3
$\begingroup$

Suppose $f:X\to Y$ which is a continuous function. Is $f\big(\mbox{int}(S)\big)\subset \mbox{int}\big(f(S)\big)$ where $S\subset X$? I try to disprove it by letting $S=(0,1)$ and $f(x)=1\: \forall x\in S$, but not sure if it really works? Can you explain why it works if it really works? Thx!

  • 0
    The property $f(\text{int}(S))\subseteq \text{int}(f(S))$ is just a characterization of $f:X\to Y$ being an open map. It need not be continuous.2012-12-11

1 Answers 1

0

That inclusion always hold if $f$ is an open map. That is, $f:X\to Y$ is open if the image of every open set in $X$ is open in $Y$. Since $int(S)$ is open, $f(int(S))$ is open so that it is contained in the largest open set in $f(S)$. Therefore, $f(int(S))\subseteq int(f(S))$.