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Let $f : A \mapsto \mathbf{R}$ with $A \subseteq \mathbf{R}$ and $f(A)$ an interval, and suppose $f$ is monotonic. Does it follow that $f$ must be continuous? If I restrict $A$ to be an interval, then the proof is not too difficult. What about arbitrary subsets of the reals?

Maxis Jaisi
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    If $f$ is a monotonic function with any domain $A \subseteq \mathbb R$, the only type of discontinuity it can have is a jump discontinuity. (See e.g. here.) If it has a jump discontinuity, then its image can't be an interval. –  Oct 23 '19 at 06:19

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Perhaps I am understanding the question wrongly, but here's my take using a simple counterexample.

Suppose $A \subseteq \mathbb{R}$ for any arbitrary $A$. Without loss of generality, let's assume $f$ is monotonically increasing. So in other words $\forall x, x+y \in A \quad x+y > x \Rightarrow f(x+y) \geq f(x)$.

Trivially, set $f(x) = \begin{cases} 1, \text{ if } x \in A \\ 0, \text {otherwise} \end{cases}$. So $f(A)$ is monotonic, and maps to the constant interval $1$.

Therefore, this is a counterexample.

zd_
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  • This is a counterexample to what? In the question it is specified that the domain of $f$ is $A$. So in your example $f$ is identical to the constant function $1$, which is continuous. –  Dec 22 '19 at 08:27
  • ok i guess i misunderstood the question. I understood it as he was asking if $f$ is continuous in general over $\mathbb{R}$. But even if it isn't, a similar example can be constructed, eg, let $f$ take on two possible values (i.e. a step function), by further partitioning $A$ into 2 disjoint sets $A_1, A_2$ such that $\forall a \in A_1, b \in A_2 \quad a < b$. – zd_ Dec 22 '19 at 13:21
  • Then $f(A)$ is not an interval –  Dec 22 '19 at 13:28
  • $f(A):\rightarrow[c,k]$ for the step values $c,k$. – zd_ Dec 22 '19 at 13:44