For bounded sequences, prove that if $x_n \leq y_n~\forall~n \in \mathbb{N}$ then $\text{lub}\{x_n\} \leq \text{lub}\{y_n\}$.

Question

Problem:

Let $\{x_n\}_{n=0}^\infty $, $\{y_n\}_{n=0}^\infty $ be bounded sequences. Prove that if $x_n \leq y_n~\forall~n \in \mathbb{N}$ then $\text{lub}\{x_n\}_{n=0}^\infty \leq \text{lub}\{y_n\}_{n=0}^\infty $.

Attempt at proof:

Suppose to the contrary that $\text{lub}\{x_n\} > \text{lub}\{y_n\}.$ Let $\epsilon = \text{lub}\{x_n\} - \text{lub}\{y_n\} > 0$. There exists $N \in \mathbb{N}$ such that $0 < \text{lub}\{x_n\} - x_N < \epsilon/2$. Then we have $$x_n \leq y_n \leq \text{lub}\{y_n\} < x_N < \text{lub}\{x_n\},$$

which contradicts the hypothesis. $\square$

Are there any logical errors in my proof? Also, I wonder if anyone has an elegant way to prove this directly? Thank you.

Answer 1

Your proof is correct. Another way to prove it is to point out that since $lub\{y_n\}$ is an upper bound for $\{y_n\}$, and for all $n$ we have $x_n<y_n$, then $lub\{y_n\}$ is also an upper bound for $\{x_n\}$ and then by definition $lub\{x_n\}\leq lub\{y_n\}$.

Answer 2

Correct. I would prefer the sentence "Then we have..." to say "Then for all $n$ we have...".

You could also (if you like) replace this sentence with "Then $x_N\leq y_N\leq $ lub $ \{y_n\}_n<x_N,$ which is a contradiction." But that is a matter of taste.