# Prove that $a_n \leq b_n \forall n \geq m \Rightarrow inf(a_n)_{n=m}^\infty \leq inf(b_n)_{n=m}^\infty$

I'm trying to prove the following:

"Suppose that $(a_n)_{n=m}^\infty$ and $(b_n)_{n=m}^\infty$ are two sequences of real numbers such that $a_n \leq b_n \forall n \geq m$. Then we have the inequalities:

(a) sup$(a_n)_{n=m}^\infty \leq$ sup$(b_n)_{n=m}^\infty$ (where sup$(a_n)_{n=m}^\infty:=$sup$\{a_n:n\geq m\}$);

(b) inf$(a_n)_{n=m}^\infty \leq$ inf$(b_n)_{n=m}^\infty$ (where inf$(a_n)_{n=m}^\infty :=$inf $\{a_n:n \geq m\}$)

I've managed to prove (a) by contradiction, and I've tried to do the same for (b) (i.e. show that if we suppose inf$(a_n)_{n=m}^\infty >$ inf$(b_n)_{n=m}^\infty$ there exists $n \geq m$ s.t. $a_n > b_n$), but I haven't been able to do so, so far.

So, I would appreciate any hint about how to deal with inequality (b).

Best regards,

lorenzo.

• You may try to consider the sequences $(-a_n)_{n=m}^{\infty}$ and $(-b_n)_{n=m}^{\infty}$ and try to apply the first exercise. – Iulia Nov 12 '15 at 9:11

Suppose $inf(a_n)_{n=m}^\infty >inf(b_n)_{n=m}^\infty$.
Then there exist an $n_0$ such that $inf(a_n)_{n=m}^\infty >b_{n0} \geq inf(b_n)_{n=m}^\infty$ and hence $a_{n0}\geq inf(a_n)_{n=m}^\infty >b_{n0}$.