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I came across this question in my test, let $f:R->R$ be defined by $f(t)=t^2$ and let $U$ be any non-empty open subset of $R$, Then

 1. f(U) is open 
 2. f^-1(U) is open 
 3. f(U) is closed 
 4. f^-1(U) is closed 

The option I thought was correct is (1.) but (2.) is correct I don't know why? I think option 1 is correct because U is an open subset of R hence the function will also map to a set which is open. So please tell me how option 2 is correct and which concept I don't know about or not getting to answer this question. I would like a hint.

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1 Answer 1

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Hint to help you consider the potential issue:

If $U =(-1,4)$ then what might $f(U)$ and $f^{-1}(U)$ represent?

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  • $\begingroup$ $f^{-1}(U)$ is not defined for -ve values of U(-1,4) as inverse of f is $\sqrt{t}$ no? thanks for help $\endgroup$
    – Onix
    Feb 3, 2016 at 7:05
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    $\begingroup$ @Abomm: If $U=(-1,4)$ then $f(U)=[0,16)$ while $f^{-1}(U) = (-2,2)$ - it is the set of $x$ such that $f(x) \in U$ - so assertions 1, 3 and 4 would be false in this case, while assertion 2 would be correct (and in fact assertion 2 is correct for all non-empty open subsets of $\mathbb{R}$) $\endgroup$
    – Henry
    Feb 3, 2016 at 14:26

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