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Are there any functions $f:\Bbb R\to \Bbb R$ with the following property:

For any $x_0\in \Bbb R$, any $\delta >0$ and any $\epsilon>0$ there is an $x$ with $|x-x_0|<\delta$ such that $|f(x)-f(x_0)|>\epsilon$.

In other words, do functions where the usual $\epsilon$-$\delta$ fails for all $x$ and $\epsilon$ exist at all?

While a proof of nonexistence or a constructive proof of existence would be preferable, something like a non-constructive proof using the axiom of choice is perfectly fine.

Arthur
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2 Answers2

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Yes split the reals into countably many disjoint dense sets $A_n$ and let your function take value $n$ on $A_n$.

Mirko
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  • Adhvaitha just posted a more constructive answer, using an enumeration of the rationals. – Mirko Nov 28 '14 at 13:55
  • @MirkoSwirko this is not as non-constructive as it might seem. Let $A_n$ for $n\geq 1$ be the set of rationals of the form $\frac m{2^n}$ with $m$ odd, and $A_0$ be all other numbers. Then this gives the same function as the "other candidate" given by Adhvaitha. – Arthur Nov 28 '14 at 14:00
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Let $f:[0,1] \mapsto \mathbb{R}$. Let $(q_n)$ be some enumeration of rationals. Define $f$ as \begin{align} f(x) & = \begin{cases} n & \text{ if }x = q_n\\ 0 & \text{ otherwise} \end{cases} \end{align}

Another candidate is \begin{align} f(x) & = \begin{cases} n & \text{ if }x = m/2^n \text{ in simplest form}\\ 0 & \text{ otherwise} \end{cases} \end{align}

Adhvaitha
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