If $x$ is a fixed point of a continuous function $f$, there is an open neighborhood $N$ of $x$ with $f(N)\subseteq N$

Question

Let $\Omega$ be an open subset of a topological space $E$ and $x\in\Omega$ be a fixed point of a continuous function $f:\Omega\to E$.

How can we show that there is an open neighborhood $N$ of $x$ with $f(N)\subseteq N$.

Clearly, by continuity, if $N_2$ is an open neighborhood of $f(x)=x$, there is an open neighborhood $N_1$ of $x$ with $f(N_1)\subseteq N_2$, but it's not immediately clear why we can take $N_1=N_2$.

Answer 1

Consider $\Omega = (-1,1)$, $E = \mathbb{R}$ and $f(x) = 2 x$. Clearly, $x=0$ is a fixed point of $f$ belonging to $\Omega$, but the condition $f(N) \subset N$ cannot be satisfied by any open set $N\subset\Omega$.