Prove: for real numbers $a$ and $b$ $$a\le b\iff (b,\infty) \subseteq (a,\infty)$$
I understand that this is an '$\rm {iff}$' statement so I should prove it both ways. The statement makes sense and seems obvious but I'm not sure where to start with this proof.
Thanks a lot for any help
$(\Longrightarrow)$ Suppose $a\leq b$. Let $c\in(b,\infty)$, that is, $c>b$. Then by hypothesis $c>b\geq a$ and so $c>a$. Hence $c\in(a,\infty)$. This shows $(b,\infty)\subseteq(a,\infty)$.
$(\Longleftarrow)$ Suppose $(b,\infty)\subseteq(a,\infty)$. Then $b+\epsilon\in(a,\infty)$ for all $\epsilon>0$, that is, $b+\epsilon>a$ for all $\epsilon>0.$ This implies $b\geq a$, that is, $a\leq b$.
$\implies\\$
Suppose $a\leqslant b$, we shall prove that $(b,\infty)\subseteq (a,\infty)$. Take any $t\in (b,\infty)$. We have that $a\leqslant b < t$, thus $t\in (a,\infty)$. Since $t$ was taken arbitrarily, the desired inclusion holds.
$\impliedby\\$
Let $a,b$ be any real numbers such that $(b,\infty)\subseteq (a,infty)$. Consider two cases:
a) $(b,\infty)$ is a proper subset of $(a,\infty)$. This implies, that $b\in(a,\infty)$. As a result $a $\lnot (a which implies $a\geqslant b \land b\geqslant a.$ As a consequence - $a=b$, thus desired inequality holds..