in our lecture we were told that a function $f$ is said to be continuous $\iff$ $\forall A$ open set in the range of the function it is verified that its inverse image $f^{-1}(A)$ is continuous. We were given to show as an exercise that $f$ is continuous $\implies$ $f^{-1}(A)$ is open for each open set $A$.

Here is my attempt:

Let $y_0$ be an element of the range of the function, and $V$ be an open set that contains $y_0 = f(x_0)$, then by definition of open set, $y_0$ is an inner point of $V$, which means that there exists a positive radius $r>0$ such that $B_r(y_0)\subset V$ ( $B_r(y_0)$ is a ball of radius $r$ with center in $y_0$).

We have that

$\forall\;y_0 = f(x_0)\in V\;\;\exists\;r>0\;\lvert\; B_r(y_0) \subset V$

By applying $f^{-1}$ to it we have that

$\forall\;x_0 \in f^{-1}(V)\;\;\exists\;r>0\;\lvert\; f^{-1}(B_r(y_0)) \subset f^{-1}(V)$

By the fact that $x_0$ is arbitrary we have that $f^{-1}(V)$ is an open set.

Am I correct?

  • 2
    $\begingroup$ See math.stackexchange.com/questions/2975566/… This applies to open sets outside of the image as well. $\endgroup$ – John Douma Nov 2 '18 at 22:01
  • $\begingroup$ @JohnDouma: Thanks, I think that the link is going to be useful. $\endgroup$ – Baffo rasta Nov 2 '18 at 22:04
  • $\begingroup$ Take $f(x)=x$ defined on $[0,1]$, it is sure continuous. Let $V=(0.9,1.1)$; $V$ is open in $\mathbb R$. Now $y_0=1$ is in the range of $f$, and the preimage of $V$ under $f$ is $(0.9,1]$, which is not open in $\mathbb R$ at all ... $\endgroup$ – Michael Hoppe Nov 2 '18 at 23:40
  • $\begingroup$ The definition is that $f^{-1}(V)$ is open in the domain of $f$, which makes all function defined on a discrete set continuous, for example. So you have to be a more careful. $\endgroup$ – Michael Hoppe Nov 2 '18 at 23:47

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