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Let $ X \subset \mathbb{R}$ be a non-empty, open set and let $f: X \rightarrow \mathbb{R}$ be a continuous function.

Prove that the inverse image of an open set is open

Here is the proof by DiegoMath.

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.

I do not understand why showing that $B(x_0,\delta)\subset f^{-1}(M)$, is sufficient to show that $f^{-1}(M)$ is open . Moreover, is it really true that $f(x)\in M$ for any x?

Michael Burr
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The definition of a set $A$ being open (in a metric space) is that for every point $a\in A$, there is some open ball around $a$ that is contained in $A$. This is what the $B(x_0,\delta)\subseteq f^{-1}(M)$ is checking. $B(x_0,\delta)$ is an open ball and it is inside $f^{-1}(M)$.

As for why $f(x)\in M$, by construction, $x\in B(x_0,\delta)$, so by continuity, $f(x)\in B(f(x_0),\varepsilon)$ and since $\varepsilon\leq r$, $f(x)\in B(f(x_0),\varepsilon)\subseteq B(f(x_0),r)$, so $f(x)\in B(f(x_0),r)$. However, $B(f(x_0),r)\subseteq M$ by assumption, so $f(x)\in M$.

Michael Burr
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