Suppose that $\{a_{n}\}$ and $\{b_{n}\}$ are bounded. Prove that $\{a_{n}b_{n}\}$ has a convergent subsequence.

In class this is how my professor argued:

By the Bolzano-Weierstrass Theorem, there exists a subsequence $\{a_{n_k}\}$ that converges to $a$. Since $\{b_n\}$ is bounded, $\{b_{n_k}\}$ is also bounded. So by the Bolzano-Weierstrass Theorem, there exists a subsequence of $\{b_{n_k}\}$ namely $\{b_{n_{{k_j}}}\}$ such that $\{b_{n_{{k_j}}}\}$ converges to $b$.

In particular $\{a_{n_{{k_j}}}\}$ will converge to $a$. And note that $\{a_{n_{{k_j}}}b_{n_{{k_j}}}\}$ is a subsequence of $\{a_{n}b_{n}\}$. So $a_{n_{{k_j}}}b_{n_{{k_j}}} \to ab$.

My question is why do we have to use so many subsequences. Is it wrong to argue as follows?

$\{a_{n}\},\{ b_{n} \}$ are both bounded, so by the Bolzano-Weierstrass Theorem, both sequences have a convergent subsequence. Namely $a_{n_k} \to a$ and $b_{n_k} \to b$. Then note that $\{a_{n_k}b_{n_k}\}$ is a subsequence of $\{a_{n}b_{n}\}$ which converges to $ab$. And we are done.

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    $\begingroup$ Did anyone ask the professor why he first didn't show that $\{a_n b_n\}$ is bounded? (Or am i missing something?) $\endgroup$ – Aryabhata Mar 20 '12 at 21:15
  • $\begingroup$ @Aryabhata: Yes, someone did ask $\{a_{n}b_{n} \}$ was bounded. Would you mind explaining why my professor had to use so many subsequences? I don't follow the other answers given. $\endgroup$ – Student Mar 20 '12 at 21:30
  • $\begingroup$ NKS has given an example. $\endgroup$ – Aryabhata Mar 20 '12 at 21:43

You can actually use less subsequences using a different proof, that Aryabhata is hinting at in his comment:

Proof. Sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are bounded, so there exist $A\geq0$ and $B\geq0$ such that $|a_n|\leq A$ and $|b_n|\leq B$ for all $n\in\mathbb{N}$. But from this we immediately see that $|a_nb_n|\leq AB$ for all $n\in\mathbb{N}$. So $(a_nb_n)_{n=1}^\infty$ is a bounded sequence as well. Therefore by the Bolzano-Weierstrass theorem it has a convergent subsequence. $\square$

If you prove it the way your professor did, however, you have to use many sequences. To see why, I suggest you try the following. First prove:

Lemma. Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are bounded sequences and $(a_n)_{n=1}^\infty$ is convergent. Then $(a_nb_n)_{n=1}^\infty$ has a convergent subsequence.

(Hint: one of the sequences is convergent, so you don't have to worry about indices properly aligning. Can you see why?)

After you have proved this, the professor's proof will look like this:

Proof. Sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are bounded, so there exists a convergent subsequence $(a_{n_k})_{k=1}^\infty$ of the sequence $(a_n)_{n=1}^\infty$. This means that $(a_{n_k})_{k=1}^\infty$ and $(b_{n_k})_{k=1}^\infty$ are bounded sequences and $(a_{n_k})_{k=1}^\infty$ is convergent. So we can use the lemma on them, telling us that some subsequence of $(a_{n_k}b_{n_k})_{k=1}^\infty$ is convergent. Since a subsequence of a subsequence is again a subsequence of the original sequence, this completes the proof. $\square$

I hope this clears things up a bit.

  • $\begingroup$ I think its all clear now. Thanks! $\endgroup$ – Student Mar 21 '12 at 0:47

Your repeated use of $k$ leads to a subtle confusion.

Suppose the subsequence of the $a$'s given by Bolzano-Weierstrass happens to use all the odd numbered $a$'s, while the subsequence from the $b$'s uses the even ones. The first term in your product sequence is thus $a_1b_2$, which is not a member of the sequence $\{a_nb_n\}$.

To avoid this potential misalignment, your professor finds a convergent subsequence in the $b$'s by using only indices that also appear in the convergent subsequence we already found for the $a$'s. This is the meaning of $b_{n_{k_j}}$ - the convergent subsequence from the $b$'s does not have to be chosen from among all indices, but only from the indices $\{n_k\}$.

Suppose, for an example, the convergent subsequence of the $a$'s given by BW happens to use all the even indices, so $n_k = 2k$. Restricting to just these indices, we look for a convergent subsequence of the $b$'s. Suppose it happens that the nonzero powers of $2$ work for $b$, so $n_{k_j} = 2^j$. The sequence of products we would concern ourselves with is $a_2b_2, a_4b_4, a_8b_8,$ and so on. Notice that all of the $a$ factors and $b$ factors are members of their respective convergent subsequences.

  • $\begingroup$ @Jon Bolzano-Weierstrass tells you there is a convergent subsequence, but it does not guarantee that any old subset of the indices will give you one. If you got so lucky that your $a$ subsequence and $b$ subsequence used the same indices, then you could avoid the double subsequence business. In general, you have no control over the functions choosing the indices. $\endgroup$ – Austin Mohr Mar 20 '12 at 22:09
  • $\begingroup$ That helps me understand why my method does not work. But why would using double subsequences resolve the problem? I still do not quite understand how they help. $\endgroup$ – Student Mar 20 '12 at 22:09

The problem with your argument is that by writing $\{a_{n_k}\}$ and $\{b_{n_k}\}$ you are implicitly (and incorrectly) assuming that the convergent subsequences of $\{a_n\}$ and $\{b_n\}$ involve the same terms. In general, there is no reason why this should be the case. We need to introduce all of the subsequences that we do to get around this problem.

EDIT: Here's a specific example.

Take $a_n$ to be $0$ for $n$ odd, and $(-1)^{n/2}$ for $n$ even. Take $b_n$ to be $(-1)^{(n+1)/2}$ for $n$ odd, and $0$ for $n$ even. These are bounded sequences. Bolzano-Weierstrass guarantees that some subsequence converges (and indeed, there are plenty of convergent subsequences, with different limit points). One possibility would be to take the terms $a_n$ for $n = 0,4,8,16,...$ and $b_n$ for $n = 0,2,4,6,...$ In other words, $a_{n_k} = a_{4k}$, and $b_{n_k} = b_{2k}$. Therefore, your candidate subsequence $\{a_{n_k} b_{n_k}\}$ of $\{a_n b_n\}$ would be the sequence $\{a_{4k} b_{2k}\}$. But this doesn't make any sense, because the terms of the product sequence $\{a_n b_n\}$ are products where the indices on the $a$'s and $b$'s match.

  • $\begingroup$ Could you elaborate? I'm still confused. $\endgroup$ – Student Mar 20 '12 at 21:25
  • $\begingroup$ Thanks for the example. But how does what my professor did make the indices match? $\endgroup$ – Student Mar 20 '12 at 21:48
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    $\begingroup$ The professor says, start with a subsequence $\{a_{n_k}\}$ that converges to $a$. In this example, you might take $a_0, a_4, a_8, a_{12}, \ldots$. Now you look at the sequence $\{b_{n_k}\}$, i.e. $b_0$, $b_4$, $b_8$, $b_{12}$, etc., and take a convergent subsequence of that. Maybe it's $b_0, b_{8}, b_{16}, \ldots$. The corresponding sequence $a_0, a_8, a_{16}, \ldots$ is a subsequence of a sequence you already know converges, so it still converges. $\endgroup$ – Robert Israel Mar 20 '12 at 22:50
  • $\begingroup$ @RobertIsrael: That was crystal clear. Could you make that an answer so I can accept it? $\endgroup$ – Student Mar 20 '12 at 22:56

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