I'm reading topology book of munkres. In exercise 8 chapter 17 solution, it is said that, if $f$ is a continuous function from topology $X$ to ordered topology $Y$, then $S = \{ x\mid f(x) > a\}$ is open without explaining anything. I know in ordered topology, the ray $(a, +\infty)$ is open. But $f$ is not surjective, so I don't think that $S = (a, +\infty)$. So all in all, why is $S$ open. Do I miss something?
Can any one explain for me why it is an open set?
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general-topology
1 Answers
10
The inverse-image of any open set under any continuous function is open. That should be a proposition stated explicitly in the textbook, where continuity is first introduced.
If $f:X\to Y$ and $A\subseteq Y$, then the inverse-image of $A$ under $f$ is the set $\{x\in X : f(x) \in A\}$.