Real Analysis: Prove every unbounded sequence contains a divergent monotone subsequence
1,285
Solution 1
If $a_n$ is unbounded, the for every $n$ , there is $a_m$ with $a_m>n$ . You can do that for each $n$...
Solution 2
If the set is unbounded above, then for each $a_n$, there exists a point in the set $x$ such that $x>a_n$. Choose $a_{n+1}=x$. Then proceed.
Similarly for the set unbounded below.
Related videos on Youtube
Author by
user115764
Updated on December 15, 2020Comments
-
user115764 almost 3 years
I have been struggling over this problem for hours, and I have no clue where to begin. Can anyone give a clear and complete proof of the following theorem.
Every unbounded sequence contains a divergent subsequence that diverges to either positive infinity or negative infinity.
Thanks
-
Eric Auld almost 10 years$a_m>n \not\Rightarrow a_{m+1}>a_m$
-
DBFdalwayse almost 10 yearsNo, but you iterate the process to find $a_k$ with $a_k >a_m$ , which is possible, since the sequence is unbounded.