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

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

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
$$\inf x_n \leq \sup y_n$$.

• $\sup x_n \le \inf y_n$ does not hold. Try some sequences with $x_n = y_n$. Nov 22 '18 at 21:41
• I would be careful with your statement. It is true that if $x_{n} \leq c$ for some fixed real number $c$, then $\sup x_{n} \leq c$ as well. But if you are comparing a sequence $x_{n}$ to another sequence $y_{n}$ and find that $x_{n} \leq y_{n}$ for all $n$, that doesn't mean $\sup x_{n} \leq y_{n}$. The left hand side of the inequality is a fixed number, but which $n$ are you choosing for the right hand side? Here is an example: Take $x_{n} = 1 - \frac{2}{n}$ and $y_{n} = 1 - \frac{1}{n}$. Clealry, $x_{n} \leq y_{n}$ for all $n$. But $\sup \{x_{n}\} = 1$ which is bigger than all $y_{n}$. Nov 22 '18 at 21:41
• It is not true that $\sup x_n \leq y_n$ and the sequences $x_n = 1/(n+1)$ and $y_n=1/n$ show for the left hand side is $1/2$ and the right hand side, when $n = 3$ is $1/3.$ Nov 22 '18 at 21:41

2 Answers

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 $$\inf x_n\leq \inf y_n.$$

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