1. ## continuous problem

could anyone help me with this problem please?
Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.

So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.

I don't know what kind of open set would have the inverse image is also open.
I don't know how to approach this problem

2. Originally Posted by namelessguy
could anyone help me with this problem please?
Prove that the definition of continuous function f: R to R is equivalent to the property that for any open set A in R, the inverse-image of A is open.

So I know the definition of a continuous function is that if for every x sub zero in R,and for every epsilon greater than 0, there exists delta such that the absolute value of f(x)-f(x_0) < epsilon if the absolute value of x-x_0 < delta.

I don't know what kind of open set would have the inverse image is also open.
I don't know how to approach this problem
Say $f$ is a continous function on $\mathbb{R}$. Let $S$ be an open set we want to show $f^{-1}(S)$ is also open. Note, if $S=\{ \}$ then the proof is complete, so it is safe to say $S$ is non-empty. Now $f^{-1}(S)$ is open if and only if every point is an interior point, so we will show every point is an interior point. Let $x_0\in f^{-1}(S)$ then $f(x_0) \in S$ and so there exists a $\epsilon > 0$ so that all $y$ satisfing $|y-f(x_0)|<\epsilon$ lie in $S$. But continuity there is a $\delta >0$ so that $|x-x_0|< \delta \implies |f(x)-f(x_0)|< \epsilon$. That means by chosing $\delta$ the set of points satisfing $|x-x_0|<\delta$ all lie in $f^{-1}(S)$. Q.E.D.

3. Now say that $f$ is a function show that the inverse-image of every open set is open. We want to show it is continous. Let $x_0 \in \mathbb{R}$. For $\epsilon > 0$ consider the set $|y - f(x_0)|<\epsilon$ (meaning all $y$ that satisfy this inequality). This set, say $S$, is clearly open. So its inverse image $f^{-1}(S)$ is too open. Now $x_0 \in f^{-1}(S)$ trivially. And so by hypothesis there is a $\delta$ such that $|x-x_0|<\delta \in f^{-1}(S)$ (meaning all $x$ that satisfy this inequality). Thus, $|x-x_0|<\delta \implies |f(x)-f(x_0)|<\epsilon$. Thus $f$ is continous at any point $x_0$. Q.E.D.