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I need help with the following two tasks:

a) Show that the sequence {$a_n$] converges to a $\in \mathbb {R}$ $\leftrightarrow$ all proper subsequence {$a_{n_k}$} of {$a_n$} converge to a.

Well the right direction $\rightarrow$ is easy to proof. If {$a_n$} converges to $a$, it is bounded. After the "Bolzano-Weierstraß-theorem" each bounded sequence has a convergent subsequence. Obviously, it converges to a too (right?). But now the problem is to proof that each real subsequence converge to $a$.

$\leftarrow$ Well thats the biggest problem for me. We are talking about proper subsequences, so we can't use {$a_n$} as a subsequence of {$a_n$}.The only thing that I could imagine is to say: Let {$a_n$} be cauchy. If the subsequence of a couchy-sequence converges to $a$ for $n_k$ $\rightarrow$ $\infty$. Then {$a_n$} converges to $a$ too.

b) Let {$b_n$} be a sequence with the following property: Each subsequence of {$b_n$} has another subsequence, that converges to $b$. Show that {$b_n$} converges to $b$.

Well I guess if we prove a), we are able to conclude b).

I am thankful for any advice.

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  • $\begingroup$ "...converges against $a$" is not English. Do you mean "converges to $a$"? $\endgroup$
    – TonyK
    Feb 18 at 22:25
  • $\begingroup$ @TonyK Yeah I am sorry for that. English is not my native language. $\endgroup$ Feb 18 at 22:27
  • $\begingroup$ Can you prove by contrapositive? That is, assume that the LHS is not true, and show then that the RHS must also be untrue. Suppose the sequence does not converge; then there must be an $\varepsilon > 0$ such that there is no $M$ so that all $a_n$ are within $\varepsilon$ of $a$ when $n > M$. Then construct a subsequence for which that is also not true. $\endgroup$
    – Brian Tung
    Feb 18 at 22:36
  • $\begingroup$ So we choose {$2^n$}_$n\in \mathbb {N}$ for example. And choose the subsequence {$2^n$}_$n\geq1$? to the right direction $\rightarrow$. How do we prove that "all" subsequences converge to a. $\endgroup$ Feb 18 at 22:53
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    $\begingroup$ I'm not sure I understand your question. This is more or less what I mean. We want to prove that if all subsequences converge to $a$, then the original sequence must likewise converge to $a$. We suppose, contrariwise, that the original sequence does not converge to $a$. Then there must exist an infinite number of elements $a_{i_1}, a_{i_2}, a_{i_3}, \ldots$ that all fall outside the interval $[a-\varepsilon, a+\varepsilon]$ for some $\varepsilon > 0$. Then choose the subsequence $a_{i_2}, a_{i_4}, a_{i_6}, \ldots$ (or some such). This shows that not-LHS implies not-RHS, thus RHS implies LHS. $\endgroup$
    – Brian Tung
    Feb 18 at 23:29

2 Answers 2

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For one of the directions in (a), if $a_n$ does not converge to $a\in\mathbb R$, then there exists $\epsilon_0 > 0$ such that $|a_n-a|>\epsilon_0$ for infinitely many bad $n$. Enumerating an increasing subsequence $n_1<n_2<\dots$ of such $n$, what can you say about the subsequence $\{a_{n_k}\}_{k=1}^\infty$?

It seems like you've got a strategy for (b).

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Concerning a):
For the "$\implies$" direction, you do not need use Bolzano-Weierstraß to get a converging subsequence. Instead, you have to take any (!) subsequence of $(a_n)$ and prove that it converges to $a$.
Hint: $a_n \to a$ means that the difference $\lvert a_n - a \rvert$ converges to $0$. (That is the definition of convergence, right?) Can you show that $\lvert a_{n_k} - a \rvert$ converges to $0$ for $k \to \infty$?

For the "$\impliedby$" direction, I think that you didn't see a subtlety which makes the proof very easy.
Hint: $(a_n)$ itself is a subsequence of $(a_n)$.

Concerning b):
As far as I'm concerned, this does not follow directly from a), but is harder. I suggest you try a contraposition: Suppose $(b_n)$ does not converge to $b$. (What does this mean, by definition of convergence?) Can you construct a subsequence of $(b_n)$ from this, which has no sub-sub-sequence converging to $b$? Then you have proven the contraposition of the statement.

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  • $\begingroup$ Tkanks. But like a said we are talking about "real subsequences". So no an is not a real subsequence of an. $\endgroup$ Feb 18 at 23:17
  • $\begingroup$ Ah, so your initial $(a_n)$ comes from a bigger set? From which set? $\endgroup$
    – NerdOnTour
    Feb 18 at 23:19
  • $\begingroup$ No it does not come from a bigger set. I don't know if "real subsequence" is the right englishterminology. But in my language real subsequence or a real subset of the set A for example is not A itself. With real subsequence I mean a sequence that contains only elements of the sequence but is not the same sequence... Mabye we are talking about genuine subsequences? $\endgroup$ Feb 18 at 23:23
  • $\begingroup$ @Analysis_Mark: I'd use something like "proper subsequence" by analogy with terms like "proper subset." $\endgroup$
    – Brian Tung
    Feb 18 at 23:24
  • $\begingroup$ Yeah, I think "proper subsequence" is what you mean. But then the sequence $(a_2, a_3, a_4, a_5, \dots)$ would be a proper subsequence of $(a_n) = (a_1, a_2, \dots)$, right? Maybe this suffices for a) "$\impliedby$" $\endgroup$
    – NerdOnTour
    Feb 18 at 23:26

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