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

Question

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.

Answer 1

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 $a_{n0}\geq inf(a_n)_{n=m}^\infty >b_{n0}$.

Didn't you use the same way to prove (1)?