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Suppose f: R->R is a continuous function. Prove that if U is an open set in R then f-1(U) is also an open set in R

For this question, what I am thinking is that we should start with assuming that f inverse is not an open set and arrive at some contradiction? But I am not sure what the contradictions are here..

ilovebulbasaur
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Joseph Rock
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1 Answers1

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Here's a direct proof. Let $x\in f^{-1}(U)$. By definition, there exists $y\in U$ so that $f(x)=y$. Since $U$ is open and that $y\in U$, there exists $\epsilon>0$ such that $B_{\epsilon}(y)\subset U$. By continuity of $f$, there exists $\delta>0$ such that $x'\in B_{\delta}(x)$ implies $f(x')\in B_{\epsilon}(y)$. Hence $B_{\delta}(x)\subset f^{-1}(U)$, i.e. there is an open ball of radius $\delta>0$ centered at $x$ completely contained in $f^{-1}(U)$. Since $x\in f^{-1}(U)$ was arbitrary, $f^{-1}(U)$ is open. $\boxed{}$

Edit: Here is a proof by contradiction. Suppose for contradiction that $f^{-1}(U)$ is not open. Then there exists $x\in f^{-1}(U)$ such that $B_{\delta}(x)\not\subset f^{-1}(U)$ for any $\delta>0$. On the other hand, $U$ is open, so there exists $\epsilon>0$ so that $B_{\epsilon}(f(x))\subset U$, then... (the ideas are similar to the direct proof, I will leave it to you to write a proof).

ilovebulbasaur
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  • Why do you use here instead of say F? Does it have a special meaning – Joseph Rock May 10 '20 at 23:06
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    Which $F$ are you referring to? $B_{\epsilon}(x)$ is one of the standard notations for "the open ball of radius ${\epsilon}$ centered at a point $x$". – ilovebulbasaur May 10 '20 at 23:08
  • Ok, but how we know that (y) here is a subset of U? – Joseph Rock May 11 '20 at 02:12
  • This follows from the definition of an open set: every point $y$ of an open set has a open ball centered around $y$ such that the open ball is completely contained in the open set. If you haven't taken a class or studied a text where this was mentioned explicitly, then you might need to find some online resources to catch up. – ilovebulbasaur May 11 '20 at 02:17
  • on the next step, your introduced x' , could you explained why we need to do that too? sorry for the trouble.. – Joseph Rock May 11 '20 at 02:27
  • Sure. $x'$ is just another variable we introduce to represent any number that is within $\delta$ of $x$. – ilovebulbasaur May 11 '20 at 02:35
  • I noticed than under the B , there is sometimes delta and sometimes Epsilon, could you elaborate on the difference between Ball delta and Ball Epsilon? thanks ilovebulbasaur – Joseph Rock May 11 '20 at 02:59
  • Yeah, the difference is that one ball has radius $\epsilon$, while the other has radius $\delta$. Notice that a "ball" on the real number line is just an interval. So for example, $B_{\delta}(x)=(x-\delta,x+\delta)$. Btw, apparently MSE frowns upon extended discussions in comments, so, feel free to message if you still need clarification. – ilovebulbasaur May 11 '20 at 03:04
  • Let us continue this discussion in chat. – Joseph Rock May 11 '20 at 03:12
  • just one last question on here, that x is an open set in R but how do we know that f(x)=y , y is in an open set in R too? – Joseph Rock May 11 '20 at 03:21