# Need to show that if a function $f$ is continuous then the inverse image of every open set is open, am I correct?

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?

• See math.stackexchange.com/questions/2975566/… This applies to open sets outside of the image as well. – John Douma Nov 2 '18 at 22:01
• @JohnDouma: Thanks, I think that the link is going to be useful. – Baffo rasta Nov 2 '18 at 22:04
• 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 ... – Michael Hoppe Nov 2 '18 at 23:40
• 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. – Michael Hoppe Nov 2 '18 at 23:47