Sequence that contains subsequences converging to every point in the infinite set $\{1/n: n\in\Bbb{N}\}$ (Abbott p 58 q2.5.3c)

Question

Show that there exists a sequence that contains subsequences converging to every point in the infinite set $\{1/n: n\in\Bbb{N} \}$.

has this property. Notice that there is also a subsequence converging to 0. We shall see that this is unavoidable.

I acquiesce to this example, but I wasn't conscious of it until I read the solution. Aren't there easier examples? Why not choose the original sequence $\{1/n \} \, \forall \, n \in N$ itself?

Answer 1

Subsequences have to be infinite, so the sequence $(1/n)_{n\in \mathbb N}$ has no subsequence converging to say, 1/2. (The sequence given has the subsequence 1/2, 1/2, 1/2,... .)

Answer 2

Take any bijection $\Bbb N\to\Bbb Q\cap(0,1]$. This is a sequence and its image has subsequences that converge to every $1/n$, and in fact, to any number in $[0,1]$.

In case you need a explicit sequence, here is one:

$$\{1,\frac12,\frac13,\frac23,\frac14,\frac34,\frac15,\ldots\}$$

Answer 3

I want to expatiate on apAULt's answer:

Subsequences have to be infinite. So the sequence $(1/n)_{n\in \mathbb N}$ , that I thought would answer this question, founders because has no subsequence converging to say, 1/2. 1/2 appears only once in it.

The solution's sequence in your picture has the subsequence 1/2 (red vertical line), 1/2 (red vertical line , 1/2 (red vertical line), ... So this behaves.