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Comparing the sup of two sequences

OhMyMarkov

Member
Mar 5, 2012
83
Hello everyone!

Let $a_n$ and $b_n$ be two sequences such that $a_n \leq b_n$ for all $n$. Let $A_n = \sup \{a_m \; | \; m \geq n\}$ and $B_n = \sup \{b_m \; | \; m \geq n\}$.

I want to prove that $A_n\leq B_n$. I attempted a proof by contradiction:

Assume $A_n > B_n$ for some $n$.
If $A_n = a_i$ and $B_n = b_j$ for some j, and $i$ not necessarily equal to $j$, then $a_i > b_j$. However, $b_i > a_i$, so that $A_n > B_n$ is not true.

But the thing is, what if the index $i$ is infinity, I'm not sure what to do there...


Any help would be appreciated! :)
 

Plato

Well-known member
MHB Math Helper
Jan 27, 2012
196
Hello everyone!
Let $a_n$ and $b_n$ be two sequences such that $a_n \leq b_n$ for all $n$. Let $A_n = \sup \{a_m \; | \; m \geq n\}$ and $B_n = \sup \{b_m \; | \; m \geq n\}$.

I want to prove that $A_n\leq B_n$. I attempted a proof by contradiction:

Assume $A_n > B_n$ for some $n$.
Assume $A_n > B_n$ for some $n$.
$\left( {\exists a_j } \right)\left[ {j \geqslant n\;\& \,B_n < a_j \leqslant A_n } \right]$.

But $a_j\le b_j\le B_n$.
 

OhMyMarkov

Member
Mar 5, 2012
83
Hello Plato, thanks for your reply!

I was thinking this way, but can't I extend this to the case where the sup is not actually in the set $\{a_n\}$, like the sequence $3, 3.14, 3.1415, 3.141592, 3.14159264, \dots$. Or an other sequence.