Lecture 11:
The completeness property of \(\R\)
The Completeness Property of \(\R\)
Every nonempty subset of real numbers that has an upper bound also has a supremum.
This property is also called the supremum property of \(\R\). This property says that if you have a nonempty subset of real numbers \(S\), and \(S\) is bounded above, then \(S\) has a supremum.
Show that the set \(A = \{x:x\geq 0\text{ and }x^2<2\}\) has a supremum.
Proof. To use the completeness property we must show that \(A\) is nonempty and bound above.
Since \(1^2=1<2\), we see that \(1\in A\), and therefore \(A\neq \varnothing\).
Next, we must show that \(A\) is bounded above. We will show that \(4\) is an upper bound for \(A\). We will prove the statement by contrapositive.
Suppose \(z>4\). By a previous theorem this implies \(z^2>16\), and hence \(z\notin A\).
Since \(A\) is nonempty and bounded above, by the completeness property we see that \(A\) has a supremum. \(\Box\)
A surprising theorem
One of the simplest sets of real numbers is \(\mathbb{N}\). Does it have an upper bound?
Theorem (the Archimedean Property). If \(x\in\R\), then there exists \(n_{x}\in\N\) such that \(n_{x}> x\).
What does it mean to say that \(\N\) is not bounded above?
No matter what real number \(x\) you pick, there is some \(n\in\N\) such that \(n>x\).
Theorem (the Archimedean Property). If \(x\in\R\), then there exists \(n_{x}\in\N\) such that \(n_{x}> x\).
We're going to prove this by contradiction. What is the negation?
Statement:
English: For all \(x\in\R\) there exists \(n_{x}\in\N\) such that \(n_{x}>x\).
Logic: \(\forall\,x\in\R,\ \exists\,n_{x}\in\N,\ n_{x}>x.\)
Negation:
Logic: \(\exists\,x\in\R,\ \forall\,n_{x}\in\N,\ n_{x}\leq x.\)
English: There exists a real number \(x\) such that \(x\geq n\) for all \(n\in\N\)
Idea of the proof:
- Assume \(\exists\, x\) such that \(x\geq n\) for all \(n\in\N\)
- \(x\) is an upper bound for \(\N\)
- Completeness Property \(\Rightarrow\) \(\N\) has a supremum \(u\).
- \(u-\frac{1}{2}\) is not an upper bound.
- there is some \(n\in\N\) such that \(n>u-\frac{1}{2}\)
- \(n+1>u\) \(\Rightarrow\Leftarrow\)
Theorem (the Archimedean Property). If \(x\in\R\), then there exists \(n_{x}\in\N\) such that \(n_{x}> x\).
Proof. Suppose to the contrary that there is some \(x\in\R\) such that \(n\leq x\) for all \(n\in\N\). This means that \(x\) is an upper bound for \(\N\). By the Completeness Property there is a number \(u\) so that \(u=\sup\N\). Since \(u-\frac{1}{2}<u\) we see that \(u-\frac{1}{2}\) is not an upper bound for \(\N\). Hence, there is a number \(n\in\N\) such that \(n>u-\frac{1}{2}\). However, this implies \(n+1>n+\frac{1}{2}>u\). This is a contradiction since \(n+1\in\N\) and \(u\) is an upper bound for \(\N\). \(\Box\)
Proof. If \(n\in\N\), then \(\frac{1}{n}>0\), and hence \(1-\frac{1}{n}<1\). This shows that \(1\) is an upper bound for the set \(\{1-\frac{1}{n} : n\in\N\}.\)
Let \(v<1\) be arbitrary. Note that \(1-v>0\), and hence
\[\frac{2}{1-v}>0.\]
By the Archimedean Property there exists \(m\in\N\) such that
\[m>\frac{2}{1-v}.\]
Multiplying both sides of this inequality by \(\frac{1-v}{2m}\) we see that \(\frac{1-v}{2}>\frac{1}{m}.\)
From this and the assumption that \(v<1\) we see
\[1-\frac{1}{m}>1-\frac{1-v}{2} = \frac{1+v}{2}>\frac{v+v}{2} = v.\]
Thus, we have produced an element of the set which is greater than \(v\), and therefore \(v\) is not an upper bound. \(\Box\)
Prove that \(\sup\{1-\frac{1}{n} : n\in\N\} = 1.\)
Practice problem.
Show that \(\inf\{\frac{1}{n} : n\in\N\}=0\).
End Lecture 11
Read Section 2.3
Proof. Set \(A = \{\frac{1}{n} : n\in\N\}\). Since \(n>0\) for all \(n\in\N\), we see that \(\frac{1}{n}>0\) for all \(n\in\N\). Therefore \(0\) is a lower bound for \(A\).
Let \(x>0\) be arbitrary. Since \(x>0\) we know that \(\frac{1}{x}>0\). By the Archimedean Property there is a number \(m\in\N\) such that \(m>\frac{1}{x}\). Multiplying both sides by the positive number \(\frac{x}{m}\) we see that
\[x>\frac{1}{m}\]
Since \(\frac{1}{m}\in A\), this shows that \(x\) is not a lower bound for \(A\). Therefore \(\inf A = 0\). \(\Box\)
Lecture 11
By John Jasper
Lecture 11
- 271
