# Inverse is not connected

Let $$f:[a, b] \rightarrow[a, b]$$ be continuous. Provide a counterexample for the following statement:

The inverse image $$f^{-1}([c, d])$$ is connected for any $$[c, d] \subset[a, b]$$.

One of the ideas that I could think of is to give an example for which inverse image $$f^{-1}([c, d])$$ would be the union of two closed intervals, concluding the disconnectedness of it. However, after trying a lot of examples, I could not come up with a solid example for this case.

Note: The values $$a,b$$ are fixed.

• Could you please exemplify what you mean by preimage? More details would be strongly appreciated! Apr 19, 2021 at 22:53
• Does anyone have ideas regarding this problem? Apr 19, 2021 at 23:05

Let $$f:[-1,1]\to[-1,1]$$ be given by $$f(x)=x^2$$. Now $$[1/4,1]\subset[-1,1]$$ is a connected subset, and $$f^{-1}([1/4,1])=[-1,\frac{-1}{2}]\cup[\frac{1}{2},1]$$ which is a disconnected subset of $$[-1,1]$$.

The same idea here works for any $$[a,b]$$ by shifting/scaling $$f$$ as desired (as noted in the comments).

• what do you mean by that Apr 19, 2021 at 22:56
• I didn't "modify" anything though. This is just a specific example for some $a,b$ Apr 19, 2021 at 22:58
• If you really want to prove it for any a, b rather than some arbitrary a, b, then just take the specific counterexample here and adjust it for the specific values you want (i.e. shift and scale the parabola until it fits in the required interval in both directions). Apr 19, 2021 at 23:09
• @Snowflake: the adjustments to Michael's answer to replace the open interval $(0, 1)$ by a closed interval are trivial. Apr 19, 2021 at 23:14
• @Snowflake: you have been given all the mathematical input you need to solve your question. Please stop your pointless bleating. Apr 19, 2021 at 23:20

Let $$f: [-2\pi, 2\pi] \to [-2\pi, 2\pi]$$, $$f(x)=\sin (x)$$.

Notice that $$[0,1] \subset [-2\pi, 2\pi]$$, try to compute $$f^{-1}([0,1])$$ to illustrate that it is a counterexample.

• How about general domain $[a,b]$? Deeply appreciate your counterexample, but it should be applicable to all $[a,b]$ though Apr 19, 2021 at 23:13
• First, first of all, this is not what is being asked in the quesiton. But if you insist, if $a < b$, try to draw a square, $[a, b] \times [a,b]$, try to sketch such a graph. Apr 20, 2021 at 2:38