Give an example of Tietze extension theorem? Give  an example of  Tietze extension theorem?
I know  the theorem definition, but i could not able to  find the  example
Pliz help me
 A: Let $\ell^\infty$ be the space of all bounded sequences of complex numbers, endowed with the metric$$d_\infty\bigl((x_n)_{n\in\mathbb N},(y_n)_{n\in\mathbb N}\bigr)=\sup\bigl\{\lvert x_n-y_n\vert\,|\,n\in\mathbb{N}\bigr\}.$$Let $C$ be the subspace of $\ell^\infty$ which consists of the convergent sequences. The map$$\begin{array}{rccc}L\colon&C&\longrightarrow&\mathbb C\\&(x_n)_{n\in\mathbb N}&\mapsto&\displaystyle\lim_{n\to\infty}x_n\end{array}$$is continuous and $C$ is a closed subspace of $\ell^\infty$. Therefore, according to the Tietze extension theorem, we can extend $L$ to a continuous function from $\ell^\infty$ into $\mathbb C$.
A: One dimensional example:
Consider $[1,2]$ as a subspace of $\Bbb{R}$ with the induced standard topology and consider $$f:[1,2] \ni x \mapsto \frac{1}{x } \in \Bbb{R}$$  Then $f$ is continuous and $[1,2]$ is closed and $\Bbb{R}$ is normal so by Tietze extension theorem, $f$ can be extended to $g$ on $\Bbb{R}$ so that $g$ is continuous and $f(x)=g(x)$ for all $x \in [1,2]$.  One such extension is  $$g(x)=\begin{cases} f(x)=\frac{1}{x} & \text{if} \;x \in [1,2]\\\\1 & \text{if} \;x \in (-\infty,1)\\\\\frac{1}{2} & \text{if}\;x\in (2, \infty)\end{cases}$$
A: An application: Let $S$ be a closed bounded subset of $\Bbb R^n$ ( with $n\in \Bbb N$) and let $f:S\to \Bbb R$ be continuous. We can extend  $f$ to a continuous $f:\Bbb R^n\to \Bbb R$  such that $\{x\in \Bbb R^n: f(x)\ne 0\}$ is bounded. 
Let $B$ be a bounded open ball of positive radius, such that $S\subset B.$ Observe that if we let $f(x)=0$ for all $x\in \Bbb R^n\setminus B$ then $f$ is continuous on the closed set $B^*=S\cup (\Bbb R^n\setminus B).$ By the Tietze Extension Theorem, the domain of ( continuous) $f$ can be extended from $B^*$ to all of $\Bbb R^n.$
Remark: The intermediate step, that $f:B^*\to \Bbb R$ is continuous, can be done by showing that if $Y$ is any closed subset of $\Bbb R$ then $(f^{-1}Y) \cap B^*$ is closed in $\Bbb R^n$ so, a fortiori, it is closed in $B^*$.
