Prove that the inverse image of an open set is open

Question

Let $ X \subset \mathbb{R}$ be a non-empty, open set and let $f: X \rightarrow \mathbb{R}$ be a continuous function. Show that the inverse image of an open set is open under f, i.e. show:

If $M \subset \mathbb{R}$ is open, then $f^{-1}(M)$ is open as well.

I think that this should basically follow from the definitions, but I'm still having some troubles with the proof.

Suppose $M \subset \mathbb{R}$ is open, then it follows that $\forall z \in M, \exists r > 0: K_r(z) \subseteq M$

Since f is continuous it follows for each $x_0 \in X$ that for each $\epsilon > 0 ,\exists \delta > 0 : |f(x)-f(x_0)|< \epsilon, \forall x \in X: |x-x_0|<\delta$

Now I need to show that $f^{-1}(M)$ is open, so that for each $z \in M, \exists r>0: K_r(f^{-1}(z_0)) \subseteq f^{-1}(M)$

I'm not sure how to prove this. I guess since $z = f(x)$ and since M is open we have that $\exists r>0:\{z:|z-z_0|<r\}\subseteq M$. And since f is continuous: $|f(x)-f(x_0)| = |z-z_0|< \epsilon$ for all $|x-x_0|<\delta$, but since $|x-x_0|=|f^{-1}(z)-f^{-1}(z_0)|<\delta$ it somehow follows from that that $f^{-1}(M)$ is open?!

Which is just horrible and probably completely wrong. But I'm really confused from all the definitions right now and need some help please.

Answer 1

Take $x_0\in f^{-1}(M)$, then $f(x_0)\in M$.

Since $M$ is open, there exists $r>0$ such that $B(f(x_0),r)\subset M$.

Now, choose any $0<\varepsilon\leq r$, since $f$ is continuous, there exists $\delta>0$ such that $f(x)\in B(f(x_0),\epsilon)$ whenever $x\in B(x_0,\delta)$.

Note that if $x\in B(x_0,\delta)$, then $f(x)\in B(f(x_0),\varepsilon)\subset M$, thus $f(x)\in M$, moreover, $x\in f^{-1}(M)$, thus $B(x_0,\delta)\subset f^{-1}(M)$, then $f^{-1}(M)$ is open.

Answer 2

Suppose $f:X \to \mathbb{R}$ is a continuous function and $M \subset \mathbb{R}$ is an open set. Let $x_0 \in f^{-1}(M)$, then $f(x_0) \in M$, which is open, so $\exists \epsilon > 0$ such that $B_{\epsilon}(f(x_0)) \subseteq M$. As $f$ is continuous, $\exists \delta_{\epsilon} > 0$ so that $|x-x_0| < \delta_{\epsilon} \Rightarrow |f(x)-f(x_0)| < \epsilon$. Claim $B_{\delta_{\epsilon}}(x_0) \subseteq f^{-1}(M)$. Indeed, let $p \in B_{\delta_{\epsilon}}(x_0)$, then $|p-x_0| < \delta_{\epsilon} \Rightarrow |f(p)-f(x_0)| < \epsilon \iff f(p) \in B_{\epsilon}(f(x_0)) \subseteq M$, so $f(p) \in M \iff p \in f^{-1}(M)$ $\\$ Thus $B_{\delta_{\epsilon}}(x_0) \subseteq f^{-1}(M)$, namely $f^{-1}(M)$ is open

Answer 3

Consider a point $f(z)$ in $U$. For clarity, the theorem assumes $U$ is open.

Since all the points in U are mapped to the set $f^{-1}(U)$, now consider the equivalent point $z$ in $f^{-1}(U)$.

Remember, we know that $f(z)$ is in $U$. All points in $U$ have a ball around them of at least $\epsilon$. Since $f$ is continuous, there must exist a $\delta>0$ such that $z$ is in a ball of radius $\delta$, and that all points in that ball are mapped by $f$ to $U$, in fact, they end up within $\epsilon$ of $f(z)$. This is true for all $z$ such that $f(z)$ is in $U$. The fact that a ball exists of non-zero radius around each point $z$ in some set $X \subset \mathbb{R}$ is equivalent to that set being open. So it is done.

Answer 4

More generally (but no fear: this is nothing more than a synthetic rewriting of DiegoMath's answer), let $f:X\to Y$ be a continuous map between two topological spaces, i.e. $$\forall x\in X\quad\forall W\text{ neighborhood of }f(x)\quad\exists V\text{ neighborhood of }x\quad f(V)\subset W.$$ (here, implicitely, $W\subset Y$ and $V\subset X$).

Let now $M$ be an open subset of $Y,$ we must prove that $N:=f^{-1}(M)$ is open in $X,$ i.e. that $N$ is a neighborhood of each of its points.

Let $x\in N.$ Since $M$ is a neighborhood of $f(x),$ there exists neighborhood $V$ of $x$ such that $f(V)\subset M,$ i.e. $V\subset N,$ hence $N$ itself is a neighborhood of $x,$ q.e.d.