Let $f:\mathbb{R}\to\mathbb{R}$ be a periodic function with period $T$ and continuous everywhere except perhaps on a countable set, and have an image that's an uncountable subset of $\mathbb{R}$. Let $a_n = cn$ for some $c > 0$. When is the subset $\{f(a_n)\}_{n \in \mathbb{N}}$ dense in the image of $f$?
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2$\begingroup$ What is a "continuous image"? $\endgroup$– abxCommented Feb 21, 2020 at 19:34

$\begingroup$ A noncountable subset of $\mathbb{R}$, like an interval. A better word escapes me, other than "noncountable subset of $\mathbb{R}$". $\endgroup$– cgmilCommented Feb 22, 2020 at 0:02

$\begingroup$ I'd reformulate as follows: let $f:\mathbf{R}/\mathbf{Z}\to\mathbf{R}$ be a function whose domain of continuity has a countable complement. Let $c$ be irrational. Is $f(cn)_{n\ge 0}$ dense in $f(\mathbf{R}/\mathbf{Z})$? $\endgroup$– YCorCommented Feb 22, 2020 at 0:28

$\begingroup$ Quite clearly there are counterexamples and you should probably rather ask whether the closure of your subset has countable complement in the closure of $\mathrm{Im}(f)$ (indeed, modify a bounded continuous function on a closed countable subset disjoint from the given arithmetic sequence...) $\endgroup$– YCorCommented Feb 22, 2020 at 0:32
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