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.

  • $\begingroup$ "...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
  • 1
    $\begingroup$ 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


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).


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.