Let $f$ be periodic with a continuous image and $a_n = cn$ for some $c > 0$. When is $\{f(a_n)\}$ dense in the image of $f$?

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 $$f$$?

• What is a "continuous image"?
– abx
Commented Feb 21, 2020 at 19:34
• A non-countable subset of $\mathbb{R}$, like an interval. A better word escapes me, other than "non-countable subset of $\mathbb{R}$". Commented Feb 22, 2020 at 0:02
• 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 $f(\mathbf{R}/\mathbf{Z})$?
– YCor
Commented Feb 22, 2020 at 0:28
• 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...)
– YCor
Commented Feb 22, 2020 at 0:32