I was searching for a function which is open but not continuous and I got the following function:
$f:[0,1]\to [0,1]$ \begin{equation} f(x)= \begin{cases} 2x& \text{$if\ 0\leq x \leq \frac{1}{2}$}\\ x& \text{$if\ \frac{1}{2}< x \leq 1$} \end{cases} \end{equation} Now I want to know how can we prove the function is open? I tried as:
Suppose $A$ is open in $[0.1]$
$\implies A=[0,1]\cap O $,for open set $O$ in $\mathbb{R}$ Now I want to show $f(A)=f([0,1]\cap O)$ is open in $[0,1]$. How should i proceed to show $f$ is a open map?? Thanks in advance!
Take $a,b\in(0,1)$ with $a<b$. Then$$f\bigl((a,b)\bigr)=\begin{cases}(2a,2b)&\text{ if }b\leqslant\frac12\\(2a,1]\cup\left(\frac12,b\right)&\text{ if }a\leqslant\frac12\text{ and }b\geqslant\frac12\\(a,b)&\text{ if }a>\frac12.\end{cases}$$So, $f\bigl((a,b)\bigr)$ is an open subset of $[0,1]$. Nw, do the same thing for $f\bigl([0,1)\bigr)$ and for $f\bigl((a,1]\bigr)$. Since every open subset of $[0,1]$ is an union of these three types of sets, you're done.