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Category:Computer-aided design software
Category:Computer-aided design software for WindowsQ:

About the existence of totally discontinuous real valued function

Q: Does there exist a function $f:\mathbb R\to\mathbb R$ such that $\;\;\;f^{ -1}(0) eq\mathbb R$?
The answer is negative but I’m not able to provide a rigorous explanation.
I know that there are functions with finitely many discontinuities.
(For example, $f(x)=|x|-2\,\mathbb {sign}(x)$ is a function with only two discontinuities).
If the target function has finitely many discontinuities, then it must be continuous (even surjective).
It is very easy to give a counter example to the existence of a function with infinitely many discontinuities.
(For example, $f(x)=x^2$ is such a function)
The explanation I have in mind is:
Let $(x_n)$ be a sequence of the reals such that $x_n\to x$ and $f(x_n)\to y$ as $n\to\infty$.
If $f^{ -1}(y)=\mathbb R$, then $x=f(y)\in f^{ -1}(y)$, contradiction.

A:

No.
Say $f$ is a function such that $f^{ -1}(0) = \mathbb R$. Then since there is an element of $f^{ -1}(0)$ that is irrational, there exists an interval $I$ such that $0 otin f(I)$.
So, choose $x_0 \in I$. Then, since $f$ is surjective, there exists $y_0 \in f(I)$. But since $f(I) \subseteq \mathbb R \setminus \{0\}$, we have \$|y_0|

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