Range of a function on subintervals

Question

Prove or disprove the existence of a function $f:[0,1] \rightarrow[0,1]$ with the following property:

for any interval $\,(a,b)\subset[0,1]\,$ with $\,a\!<\!b,f\big((a,b)\big)\!=\![0,1]\,.$

It turns out that such a function does not exist. The proof is a consequence of the Baire Category Theorem.

Proof by Contradiction: Assume that such a function $f$ does exist.

Consider the set $A=\{x \in[0,1]: f(x)=1\}$. The set $A$ is non-empty because $f((0,1))=[0,1]$ by assumption. That is, there must exist some $x \in(0,1)$ such that $f(x)=1$.

Now, let's use the Baire Category Theorem. For each $x \in A$, consider the preimage of an open interval $(1-\epsilon, 1]$ under $f$, where $0<\epsilon<1$. The preimage $f^{-1}((1-\epsilon, 1])$ must contain an interval around $x$ because $f(x)=1$. Let this interval be $\left(a_x, b_x\right)$.

The set $A$ can thus be expressed as a countable union of open intervals $\left(a_x, b_x\right)$ around each $x \in A$. According to the Baire Category Theorem, a complete metric space cannot be represented as a countable union of nowhere-dense sets. However, each $\left(a_x, b_x\right)$ has an image under $f$ that is the entire $[0,1]$, making it nowhere-dense in $[0,1]$.

Thus, our assumption that such a function $f$ exists leads to a contradiction with the Baire Category Theorem. Therefore, such a function $f$ does not exist.

Note: The proof utilizes the Baire Category Theorem and the completeness of $[0,1]$ in the metric of real numbers.

Can anyone check whether there are no mistakes in the above solution?

There must be something wrong because such functions exist, see for example https://math.stackexchange.com/q/790927/42969 or https://math.stackexchange.com/q/3845483/42969. — Martin R, Sep 01 '23 at 11:40
You said nothing about continuity of $f$ but used it freely in the proof. — Kavi Rama Murthy, Sep 01 '23 at 11:40
Why do you need Baire? Aren’t you done after you have that $f((a_x,b_x))\subset (1-\epsilon,1]$ and so isn’t $[0,1]$. Also, you use that f is continuous… — Eric, Sep 01 '23 at 12:12
@geetha290krm I am sorry about the continuity part, you are right! — Snowball, Sep 01 '23 at 12:51
@MartinR Do you have specific functions for the stated problem, not the ones you suggested?! — Snowball, Sep 01 '23 at 12:52
@Snowball, the solution you are looking for is here. You should remove your question because it is a duplicate. — Angelo, Sep 01 '23 at 17:48
I don't think your question is a duplicate, but it needs to be corrected by adding the continuity hypothesis. — Anne Bauval, Sep 01 '23 at 19:09

score -2 · Accepted Answer · answered Sep 01 '23 at 13:52

Are you sure? I think your claim is there exists no continuous function with given conditions. Because I can easily give a non-continuous function of the required kind... I will just choose a closed interval say $[a_1,b_1] \subset (a,b)$ and it is now a easy job to give a function from $[a_1,b_1] \to [0,1]$ now for $x \in (a,b)$ such that $x<a_1, f(a)=0 $ and similarly $x \in (a,b)$ such that $x>b_1, f(x)=1$

Range of a function on subintervals

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