# Prove that a sequence either: 1) converges 2) is unbounded or 3) has more than one limit point

Prove that a sequence either

1) converges
2) is unbounded
3) has more than one limit point

Well i think that this is equivalent to looking at the limit set of the sequence and showing that sequence has either:

1) limsup = liminf
2) at least limsup or liminf is infinity
3) limsup > liminf

Just for clarification, a limit point is a limit of some subsequence. The limit set is the set of all limit pts.

Not so sure how to show this. Any ideas?

• Yes, what you’ve written is correct. You have two things to do. (a) Show that the problem’s (1)-(3) are equivalent, in that order, to your (1)-(3). (b) Show that every sequence fits exactly one of your (1)-(3). I understand that you need some help with (b). Are you mostly okay with (a)? Apr 11 '13 at 23:27
• actually, could you show maybe one part? like show why if a sequence converges, its limsup = liminf. I know that is the definition Apr 11 '13 at 23:33
• I didn't read the answers, but I would assume (2) and (3) are false, and show it implies (1) Apr 12 '13 at 4:02

You're correct, maria. Assuming you mean the "inclusive or" for $(2)$: one or both of $\limsup$ or $\liminf$ is $\infty$ (by $\infty$ we have $\pm \infty$).

Either $\liminf = -\infty$ and $\limsup = +\infty$, or $\liminf = -\infty$ or $\limsup = +\infty$.

For $3$ you want some $x_s = \limsup \lt +\infty$, some $x_i = \liminf > -\infty$, with $\liminf = x_i \lt x_s = \limsup$

For $(1)$ By definition of convergence, if a sequence converges, then there exists a point $x$ such that $x_n \to x$ and $\liminf =\limsup = x$.

The equivalence of $1$ with $1$, $2$ with $2$, $3$ with $3$ can be shown, similarly, by definition.

Can you also show that every sequence satisfies one and only one of the three categories?

• Nice - always clean! +1 Apr 12 '13 at 0:22

Let $\langle x_n:n\in\Bbb N\rangle$ be a sequence of real numbers, and let $E$ be the set of all limits of subsequences of $\langle x_n:n\in\Bbb N\rangle$; then $\limsup_{n\to\infty}x_n=\sup E$ and $\liminf_{n\to\infty}x_n=\inf E$. If $\langle x_n:n\in\Bbb N\rangle$ is convergent, say with limit $x$, then all of its subsequences converge to $x$. But then $E=\{x\}$, so $\inf E=x=\sup E$, and therefore $\liminf_{n\to\infty}x_n=\limsup_{n\to\infty}x_n=x$.

Conversely, if $\liminf_{n\to\infty}x_n=\limsup_{n\to\infty}x_n=x$, say, then $\inf E=\sup E=x$, which is possible if and only if $E=\{x\}$. But then every subsequence of $\langle x_n:n\in\Bbb N\rangle$ converges to $x$, and since $\langle x_n:n\in\Bbb N\rangle$ is certainly a subsequence of itself, it must converge to $x$.

By the way, instead of working with liminfs and limsups, which can get a little messy, you could use this earlier question of yours.

• Great, so we use the fact that if a sequence converges, so does its subsequences Apr 12 '13 at 0:29
• So if a sequence is unbounded, either +/-infinity is a limit point right, since a sequence is a subsequence of itself. How do I show the converse? If a subsequence does not converge, then the sequence cannot converge? I took the negation from the first comment Apr 12 '13 at 0:34
• @maria: You don’t really need to deal with the converse. If a sequence is unbounded, you’re in case (2). Otherwise, it has a convergent subsequence. If the sequence itself is convergent, you’re in case (1). And if it’s not, you’re in case (3) by your earlier question. Apr 12 '13 at 0:43
• Right, so we used BW theorem- every bounded sequence contains a converging subsequence Apr 12 '13 at 0:46
• @maria: Yep. B-W is a very useful tool. Apr 12 '13 at 0:47

Looks good (in light of the fact that by limit point you mean cluster point), but I would correct (2) slightly, to say that at least one of limsup or liminf is infinite.

Now, you should know that for all sequences of points $x_n$, $\liminf x_n\le\limsup x_n$. The sequence is bounded below if and only if $\limsup x_n$ is real, and bounded above if and only if $\liminf x_n$ is real (use monotone convergence theorem to prove these). The sequence converges if and only if $\liminf x_n=\limsup x_n$ (squeeze theorem to prove this). Can you take it from there?

• limit point = cluster pt = accumulation pt= partial pt Apr 11 '13 at 23:28
• "Limit point" here obviously means "limit point of the set formed by elements of the sequence". Apr 11 '13 at 23:28
• Ah! Yes, if by limit point they mean a limit point of the set of points (rather than limit of the sequence), then that's fine. I will correct. Apr 11 '13 at 23:29
• @Brian: Thanks. Apr 11 '13 at 23:31
• @Sharkos: Thanks. Apr 11 '13 at 23:32

Your approach works! (Though in case 2 you should make clear you only need to have at least one of $\limsup,\liminf$ being either $\pm\infty$, and in case 3 you should emphasize that the finiteness of the $\limsup,\liminf$ is important, and explain why these are two limit points.)

It sounds like you're not sure how to show that one of $$\limsup = \liminf, \qquad \{\limsup,\liminf\}\cap\{\pm\infty\} \neq \emptyset, \qquad \infty > \limsup > \liminf > -\infty$$ holds? There's not much to say here; this is just a property of two numbers guaranteed to lie in $\mathbb{R}\cup\{\pm\infty\}$ with one larger than the other.

If proving the equivalence is the issue, then note that e.g.

1) $x_n \to x$ holds iff $\limsup x_n = \liminf x_n = x$.

2) A sequence is unbounded iff you cannot choose finite upper and lower bounds, which holds iff the $\limsup,\liminf$ are bounded.

3) As I say above, in this case note that holding $\limsup,\liminf$ apart means that there are infinitely many points (arbitrarily late in the sequence, since the $\lim$ bit comes into play) near these two points, so they are limit points.