4

How to prove that the sequence $\{a_n\}$ has no convergent subsequence if and only if $|a_n|$ approaches infinity? The forward direction is obvious, how to prove the other direction?

Mark Bennet
  • 100,194
ysen07
  • 49

5 Answers5

6

Use these two facts:

  • Every sequence has a monotone subsequence.
  • Every bounded monotone sequence converges.
lhf
  • 216,483
  • The first of these facts requires proof. – Mark Bennet Jul 04 '13 at 14:46
  • 2
    @MarkBennet, I'd say both "facts" require proof. I think lhf meant that there's a good chance the OP already covered this and thus he can comfortably use them to attack his problem. – DonAntonio Jul 04 '13 at 15:06
  • 1
    @DonAntonio - depending on whether we are dealing with sequences over the reals, the second is quite an easy consequence if the reals are defined as an ordered field in which every bounded set has a least upper bound (which is not the only definition). The first is not obvious in the same way. I agree that both need proof. – Mark Bennet Jul 04 '13 at 15:18
  • 3
    Each time I see this approach being recommended, it leaves me with a strange feeling: of course it works, but the proof uses so much the specifics of the real line while the result holds in such a vastly more general setting! – Did Jul 04 '13 at 16:31
3

First show that, if $|a_n|$ doesn't approach $\infty$, then $a_n$ has a bounded subsequence; then invoke the fact that a bounded sequence of real numbers has a convergent subsequence.

Andreas Blass
  • 71,833
3

Assume $\lim_{n\to\infty}|a_n|=\infty$ and consider an $\alpha\in{\mathbb R}$. There is an $n_0$ such that $|a_n|>|\alpha|+1$ for all $n>n_0$. Therefore $|a_n-\alpha|\geq|a_n|-|\alpha|>1$ for all $n>n_0$, and this implies that the sequence $(a_n)_{n\geq1}$ cannot have a subsequence converging to $\alpha$.

For the converse, assume that $\lim_{n\to\infty}|a_n|=\infty$ does not hold. This means that there is an $M>0$ such that there are arbitrary large $n$ with $|a_n|\leq M$. This can be exploited as follows: Put $n_0:=0$. Then for each $k\geq1$ we can find and $n_k>n_{k-1}$ such that $|a_{n_k}|\leq M$. The subsequence $(a_k')_{k\geq1}$ defined by $a_k':=a_{n_k}$ $\>(k\geq1)$ is bounded, therefore it has a subsequence $(a_l'')_{l\geq1}$ converging to some $\alpha\in{\mathbb R}$. This sequence $(a_l'')_{l\geq1}$ can be considered as a subsequence of the originally given sequence $(a_n)_{n\geq1}$.

Christian Blatter
  • 226,825
0

Hint If this result not true then weierstrass sequence theorem is false.

Up date Imitate this proof of weierstrass sequence theorem (with necessary adaptations of course) to get the proof of this fact. What I mean is that the same idea of this proof of weierstrass sequence theorem can be used to prove this result.

Elias Costa
  • 14,658
0

As others have stated, you need a theorem of Bolzano (proved below), usually known as the Bolzano$-$Weierstrass theorem, that a bounded sequence has a cluster point: a point such that an infinite number of terms of the sequence lie within any given neighbourhood of the point. First note that, if the terms do not approach $\infty$ in magnitude, then infinitely many of them must be bounded away from $\pm\infty$ and so lie within a finite interval, say $[a\,,b]$, with $a<b$.

For convenience, scale the terms by $x\mapsto (x-a)/(b-a),$ so that we are dealing with infinitely many terms in $[0\,,1].$ Cut this interval in half. Then at least one half has infinitely many terms. Choose such a half, and record $s_1=0$ if it is the left half or $s_1=1$ if the right half. Cut that reduced interval in half and, as before, choose a half with infinitely many terms, and record a corresponding $s_2=0$ or $s_2=1$ again. Continue this process indefinitely, to produce an infinite nest of intervals, each being half of its predecessor and containing infinitely many terms of the original scaled sequence, along with a corresponding string $s_1s_2\dots$ of $0$s and $1$s. Let $s$ be the real number represented as $\cdot s_1s_2\dots$ in binary notation. Then, by reverting to our original number scale, the required cluster point is $a+(b-a)s$.

John Bentin
  • 18,454