How can I proceed with this proof?
Prove that a sequence that contains subsequences that converges to every point in the set $\{1/n: n\in\Bbb{N}\}$ contains a subsequence that converges to $0$.
Suppose that such a sequence, $a_n$, existed. Fix $\epsilon \gt 0$. By the Archimedean Property we can find a large enough $m$ such that we have a subsequence converging to $1/m$.
Is it correct to say that: So, since $1/m$ tends to zero, we have a subsequence that tends to zero?
I think something is missing.
Best Answer
Consider the following sequence OEIS A002260
$$1,1,2,1,2,3,1,2,3,4,1,2,3,4,5,\ldots$$
and then its reciprocal
$$\frac11,\frac11,\frac12,\frac11,\frac12,\frac13,\frac11,\frac12,\frac13,\frac14,\frac11,\frac12,\frac13,\frac14,\frac15,\ldots$$
which has subsequences converging to every unit fraction while the whole sequence does not converge to $0$ as it does not converge at all.
It does have a subsequence which converges to $0$. Was that what you had intended to ask?
To find such a subsequence, look for a sequence of ${N_1},{N_2},{N_3},\ldots$ where $N_m>N_{m-1}$ and $0 < a_{N_m} < \frac2m$. Given $N_{m-1}$, such an $N_m$ must exist as you have a subsequence converging to $\frac1m$ and so an infinite number of terms strictly between $\frac1m-\frac1m$ and $\frac1m + \frac1m$. Then $a_{N_1},a_{N_2},a_{N_3},\ldots$ converges to $0$ so is an example of the desired subsequence.