User:Anghel/Sandbox

Theorem
Let $\struct{ X, d_X }$ be a compact metric space.

Let $\struct{ Y, d_Y }, \struct{ Z, d_Z }$ be metric spaces.

Let $\phi : Y \to Z$ be a continuous mapping.

For all $n \in \R_{\ge 0}$, let $f_n : X \to Y$ be a continuous mapping.

Let $f: X \to Y$ be a mapping such that $\sequence {f_n}_{n \in \R_{\ge 0} }$ converges to $f$ uniformly on $X$.

Then the sequence $\sequence {\phi \circ f_n}_{n \in \R_{\ge 0} }$ converges to $\phi \circ f$ uniformly on $X$.

Proof
that $\sequence {\phi \circ f_n}_{n \in \R_{\ge 0} }$ do not converge to $\phi \circ f$ uniformly.

This implies that there exists $\epsilon \in \R_{>0}$ such that for all $n \in \N$, there exist $N \ge n$ and $x_n \in X$ such that:


 * $\map {d_Z}{ \map { \phi \circ f_N }{ x_n }, \map { \phi \circ f }{ x_n} } \ge \epsilon$

From Compact Subspace of Metric Space is Sequentially Compact in Itself, it follows that $\struct{ X, d_X }$ is sequentially compact in itself.

This implies that there exist $x' \in X$ and a subsequence $\sequence {x_{n_m} }_{m \in \N}$ such that $x_{n_m} \to x'$ for $m \to \infty$.

We calculate:

which is a contradiction, as $\epsilon > 0$.

Category:Uniform Convergence]]

Theorem
Let $\gamma : \closedint 0 1 \to \R^2$ be a Jordan curve.

Let $t_1, t_2 \in \closedint 0 1$ with $t_1 < t_2$.

Let $h \in \R_{>0}$.

Let $\Img {\gamma}$, $\Int { \gamma }$ and $\Ext { \gamma }$ denote the image, interior, and exterior of $\gamma$.

Then there exists $r \in \R_{>0}$ such that $r \le h$, and for all $p \in \map {B_r}{ \map \gamma {t_1} }, q \in \map {B_r}{ \map \gamma {t_2} }$:


 * if $p, q \in \Int \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Int \gamma$ with $\map \sigma 0 = p, \map \sigma 1 = q$ such that $\map d {\Img \sigma, \Img \gamma} < h$;


 * if $p, q \in \Ext \gamma$, then there exists a path $\sigma : \closedint 0 1 \to \Ext \gamma$ with $\map \sigma 0 = p, \map \sigma 1 = q$ such that $\map d {\Img \sigma, \Img \gamma} < h$.

Here $\map {B_r}{ x }$ denotes the open ball with radius $r$ and center $x$, and $\map d {X,Y}$ denotes the distance between the sets $X$ and $Y$.

Proof
Let $\phi : \R^2 \to \R^2$ be the homeomorphism defined in the Jordan-Schönflies Theorem such that:


 * $\phi \sqbrk {\Img {\gamma} } = \mathbb S_1$
 * $\phi \sqbrk {\Int {\gamma} } = \map {B_1}{ \mathbf 0 }$
 * $\phi \sqbrk {\Ext {\gamma} } = \R^2 \setminus \map {B_1^-} { \mathbf 0 }$

where $\mathbb S_1$ denotes the unit circle in $\R^2$.

For all $\epsilon \in \openint 0 {\dfrac 1 2}$, let $\sigma_{\epsilon, \operatorname {Int} }, \sigma_{\epsilon, \operatorname {Ext} } : \closedint {t_1}{t_2}$ be the paths defined by:


 * for all $s \in \closedint {t_1}{t_2}$: $\map { \sigma_{\epsilon, \operatorname {Int} } } s = \paren { 1 - \epsilon } \paren{ \map {\phi \circ \gamma} s }$
 * for all $s \in \closedint {t_1}{t_2}$: $\map { \sigma_{\epsilon, \operatorname {Ext} } } s = \paren { 1 + \epsilon } \paren{ \map {\phi \circ \gamma} s }$

As $\norm{ \map {\phi \circ \gamma} s } = 1$, it follows that $\norm{ \map { \sigma_{\epsilon, \operatorname {Int} } } s } = 1 - \epsilon$, where $\norm {\mathbf x}$ denotes the Euclidean norm of $\mathbf x$ on $\R^2$.

It follows that $\sigma_{\epsilon, \operatorname {Int} }$ is a path in $\map {B_1}{ \mathbf 0 }$.

From Composite of Continuous Mappings between Metric Spaces is Continuous, it follows that $\phi^{-1} \circ \sigma_{\epsilon, \operatorname {Int} } : \closedint 0 1 \to \R^2$ is continuous.

For all $s \in \closedint 0 1$, we have:

The set $\map {B_h}{ \map { \phi \circ \gamma_0 } t } \setminus \mathbb S^1$ is split into these two path components:


 * $B_{\operatorname{Int} } = \set { p' \in \map {B_h}{ \map { \phi \circ \gamma_0 } t } : \norm {p'} < 1 }$
 * $B_{\operatorname{Ext} } = \set { p' \in \map {B_h}{ \map { \phi \circ \gamma_0 } t } : \norm {p'} > 1 }$

Category:Jordan Curves]]