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I understand that $f:X\rightarrow Y$ is continuous if the pre-image of an open in $Y$ is open in $X$.

Take $f(x)=x²$ as a function $\mathbb{R}\rightarrow \mathbb{R}$.

Then $f^{-1}((-1,1))=[0,1)$ so the square function is not continuous(?).

However if I consider $f$ as a function $\mathbb{R}\rightarrow \mathbb{R^+_0}$ then the conclusion seems different.

Does it mean that continuity depends on the range of the function?

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You've just got your images and inverse images muddled. $f^{-1}((-1,1)) = f^{-1}([0,1)) = (-1,1)$, whereas $f((-1,1)) = [0,1)$. So the inverse image of $(-1,1)$ is an open set in $\mathbb{R}$, as you'd expect. Easy mistake to make! –  HTFB Nov 22 '12 at 17:39
    
Yes of course, that was a silly mistake...thanks! –  John Nov 22 '12 at 17:43
    
@HTFB has answered your question (and should perhaps make it an answer). Not to confuse you , but note that if you defined the domain as $[0,\infty)$, ie, $f:[0,\infty) \to \mathbb{R}$, then $f^{-1} (-1,1) = [0,1)$. However, $[0,1)$ is open in $[0,\infty)$. –  copper.hat Nov 22 '12 at 18:39

1 Answer 1

You've just got your images and inverse images muddled. $f^{−1}(\enspace(−1,1)\enspace)=f^{−1}(\enspace[0,1)\enspace)=(−1,1)$, whereas $f(\enspace(−1,1)\enspace)=[0,1)$. So the inverse image of $(−1,1)$ is an open set in $\mathbb{R}$, as you'd expect. Easy mistake to make!

(Now in answer format, by popular demand.)

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