For the definitions of an upper bound and the least upper bound of a set, see this.
Proof
Let \(A\) and \(B\) be nonempty and bounded sets of real numbers sets such that \(A \subseteq B\). We will prove the following
- \(\sup A \leq \sup B\)
- \(\inf A \leq \sup A\)
- \(\inf B \leq \inf A\)
To show \(\sup A \leq \sup B\), let \(a \in A\). We know that \(a \leq \sup A\). But since \(A \subseteq B\) then \(a \in B\). Therefore, \(\sup B\) is an upper bound on \(A\). But \(\sup A\) is the least upper bound of \(A\) and by the definition of the least upper bound, this means that \(\sup A \leq \sup B\). (All we needed to do here is to show that \(\sup B\) is an upper bound on \(A\). Then everything will follow from the defintion of a least upper bound.)
To show that \(\inf A \leq \sup A\). By definition we know \(\inf A \leq a\) and \(a \leq \sup A\). So \(\inf A \leq a \leq \sup A\) and \(\inf A \leq \sup A\).
To show that \(\inf B \leq \inf A\), let \(a \in A\). We know \(\inf A \leq a\). But \(A \subseteq B\) so \(a\) is also an element of \(B\) and \(\inf B \leq a\). This means that \(\inf B\) is a lower bound on \(A\). But \(\inf A\) is the greatest lower bound on \(A\). So by the the definition of the greatest lower bound, we must have \(\inf A \geq \inf B\). We can finally concluded that \(\inf B \leq \inf A \leq \sup A \leq \sup B\).
\(\blacksquare\)