Does there exist a closed nowhere dense set $E\subseteq [0,\infty)$ so that for every $\epsilon>0$, there is $t>0$ such that $t+i\epsilon\in E$ for all $i\in \mathbb N$?
This is not a homework problem, but a question that arises in my research. I do not know if Baire Category theorem helps me.
My thoughts: I guess this is wrong. We may consider taking a sequence $1/n$ so that there is $t_n$ such that $t_n+i/n\in E$ for all $i$. If $t_n$ is bounded, then $t_{n_k}\to t$ for a subsequence, so for any $\delta>0$, if $K$ is large enough, then $t\in E_\delta-i/n_k$ for all $i$ and for all $k\geq K$, where $E_\delta$ is the $\delta$-neighbourhood of $E$. Then we use the density of $\{i/n_k:i,k\}$ to conclude that $E_\delta-t$ contains a fixed interval for all $\delta$. Letting $\delta\to 0$ we have $E$ contains an interval, since $E$ is closed.
The problem is that $t_n$ may go to infinity. However, since the above holds for all $\epsilon>0$, taking a countable sequence $\epsilon_n>0$ does not fully utilize the assumption.
Yes, this is possible. First, note that it suffices to construct such a set $E$ which has the desired property for all $\epsilon>1$, since then $\bigcup_{n\in\mathbb{N}}n+\frac{1}{n}E$ will work for all $\epsilon>0$ (the scalings by $\frac{1}{n}$ are to get arbitrarily small $\epsilon$, and the translations by $n$ are to make sure the union is still closed).
Now let $(U_n)$ be a countable basis for the topology of $[0,\infty)$. For each $n$, let $V_n$ be some interval $(a_n,b_n)\subseteq U_n$ such that $(b_n+1)(b_n-a_n)<2^{-n}$ (this is obviously always possible, by just fixing $b_n\in U_n$ and choosing $a_n<b_n$ that is sufficiently close to $b_n$). We define $E=[0,\infty)\setminus\bigcup_n V_n$. Then clearly $E$ is closed, and it is nowhere dense because by construction it cannot be dense in any $U_n$.
Now fix any $\epsilon>1$. Let $A_n$ be the set of $t\in[0,\infty)$ such that $t+i\epsilon\in V_n$ for some $i\in\mathbb{N}$. Note that $A_n$ is just a union of translates of $V_n$ by multiples of $-\epsilon$, and in fact there are at most $b_n+1$ such translates which intersect $[0,\infty)$ since $\epsilon>1$. So the measure of $A_n$ is at most $(b_n+1)(b_n-a_n)<2^{-n}$. In particular, this means the measure of the union $\bigcup_n A_n$ is finite, so there exists some $t\in [0,\infty)$ which is not in any $A_n$. This $t$ then satisfies $t+i\epsilon\in E$ for all $i\in\mathbb{N}$.