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.

Share:
1,285

Related videos on Youtube

user115764
Author by

user115764

Updated on December 15, 2020

Comments

  • user115764
    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
    Eric Auld almost 10 years
    $a_m>n \not\Rightarrow a_{m+1}>a_m$
  • DBFdalwayse
    DBFdalwayse almost 10 years
    No, but you iterate the process to find $a_k$ with $a_k >a_m$ , which is possible, since the sequence is unbounded.