- #1

Eclair_de_XII

- 1,083

- 91

- TL;DR Summary
- Tried copying math proof from this forum into LaTeX. Need feedback on how to better structure it.

style:

```
\newcommand\func{\(f\)}
\newcommand\myset[1][math]{\ifthenelse{\equal{math}{#1}}{\(K\)}{K}}
\newcommand\diff[4]{\(|#1-#2|#4#3\)}
\newcommand\ball[3]{\diff{#1}{#2}{#3}{<}}
\newcommand\Fllab[2]{\diff{f(#1)}{f(#2)}{\epsilon}{\geq}}
\newcommand\term[2][n_k]{\({#2}_{#1}\)}
\newcommand\setdelta[1]{\(\delta=#1\)}
\newcommand\ptsinset[1]{\(x_{#1},y_{#1}\in\myset[]\)}
\newcommand\xyball[1]{\ball{x_{#1}}{y_{#1}}{\ifthenelse{#1>1}{\frac{1}{#1}}{1}}}
\newcommand\seq[2][n]{\(\{{#2}_{#1}\}\)}
\newcommand\ucontinv{there is \(\epsilon>0\) such that for any \(\delta>0\), there are points \(x,y\in \myset[]\) such that \ball{x}{y}{\delta} but \Fllab{x}{y}}
\newcommand\seqconverge[2]{There is an integer \(#1\) such that if \(n_k\geq#1: |{#2}_{n_k}-z|<\delta'\)\par}
```

proof:

```
Let \func{} be a continuous function defined on a sequentially compact set \myset. Suppose \func is not uniformly continuous. By definition, \ucontinv.
In particular, choose \setdelta{1} and find points \ptsinset{1} with the property that \xyball{1} and \Fllab{x_1}{y_1}. Now choose \setdelta{2} and then find points \ptsinset{2} with the property that \xyball{2} and \Fllab{x_2}{y_2}.
Continuing in this fashion, we obtain two sequences \seq{x}, \seq{y} with the property that for any \(\delta>0\), we can choose an integer \(N>\frac{1}{\delta}\) in order to ensure that \(|x_N-y_N|<\frac{1}{N}<\delta\).
Since \myset{} is sequentially compact, it follows that there exists a subsequence of \seq{x}, which we shall denote as \seq[n_k]{x} that converges to some point \(z \in \myset[]\). Now for each \seq[n_k]{x}, choose \seq[n_k]{y} such that \ball{x_{n_k}}{y_{n_k}}{\frac{1}{m}} where \(m\) is the index of \term{x}, \term{y} in their respective parent sequences.
This gives us a sequence \seq[n_k]{y} with the property that \(|y_{n_k}-x_{n_k}|\rightarrow0\). Moreover, since a given \term{x} gets arbitrarily close to \(z\), it follows that \term{y} must get close to \(z\) as well.
There is \(\delta'>0\) such that for all \(x\in \myset[]\), in particular the terms in the sequence constructed, with the property that whenever \ball{x}{z}{\delta'}, it follows that \ball{f(x)}{f(z)}{\frac{\epsilon}{2}}.
\seqconverge{N_1}{x}
\seqconverge{N_2}{y}
Choose \(N\equiv \sup\{N_1,N_2\}\) such that if \(n_k\geq N\):
\begin{align*}
|f(x_{n_k})-f(z)|<\frac{\epsilon}{2}\\
|f(z)-f(y_{n_k})|<\frac{\epsilon}{2}
\end{align*}
It follows that:
\begin{align*}
\epsilon&>&|f(z)-f(y_{n_k})|+|f(x_{n_k})-f(z)|\\
&\geq&|[f(z)-f(y_{n_k})]+[f(x_{n_k})-f(z)]|\\
&=&|-f(y_{n_k})+f(x_{n_k})|
\end{align*}
contrary to the assumption that \(|f(x_n)-f(y_n)|%
\nrightarrow0\) for all \(x_n,y_n\).
```

main:

```
\documentclass{minimal}
\usepackage{ifthen}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{style}
\begin{document}
\input{proof}
\end{document}
```