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This question has been asked. For example: here. I am interested in a different way of proving this statement.

A subsequence $(b_k)$ of a sequence $(a_n)$ is defined as $ b_k = a_{n_k}$ such that $n_k < n_{k+1}$.

(1) With this definition choosing $k = n_k, k = 1,2,3,...$ results in $b_1 = a_1, b_2 = a_2, ..., b_k = a_k$. So by this definition any sequence $a_n$ contains itself as a sub-sequence.

Now if every subsequence $b_k$ converges to $a$, so does the specific subsequence $b_k$ when $k = n_k$. From (1), if every $(b_k)$ converges to $a$ we can choose $k = n_k$, which leads to $(b_k) = (a_k)$ and $(a_k)\rightarrow a$.

Is this proof valid?

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tmaric
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    This would be valid, except the question linked to is more subtle: if every convergent subsequence (of a bounded sequence!) converges to the same $a$, so does the whole sequence. So it is not assumed that all subsequences converge. Only those that do, converge to $a$. – Henno Brandsma Dec 22 '15 at 12:10
  • @HennoBrandsma: thank you for pointing that out. – tmaric Dec 22 '15 at 12:14

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This would be valid, except the question linked to is more subtle: if every convergent subsequence (of a bounded sequence!) converges to the same $a$, so does the whole sequence. So it is not assumed that all subsequences converge. Only those that do, converge to $a$. So convergence of the whole sequence is still unknown.

The boundedness of the sequence is essential (consider $0,1,0,2,0,3,0,4,\ldots$, where the convergent subsequences all converge to $0$, but the whole sequence does not converge at all).

Another more subtle true statement: if every subsequence of $(a_n)$ has itself a convergent subsequence that converges to $a$ (same $a$ throughout) than $(a_n)$ converges to $a$ as well.

Henno Brandsma
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