Supremum of Subset of Real Numbers May or May Not be in Subset

Theorem

Let $S \subset \R$ be a proper subset of the set $\R$ of real numbers.

Let $S$ admit a supremum $M$.

Then $M$ may or may not be an element of $S$.

Proof

Consider the subset $S$ of the real numbers $\R$ defined as:

$S = \set {\dfrac 1 n: n \in \Z_{>0} }$

It is seen that:

$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$

and hence $\sup S = 1$.

Thus $\sup S \in S$.

Consider the subset $T$ of the real numbers $\R$ defined as:

$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$

It is seen that:

$T = \set {-1, -\dfrac 1 2, -\dfrac 1 3, \ldots}$

and hence $\sup T = 0$.

Thus $\sup T \notin T$.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.35$. Example $\text{(a)}$