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I know the above quoted statement is correct but I have a confusion in one example.... f(x)=sin x f: (π/4 , 7π/4)--> [-1,1]…… If we take the closed set in the range as the whole set [-1,1] then its inverse image will be the whole domain....(according to me)... i.e. inverse image of a closed set is open....(f is continuous also...)

Please solve my confusion and correct me where I am going wrong....

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2 Answers 2

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The whole domain is both open and closed, so the inverse image is still closed.

To see that the whole domain is closed, consider the domain's complement (the empty set). The empty set is open, hence its complement is closed.

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  • $\begingroup$ The same question posted below.....if domain is closed and inverse image is open then what? $\endgroup$
    – user364168
    Oct 22, 2016 at 7:30
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    $\begingroup$ @user364168: you are thinking of a domain as being a subset of some larger space, but it isn't correct to think of it that way. Take for example, $f(x)=1/x$. It has domain all real numbers except $x=0$. Now, it is natural to think in this setting that the set isn't closed because $0$ is a limit point, but when we talk about the domain, that is our universe...there isn't a $0$ in the domain. Put in other notation, call the domain $X$. Do we ever talk about elements outside of $X$? No, because they have no meaning relative to the function. $\endgroup$
    – Clayton
    Oct 22, 2016 at 7:47
  • $\begingroup$ Sry I might be irritating now... In your example the function is not defined at x=0.... But in my case this is not so.... $\endgroup$
    – user364168
    Oct 22, 2016 at 8:04
  • $\begingroup$ Let me rewrite..... If f(x)= sinx and f:[π/4,7π/4]–>[-1,1] then if I consider inverse image of [-1,1] to be (π/4,7π/4) then it is not wrong...also the above quoted statement is not satisfied...... $\endgroup$
    – user364168
    Oct 22, 2016 at 8:09
  • $\begingroup$ Subspace topology $\endgroup$
    – Clayton
    Oct 22, 2016 at 14:47
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A set can be both open and closed. In particular, in any space $X$, $X$ itself is always both open and closed.

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  • $\begingroup$ Okay....but if I take domain as the closed interval and imverse as the open interval.....then what? Actually in my problem I have not taken domain as the principal value of sin inverse x... $\endgroup$
    – user364168
    Oct 22, 2016 at 7:17
  • $\begingroup$ An open interval is still closed as a subset of itself. When you say "the inverse image of a closed set is closed", you mean as a subset of the domain. $\endgroup$ Oct 22, 2016 at 7:34
  • $\begingroup$ But this tym I took domain as the closed interval [π/4,7π/4]…… $\endgroup$
    – user364168
    Oct 22, 2016 at 7:41
  • $\begingroup$ If the domain is $[\pi/4,7\pi/4]$, then the inverse image of $[-1,1]$ will be $[\pi/4,7\pi/4]$, which is indeed closed in the domain. $\endgroup$ Oct 22, 2016 at 7:52

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