all these balls covers $X$, since each point $x_0$ in $X$ is contained in its own ball $B_{(X, d_X)}(x_0, \delta(x_0) / 2)$.

Since $X$ is compact, there exist a finite number of points $x_1, x_2, …, x_n $ such that the balls $B_{(X, d_X)}(x_j, \delta(x_j) / 2) $ for $j = 1, …, n $ cover $X$:

$$X \subseteq \bigcup_{ j = 1 }^n B_{(X, d_X)}(x_j, \delta(x_j) / 2)$$

Now let $\delta := \min{ \delta(x_j) / 2} $.

Since each of the $\delta(x_j) $ are positive and there are only a finite number of $j$, we see that $\delta > 0$.

Now let $x, x'$ be any two points in $X$ such that $d_X(x, x') < \delta$.

Since the balls $B_{(X, d_X)}(x_j, \delta(x_j) / 2)$ cover $X$, we see that there must exist $1 \leq j \leq n $ such that $x \in B_{(X, d_X)}(x_j, \delta(x_j) / 2)$.

Since $d_X(x, x') < \delta$, we have $d_X(x, x') < \delta(x_j) / 2 $, and so by the previous discussion we have $d_Y(f(x), f(x')) < \epsilon$.

Thus, we found a $\delta$ that $d_Y(f(x), f(x')) < \epsilon$ whenever $d(x, x')< \delta$ and this proves uniform continuity as desired.

Theorem : Uniform Continuity Preserved by Composition

$(X, d_X), (Y, d_Y), (Z, d_Z)$. $f: X \to Y, g: Y \to Z$.

$$f, g \text{ uniformly continuous }\implies g \circ f : X \to Z\text{ uniformly continuous}$$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

Since $f$ is uniformly continuous,

$\forall \delta > 0, \exists \alpha > 0$ such that $\forall x, x' \in X, d_X(x, x') < \alpha$ implies $d_Y(f(x), f(x)^\prime) <\delta$.

Since $g$ is also uniformly continuous,

Then fix $\epsilon > 0, \exists \delta > 0$ such that $\forall x, x' \in X$, $d_Y(f(x), f(x')) < \delta$ implies $d_Z(g(f(x)), g(f(x'))) < \epsilon$.

That means for $\epsilon$, there exists some $\alpha$ such that $\forall x, x' \in X, d_X(x, x') < \alpha $ implies $d_Z(g(f(x)), g(f(x'))) < \epsilon$.

Theorem : Direct Sum Operation Preserves Uniform Continuity

$(X, d_X) $, $f: X \to \R, g: X \to \R, f \oplus g : X \to \R^2$. $d_{l^2} $ metric.

$$f, g \text{ uniformly continuous }\implies f \oplus g : X \to \R^2 \text{ uniformly continuous}$$

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

If $f$ and $g$ are both uniformly continuous.

$\forall \epsilon > 0, \exists \delta_1 > 0, \delta_2 > 0 $ such that

$\forall x, x' \in X,$

$d_X(x, x') < \delta_1$ implies $d(f(x), f(x')) < \frac{ 1 }{ \sqrt[ ]{ 2 }} \epsilon$

$d_X(x, x') < \delta_2$ implies $d(g(x), g(x')) < \frac{ 1 }{ \sqrt[ ]{ 2 }}\epsilon$

choose $\delta = \min{ \delta_1, \delta_2 }$,

if $d_X(x, x') < \delta $

$d_{l^2}((f(x), g(x)), (f(x'), g(x'))) = \sqrt[ ]{ d(f(x), f(x'))^2 + d(g(x), g(x'))^2 } < \sqrt[ ]{ \frac{ 1 }{ 2 }\epsilon^2 + \frac{ 1 }{ 2 }\epsilon^2 } < \epsilon $

$\therefore $, $f \oplus g $ is still uniformly continuous.

Theorem

$x, y \in \R^2 $

The addition function $(x, y) \mapsto x + y $, are uniformly continuous from $\R^2$ to $\R$.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

$\forall (x, y), (x', y') \in \R^2$, fix $\epsilon > 0$, choose $\delta = \epsilon$.

Then,

$d_{l^2}((x, y), (x', y')) < \delta $ implies that $\sqrt{(x- x')^2 + (y - y')^2 } < \delta$

$$\begin{align*} (x- x')^2 + (y - y')^2 &< \delta^2 \br 2(x - x')(y - y') &< \delta^2 \br (x- x')^2 + (y - y')^2 + 2(x - x')(y - y') &< 2 \delta^2 \br ((x - x') + (y - y'))^2 &< 2 \delta^2 \br \abs{(x + y) - (x' + y')} &< \sqrt{ 2 }\delta \end{align*}$$

Hence, given $\epsilon > 0 $, choose $\delta = \frac{ \epsilon }{ \sqrt{ 2 }}$.

We have $d_{l^2}((x, y), (x', y')) < \delta $ implies $d((x + y), (x' + y')) < \epsilon$.

Hence the addition function $(x, y) \mapsto x + y$ is uniformly continuous.

Theorem

$x, y \in \R^2 $

The subtraction function $(x, y) \mapsto x - y$, are uniformly continuous from $\R^2$ to $\R$.

Proof 
  </span>
</span>
<span class="proof__expand"><a>[expand]</a></span>

We first prove that $(x, y) \mapsto (x, -y)$ are uniformly continuous.

fix $\epsilon > 0 $

$\forall (x, y), (x', y') \in \R^2$