Suppose $f:X \to Y$ and some "not open set" in $X$ is the inverse image of an open set in $Y$. then the function is not continuous as there is an open set whose inverse image is not open. But, intuitively thinking, what is the reason behind this ? why the function is not continuous at the point (say $p$) which is not an interior point. What is the value of $\epsilon$ such that for any neighbourhood around $p$ there exists a point $x$ (this point is nothing but the point which is not an interior point) in that neighbourhood such that the distance between the images of $f(p)$ and $f(x)$ is greater than $\epsilon$.
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Your first sentence is wrong. You can have $f(A) = U$, but $f^{-1}(U) \neq A$. In fact, since $f$ is continuous you must have $f^{-1}(U) = V \supsetneq A$, and $V$ is open in $X$. – Daniel Donnelly Oct 12 '13 at 05:05
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@EnjoysMath: corrected. but what is the answer to my question ? – user96000 Oct 12 '13 at 05:13
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Suppose $U \subset Y$ is open but $f^{-1}(U)$ is not open, so there is a point $p \in f^{-1}(U)$ which is not an interior point of $f^{-1}(U)$. Now we have $f(p) \in U$, and $U$ is open, so there is an $\epsilon$ such that the ball $B(f(p), \epsilon) \subset U$. This is the $\epsilon$ you requested: since $p$ is not an interior point of $f^{-1}(U)$, every neighborhood of $p$ contains a point $x$ with $x \notin f^{-1}(U)$. That is, $f(x) \notin U$, so in particular, $d_Y(f(x), f(p)) \ge \epsilon$.
Nate Eldredge
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Intuitively if $f^{-1}((a,b))$ isn't open, for example if $f^{-1}((a,b))=[c,d]$, then for any $x$ arbitrarily close to $c$ but less than $c$, $f(x)\notin (a,b)$. Hence $f(x)$, no matter how close $x$ it is to $c$, cannot get arbitrarily close to $f(c)$ (since if $f(c)=a+\epsilon$, then $f(x)$ will always be at least $\epsilon$ away from $f(c)$). So $f$ isn't continuous.
JLA
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