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?
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?
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.
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?
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.
