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?
Best Answer
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,... .)