# Show that the sequence {$a_n$} converges to $a \in \mathbb {R}$ $\leftrightarrow$ all real subsequence {$a_{n_k}$} of {$a_n$} converge to $a$

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.

• "...converges against $a$" is not English. Do you mean "converges to $a$"? Feb 18 at 22:25
• @TonyK Yeah I am sorry for that. English is not my native language. Feb 18 at 22:27
• 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. Feb 18 at 22:36
• 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. Feb 18 at 22:53
• 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. Feb 18 at 23:29

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 $$n_1 of such $$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.

• Tkanks. But like a said we are talking about "real subsequences". So no an is not a real subsequence of an. Feb 18 at 23:17
• Ah, so your initial $(a_n)$ comes from a bigger set? From which set? Feb 18 at 23:19
• 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? Feb 18 at 23:23
• @Analysis_Mark: I'd use something like "proper subsequence" by analogy with terms like "proper subset." Feb 18 at 23:24
• 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) "$\impliedby$" Feb 18 at 23:26