$X_n\leq Y_n \implies \inf X_n \leq \inf Y_n$

Question

$x_n\leq y_n\ \forall n \in N \implies \inf x_n \leq \inf y_n$

It is obvious from the definition of infimum and supremum, $\sup x_n \leq y_n$ and $\inf x_n \leq x_n \leq y_n$. However I do not know how to use the definition to prove formally that $\sup x_n \leq \inf y_n$ and conclude that
$\inf x_n \leq \sup y_n$.

Answer 1

Note that $$ \inf x_n\leq x_n\le y_n $$ for all $n$. So $\inf x_n$ is a lower bound for $y_n$ whence $$ \inf x_n\leq \inf y_n. $$

Answer 2

Assume by contradiction $$\inf x_n=x>\inf y_n=y$$therefore$$\forall 0<\epsilon <x-y,\exists N\qquad \forall n>N\to 0\le y_n-y<\epsilon<x-y$$which means that the exists $n\in \Bbb N$ such that $$y\le y_n<x\le x_n$$or equivalently $$y_n<x_n$$which is a contradiction. Therefore $x\le y$ and the proof is complete $\blacksquare$