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Consider the following two statements :

$S_1$: If $(a_n)$ is any real sequence , then $(\frac{a_n}{1+|a_n|})$ has a convergent subsequence .

$S_2$ : If every subseqeunce of $(a_n) $ has convergent subsequence , then $(a_n)$ is bounded .

Which is of the follwoing statement is true ?

(A) Both $S_1$ and $S_2$ are true .

(B) Both $S_1$ and $S_2$ are false .

(c) $S_1 $ is false but $S_2$ is true .

(D) $S_2$ is false and $S_1$ is true .

My work : $1+|a_n|\geq 1+a_n>a_n$ . So $|(\frac{a_n}{1+|a_n|})|\leq 1$ . So it is a bounded sequence and via Bolzano-wietrass theorem it has a convergent subsequence .

$a_{2n}$ and $a_{2n-1}$ are convergent . So they are bounded . Let the bounds be $M $ and $L$ respectively .

Hence $a_n \leq M+L $ .So $a_n$ is bounded .

So both $S_1$ and $S_2$ are true .

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You are right about $S_1$, and your proof is correct.

Your proof for $S_2$ is wrong, since $(a_{2n})$ and $(a_{2n+1})$ does not represent every subsequence.

Plus, they can not be bounded in general, since $(a_n)$ can be anything.

What you can do instead:

If $(a_n)$ is not bounded, then there exist a subsequence $(b_n)$ such that

$$\vert b_n\vert \to \infty.$$

And this sequence $(b_n)$ can not have a convergent subsequence.

So $S_2$ is true.

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  • $\begingroup$ Well, you can guarantee that $|b_n| \to \infty$. $\endgroup$ – Omnomnomnom Jun 3 '17 at 20:35
  • $\begingroup$ @Omnomnomnom I think you can choose $(b_n)$ such that either $b_n\to +\infty$ or $b_n\to -\infty$. But you are right, may be it is more clear that way. $\endgroup$ – E. Joseph Jun 3 '17 at 20:38
  • $\begingroup$ I'm not sure about clarity, but using $|b_n|$ means you don't have to consider two separate cases. $\endgroup$ – Omnomnomnom Jun 3 '17 at 20:42
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Your assessment of $S_1$ is correct: the statement is true.

Note that your approach for $S_2$ is incorrect: we cannot assume that the particular subsequences $(a_{2n}), (a_{2n-1})$ are convergent; all we know is that these subsequences have their own convergent subsequences in turn.

However, the statement is indeed true. I find it's easiest to prove by contrapositive: suppose that $(a_n)$ is not bounded. Then for every $k \in \Bbb N$, there exists an $n_k \in \Bbb N$ such that $|a_{n_k}| > k$ (that is, we can construct a sequence with $|a_{n_k}| \to \infty$). If we construct such a subsequence $(a_{n_k})$, then this subsequence can have no convergent subsequence (why is this the case?).

So, if $(a_{n})$ is such that every subsequence has a convergent subsequence, $(a_n)$ must be bounded.

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Your work on the first statement looks good.

On the second, the problem statement says

S2 : If every subseqeunce of $(a_n)$ has convergent subsequence , then $(a_n)$ is bounded.

So we don't know at $a_{2n}$ and $a_{2n+1}$ are convergent, only that they have a convergent subsequence.

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  • $\begingroup$ You are right . But how about this approach : $a_{2n}$ is a subsequence of $a_n$ .So $a_{4n}=a_{2(2n)} $ and $a_{2(2n+1)} $ are subsequences of $a_{2n}$ and hence bounded by $M_1 $ and $M_2$ respectively . Similarly $a_{4n+1}=a_2{2n}+1$ and $a_{4n+3}=a_{2{2n+1}} $ are bounded by $M_3$ and $M_4$ . So $a_n$ is bounded by $M_1+M_2+M+M_3+M_4$ . Does it work ? $\endgroup$ – Suman Kundu Jun 3 '17 at 20:39
  • $\begingroup$ @SumanKundu now you're assuming that $a_{2(2n)}$ is convergent, which isn't necessarily the case. $\endgroup$ – Omnomnomnom Jun 3 '17 at 20:43
  • $\begingroup$ That still doesn't work because, for $a_{2n}$, for example, we know that it has some convergent subsequence, not that every subsequence is convergent. So it may happen at $a_{4n}$ is not convergent, but some other subsequence is. $\endgroup$ – idontseethepoint Jun 3 '17 at 20:43
  • $\begingroup$ Ahh ! Now i understand . Thank you . $\endgroup$ – Suman Kundu Jun 3 '17 at 20:45

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