Does the set given by $\{(\frac 1n)\}_{n=1}^\infty$ include $0$?

Question

Is there some sort of consensus on whether or not $$0 \in \{(\frac 1n)\}_{n=1}^\infty?$$

Answer 1

Yes. It's not there. The only elements of that set are $\frac1n$ for $n=1,2,3,\ldots$. Since $0$ is not of the form $\frac1n$, it's not there.

Answer 2

No. $0$ will be in the closure of the set if it is taken in the suitable topology in $\Bbb R$

Answer 3

Perhaps some confusion arises from the notation $\{1/n\}_{n=1}^\infty$. The index $n$ is a member of the set $\{1,2,3,4,\ldots\}$ in which each number is a finite integer. There is no finite integer whose reciprocal is $0$. Therefore $0$ is not a member of this set. What I suspect sometimes causes confusion is the notation's use of the symbol "$\infty$", perhaps suggesting that something called $\infty$ occurs as one of the values of the index. But it is actually intended only to indicate that the sequence $1,2,3,4,\ldots$ does not come to an end.