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!
Is $f\big(\mbox{int}(S)\big)\subset \mbox{int}\big(f(S)\big)$
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general-topology
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1Your example works fine. An even simpler version of the same idea: let $f(x)=0$ for every $x\in\Bbb R$. Then $f$ is a continuous function from $\Bbb R$ to $\Bbb R$, but $$f[\operatorname{int}\Bbb R]=f[\Bbb R]=\{0\}\nsubseteq\varnothing=\operatorname{int}\{0\}\;.$$ – 2012-12-11
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0@BrianM.Scott what about $Y=(\mathbb{R},d)$ where d is a discrete metric? – 2012-12-11
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0@BrianM.Scott Is int{0} still empty? – 2012-12-11
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0In that case $\operatorname{int}f[S]=f[S]\supseteq f[\operatorname{int}S]$ for all $S\subseteq\Bbb R$; in particular, the interior of $\{0\}$ is $\{0\}$ itself. – 2012-12-11
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0@BrianM.Scott ya so do we need to specifies the metric defined on $\mathbb{R}$ when solving this question? – 2012-12-11
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1The inclusion doesn’t always hold, so if you want to use it in a particular instance, you have to show that it does hold in that instance. That will in general depend on all three of $X$, $Y$, and $f$. – 2012-12-11
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0@BrianM.Scott From your comment is the statement holds in $\mathbb{R}$ with discrete metric? – 2012-12-11
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0It will always hold if $Y$ has the discrete metric. – 2012-12-11
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0@BrianM.Scott I mean at the step when we claim that {0}=$\emptyset$ should we state any reason like what is the metric or what? because someone may argue why. – 2012-12-11
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0I’m not sure what you mean. In the usual (Euclidean) metric on $\Bbb R$, $\operatorname{int}\{0\}=\varnothing$; this is obvious, since $\{0\}$ does not contain any non-empty open interval. In the discrete metric, on the other hand, it’s equally obvious that $\operatorname{int}\{0\}=\{0\}$. You can’t make a general argument: you can only investigate particular metrics on $X$ and $Y$ and particular functions $f$. – 2012-12-11
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0The 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
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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, $f(int(S))\subseteq int(f(S))$.