Let $λ ∈ \mathbb I$ and $E: = \{nλ + m: m ∈ \mathbb Z, n ∈ \mathbb N\}$. Prove that given $x ∈ \mathbb R$ and $ε> 0$, then there exists $t ∈ E$ such that $| x − t | <ε$. Hint: Prove that given $N ∈ \mathbb N$ exists $a ∈ E$ such that $| a | < 1/N .$
I tried and tested the suggestion with the loft principle as follows,
Let $N ∈ \mathbb N$ be given.
Nests: The intervals $(0, 1 / N), (1 / N, 2 / N), (2 / N, 3 / N),. . . , ((N − 2) / N, (N − 1) / N), ((N− 1) / N, 1)$. There are $N$ "nests".
Pigeons: $λ− \lfloorλ\rfloor, 2λ− \lfloor2λ\rfloor, 3λ− \lfloor3λ\rfloor,. . . , (N + 1) λ− \lfloor (N + 1) λ\rfloor$. There are $N + 1$ "pigeons".
Suppose that $kλ - \lfloor kλ\rfloor = i / N$, where $1 ≤ k ≤ N + 1, 0 ≤ i ≤ N.$ If we solve for $λ$, it turns out that $λ$ is rational. Contradiction. By the loft principle, there is $1 ≤ i ≤ N$ and at least two different integers $1 ≤ k, l ≤ N + 1$, such that $kλ - \lfloor kλ\rfloor, lλ - \lfloor lλ\rfloor ∈ ((i - 1) / N, i / N)$. It follows that $$1 / N = | i / N - (i - 1) / N | > | (kλ - \lfloor kλ\rfloor) - (lλ - \lfloor l\lambda \rfloor) | = | (k - l) λ + (\lfloor lλ\rfloor - \lfloor kλ\rfloor) |$$ Let $n = k - l$ and $m = \lfloor lλ\rfloor - \lfloor kλ\rfloor$. So $$1 / N> | nλ + m |.$$ If $n> 0$, $a: = nλ + m ∈ E.$
If $n <0$, $$1 / N> | nλ + m | = | (−n) λ + (−m) |,$$ y $a: = (−n) λ + (−m) ∈ E$.
That way the suggestion is tested. Proved the suggestion, how to conclude that there exists $ t \in E $ such that $ | x-t | <\epsilon $.