Filling the Holes in the Number Line
The Completeness Axiom is what distinguishes the Real Numbers (\(\mathbb{R}\)) from the Rational Numbers (\(\mathbb{Q}\)). It guarantees that there are "no holes" in the number line. If a set of real numbers has an upper bound, it must have a least upper bound (supremum) that is also a real number.
Supremum (sup S)
The smallest upper bound. Think of it as the tightest possible ceiling for the set.
Infimum (inf S)
The largest lower bound. Think of it as the tightest possible floor for the set.
Interactive Visualizer: Max/Min vs Sup/Inf
Select a set to see its boundaries. Notice how max and min don't always exist, but sup and inf (for bounded sets) always do.
The Epsilon Definition of Supremum
To prove \( M = \sup S \), you must show two things:
- Upper Bound: \( s \le M \) for all \( s \in S \).
- Least Upper Bound: For any \( \epsilon \gt 0 \), there exists some \( s \in S \) such that \( s \gt M - \epsilon \).
Basically: If you lower the ceiling by even a tiny amount (\(\epsilon\)), you will hit an element of the set.
Try it: Lower the ceiling!
Set \( S = (0, 1) \). We claim \( \sup S = 1 \). Pick an \( \epsilon \) to lower the bound to \( 1 - \epsilon \).
Practice Problems
True/False
1. If a set \( S \) has a maximum, then \( \sup S = \max S \).
Show Solution
TRUE - If max S exists, it is an upper bound and belongs to S, so it must be the least upper bound.
2. Every finite nonempty subset of \( \mathbb{R} \) has both a supremum and an infimum.
Show Solution
TRUE - Finite sets always have a maximum and minimum, which are the sup and inf.
3. The set \( \mathbb{Q} \) (rational numbers) satisfies the Completeness Axiom.
Show Solution
FALSE - The set \( \{r \in \mathbb{Q} : r^2 \lt 2\} \) is bounded above in \( \mathbb{Q} \) but has no supremum in \( \mathbb{Q} \) (since \( \sqrt{2} \notin \mathbb{Q} \)).
4. The Archimedean Property guarantees that for any \( \epsilon \gt 0 \), there exists \( n \in \mathbb{N} \) such that \( 1/n \lt \epsilon \).
Show Solution
TRUE - This is a standard corollary of the Archimedean Property.
Fill in the Blank
5. Consider the set \( S = \{1 - 1/n : n \in \mathbb{N}\} \).
\( \sup S = \) _____
\( \inf S = \) _____
\( \max S = \) _____ (or DNE)
\( \min S = \) _____ (or DNE)
Show Solution
\( \sup S = 1 \)
\( \inf S = 0 \)
\( \max S = \) DNE
\( \min S = 0 \)
6. For the interval \( (2, 7] \):
\( \sup(2, 7] = \) _____
\( \inf(2, 7] = \) _____
Is \( (2, 7] \) bounded? _____
Show Solution
\( \sup = 7 \)
\( \inf = 2 \)
Bounded = YES
7. The Completeness Axiom states: Every nonempty subset of \( \mathbb{R} \) that is _____ has a _____.
Show Solution
"bounded above" ... "least upper bound (supremum)"
Full Problems
8. For \( S = [0, 1] \cup [2, 3] \), determine sup, inf, max, min.
Show Solution
\( \sup = 3 \), \( \inf = 0 \), \( \max = 3 \), \( \min = 0 \)
9. For \( S = \{1/n : n \in \mathbb{N}\} \), determine sup, inf, max, min.
Show Solution
\( \sup = 1 \), \( \inf = 0 \), \( \max = 1 \), \( \min = \) DNE
10. For \( S = \{r \in \mathbb{Q} : r^2 \lt 2\} \), determine sup, inf, max, min.
Show Solution
\( \sup = \sqrt{2} \), \( \inf = -\sqrt{2} \), \( \max = \) DNE, \( \min = \) DNE
11. Let \( S = \{n/(n+1) : n \in \mathbb{N}\} \). Prove \( \sup S = 1 \).
Show Solution
1. Upper bound: \( n \lt n+1 \implies n/(n+1) \lt 1 \). So 1 is an upper bound.
2. Least upper bound: Let \( \epsilon \gt 0 \). We need \( n/(n+1) \gt 1 - \epsilon \). This simplifies to \( n \gt (1-\epsilon)/\epsilon \). By Archimedean Property, such \( n \) exists.
12. Find a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Show Solution
Example: \( r = 1.5 \) (since \( 1.414 \lt 1.5 \lt 1.732 \)) or \( r = 3/2 \).