Mathematics Asked on December 25, 2021
Let $Omega$ be an open subset of a topological space $E$ and $xinOmega$ be a fixed point of a continuous function $f:Omegato E$.
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 $N_1=N_2$.
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 $NsubsetOmega$.
Answered by Rigel on December 25, 2021
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