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?
Best Answer
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?