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

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
$\inf x_n \leq \sup y_n$.

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  • $\begingroup$ $\sup x_n \le \inf y_n$ does not hold. Try some sequences with $x_n = y_n$. $\endgroup$
    – mschauer
    Nov 22 '18 at 21:41
  • $\begingroup$ 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 $y_{n}$. $\endgroup$
    – layman
    Nov 22 '18 at 21:41
  • $\begingroup$ 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 $1/3.$ $\endgroup$
    – William M.
    Nov 22 '18 at 21:41
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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 $$ \inf x_n\leq \inf y_n. $$

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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 $\blacksquare$

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