Let $I, J\subseteq \Bbb R$ be open intervals, and let $f:I \to\Bbb R$ be a function.

Suppose that $f$ is continuous. Let $x \in f^{-1} (J)$.

Prove that there is an open interval $K\subseteq \Bbb R$ such that $x \in K \cap I \subseteq f^{-1} (J)$

What I have tried so far:

Since $f$ is continuous it follows that there exists a point $c\in\Bbb R$ such that $\lim_{x→c}f(x)=f(c)$. Since $x\in f^{-1}(J)$, then there exists $y \in J$ such that $f(c)=y$.

Also, $x$ is continuous for any $\epsilon>0$, it also follows that $|f(x)-f(c)|<\epsilon$.

It then follows that when $|x-y|<\delta$, $c$ must be an element of $f^{-1} (J)$.

It can then be seen that $–\delta<|x-c|<\delta-\epsilon<|f(x)-f(c)|<\epsilon$, which can be manipulated to show that $x$ falls somewhere between $c-δ$ and $c+δ$.

If we then let $x\in(c-\delta,c+\delta)=k\cap I\subseteq f^{-1}(J)$, and therefore $x$ must be a subset of $k\cap I\subseteq f^{-1}(J)$.

Therefore: $x\subseteq k∩I\subseteq f^{-1}(J)$

My instructors reply was:

Try expounding on this:
If $x$ is in $f^{-1}(J)$ then $f(x) = y$ for some $y$ in $J$ . Since $J$ is an open interval, there exists some $\epsilon$ such that the interval $(y - \epsilon, y + \epsilon)$ is completely contained in $J$.
Now, since $f$ is continuous at $x$ we know that given this $\epsilon$, there exists a $\delta$ such that $|z - x| < \delta$ implies $|f(z) - f(x)| = |f(z) - y| < \epsilon$. Or, in other words, $|z - x| < \delta$ implies $f(z)$ is an element of $(y - \epsilon, y + \epsilon)$. How does this help us?


1 Answer 1


You suppose $f: I \rightarrow \mathbb{R}$, and let $x \in I$ be such that $f(x) \in J$. We must find an open interval $I_0 \subset I$ such that $x \in I_0 \subset f^{-1}(J)$. (This equivalent to your claim: the intersection of two open intervals is again an open interval, if not empty).

But $J$ is an interval, so there is some small interval $J_0:= (f(x)-\epsilon, f(x)+ \epsilon) \subset J$. And $f$ is continuous. So there is some small $\delta$ such that $|y-x|< \delta \Rightarrow f(y) \in J_0$. Let $I_0 := (x-\delta, x + \delta)$. (Note: without loss of generality, $I_0 \subset I$; if not, just pick a smaller $\delta$).

We know that $f(I_0) \subset J_0 \subset J$. Also, $x \in I_0$. So $x \in I_0 \subset f^{-1}(J_0) \subset f^{-1}(J)$.

If you want you can let $K:= I_0$.


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