If \(A\) and \(B\) are nonempty and bounded sets of real numbers such that \(A \subseteq B\) then \(\inf B \leq \inf A \leq \sup A \leq \sup B\).


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

  • \(\sup A \leq \sup B\)
  • \(\inf A \leq \sup A\)
  • \(\inf B \leq \inf A\)

From these inequalities we can then conclude that \(\inf B \leq \inf A \leq \sup A \leq \sup B\). To prove \(\sup A \leq \sup B\), let \(a \in A\). We know that \(a \leq \sup A\) by the definition of least upper bound. But since \(A \subseteq B\) then each element of \(A\) is also in \(B\) so \(\sup B\) is an upper bound on the set \(A\) and \(a \leq sup B\). Finally, we know that for any other upper bound \(y\) on the set \(A\) that \(\sup A\) is less than or equal \(y\) since \(\sup A\) is the least upper bound. This means that \(\sup A \leq \sup B\). (So the idea is just to prove that \(\sup B\) is an upper bound on \(A\) here and then everything will come together.)

To prove 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 prove the last piece \(\inf B \leq inf A\), let \(a\) be an arbitrary element of \(A\). We know \(\inf A \leq a\) by the defintion of the greatest lower bound. 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 \(A\). By the definition of the greatest lower bound we know that for any other lower bound \(x\), we must have \(\inf A \geq x\). Therefore we can conclude that \(\inf A \geq \inf B\). This is the last piece of the proof and therefore \(\inf B \leq \inf A \leq \sup A \leq \sup B\). \(\blacksquare\)

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