Supremum of Subset of Real Numbers May or May Not be in Subset
Jump to navigation
Jump to search
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 $\text{(a)}$