If the sets $A$ and $B$ are bounded above and $A\subseteq B$ and $A$ and $B$ both have supremums, then $sup(A)\le sup(B)$
Came across it in my textbook and was wondering how to prove it. It looks pretty simple. Thanks!
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assume $sup(A)>sup(B)$. That means there are elements of the set $A$ that between $sup(A)$ and $sup(B)$ ($x \in A$, $sup(B)<x<sup(A)$) – T_O Apr 23 '14 at 14:32
