# Real Analysis

Real Numbers

# The Set of Real Numbers \(\mathbb{R}\)

*Theorem:* \(\mathbb{R}\) is an ordered field, with the *least upper bound property*, and \(\mathbb{R}\) contains \(\mathbb{Q}\) as a *subfield*.

## Construction of \(\mathbb{R}\)

Dedekind: A *cut* \(\alpha\) is a subset of \(\mathbb{Q}\) satisfying

- (nontriviality) i.e. \(\alpha \neq \emptyset, \mathbb{Q}\),
- (closed downwards or closed to the left) if \(p \in \alpha\), \(q \in \mathbb{Q}\), and \(q < p\), then \(q \in \alpha\), and
- (no largest member) if \(p \in \alpha\) then \(p < r\) for some \(r \in \alpha\).

For example, the set \(A = \{x : x^2 < 2\}\) is not a cut because it does not satisfy (2). The set \(\mathbb{Q}_{-}\) is a cut. The set \(\beta = \{x \in \mathbb{Q} : x \leq 2\}\) is not a cut because it does not satisfy (3).

Let \(\mathbb{R} := \{\alpha : \alpha \text{ is a cut}\}\) (\(:=\) means defined to be).

## Properties of \(\mathbb{R}\)

### Order on \(\mathbb{R}\)

\(\alpha < \beta\) is defined as \(\alpha \subsetneq \beta\), or \(\alpha\) is a proper subset of \(\beta\).

*Proof of well-definedness.* This definition is well-defined because it satisfies

- trichotomy: for all cuts \(\alpha\) and \(\beta\), either \(\alpha \subsetneq \beta\), \(\alpha = \beta\), or \(\beta \subsetneq \alpha\), and thus either \(\alpha < \beta\), \(\alpha = \beta\), or \(\beta < \alpha\).
- transitivity: let \(\alpha, \beta, \gamma\) be cuts, \(\alpha < \beta\), and \(\beta < \gamma\). Since \(\alpha < \beta\), \(\alpha \subsetneq \beta\). Similarly, \(\beta \subsetneq \gamma\). Thus, \(\alpha \subsetneq \gamma\), due to transitivity of inclusion. Hence, \(\alpha < \gamma\). QED.

### Operations on \(\mathbb{R}\)

Let \(\alpha + \beta := \{r + s : r \in \alpha, s \in \beta\}\).

*Proof of well-definedness.* Let \(\alpha\) and \(\beta\) be cuts. \(\alpha + \beta\) is a cut because

- it is nontrivial: since \(\alpha\) is a cut, there exists \(r \in \alpha\). Similarly, there exists \(s \in \beta\). So, there exists \(r + s\). Hence, \(\alpha + \beta \neq \emptyset\). Since \(\alpha\) is a cut, there exists \(p \in \mathbb{Q} \setminus \alpha\) such that \(p > r\) for all \(r \in \alpha\). Similarly, there exists \(q \in \mathbb{Q} \setminus \beta\) such that \(q > s\) for all \(s \in \beta\). Thus, \(p + q > r + s\) for all \(r \in \alpha\) and \(s \in \beta\). Hence, \(\alpha + \beta \neq \mathbb{Q}\).
- it is closed downwards: let \(p \in \alpha + \beta\), and \(q < p\). Since \(p \in \alpha + \beta\), there exist \(r \in \alpha\) and \(s \in \beta\) such that \(p = r + s\). Since \(q < p = r + s\), \(q - s < r\), so \(q - s \in \alpha\). Then \(q = (q - s) + s\), as desired.
- it has no largest member: Fix arbitrary \(p \in \alpha\) and \(q \in \beta\). Since \(\alpha\) is a cut, there exists \(r \in \alpha\) such that \(p < r\). Similarly, there exists \(s \in \beta\) such that \(q < s\). Hence, \(p + q < r + s \in \alpha + \beta\), as desired. QED.

Additive identity \(0^\star := \mathbb{Q}_{-}\).

*Proof of well-definedness.*

\(0^\star\) is a cut because

- it is nontrivial: \(0^\star \neq \emptyset\) since there exist negative rationals. \(0^\star \neq \mathbb{Q}\) since there exist positive rationals.
- it is closed downward: let \(p \in 0^\star\) and \(p \in \mathbb{Q}\) with \(q < p\). Since \(p \in 0^\star\), it is less than \(0\). Hence, \(q < 0\) and thus \(q \in 0^\star\).
- it has no largest element: Let \(p \in 0^\star\). Consider \(\frac{p}{2}\). We have \(p < \frac{p}{2} < 0\), as desired.

\(\alpha + 0^\star = \alpha\) because

- \(\alpha + 0^\star \subset \alpha\): let \(p \in \alpha + 0^\star\). Hence, there exist \(r \in \alpha\) and \(s \in \beta\) such that \(p = r + s\). We have \(p = r + s < r + 0 < r \in \alpha\), as desired.
- \(\alpha \subset \alpha + 0^\star\): let \(p \in \alpha\). Since \(\alpha\) is a cut, there exists \(q \in \alpha\) such that \(q > p\). Let \(r = p - q\). \(r \in 0^\star\) since \(r < 0\). We have \(p = q + r\), as desired. QED.

Additive inverse for \(\alpha\) is \(\beta := \{p : \exists r > 0 \text { such that } -p - r \notin \alpha\}\).

Definition of multiplication: if \(\alpha, \beta \in \mathbb{R}_{+}\) (\(\alpha, \beta > 0^\star\)), then \(\alpha\beta := \{p : p < rs \text{ for some } r \in \alpha, s \in \beta, r > 0, s > 0\}\).

Multiplicative identity \(1^\star := \{q \in \mathbb{Q} : q < 1\}\).

Given a set of cuts \(A\), let \(\gamma = \bigcup \{x : x \in \alpha\}\). \(\gamma\) is a cut and is \(\sup A\).

### \(\mathbb{R}\) is an Extension of \(\mathbb{Q}\)

\(\mathbb{R}\) contains \(\mathbb{Q}\) as a *subfield*. Associate \(q \in \mathbb{Q}\) with the cut \(q^\star = \{r \in \mathbb{Q} : r < q\}\).

Formally, \(f : \mathbb{Q} \rightarrow \mathbb{R}\) is \(f(q \in \mathbb{Q}) = q^\star\), where \(q^\star = \{r \in \mathbb{Q} : r < q\}\).

Then \(\mathbb{Q}' = \{q^\star : q \in \mathbb{Q}\}\) is a subfield of \(\mathbb{R}\).

Notice “\(\sqrt{2}\)” sits in \(\mathbb{R}\) as the cut \(\gamma = \{q : q^2 < 2 \text{ or } q < 0\}\).

### \(\mathbb{R}\) has the Least Upper Bound Property

Let \(A\) contain cuts with upper bound \(\beta\), and \(\gamma = \bigcup\{\alpha : \alpha \in A\}\), a subset of \(\mathbb{Q}\).

Check that \(\gamma\) is a cut and \(\gamma = \sup A\).

*Proof.* \(\gamma\) is a cut because

- it is nontrivial since it contains cuts that are bounded above,
- it is closed downwards since it is a union of cuts which are closed downwards, so for all \(q \in \gamma\), there exists \(\alpha \in A\) such that \(\gamma \in \alpha\) and for all \(r < q\), \(r \in \alpha\), hence \(r \in \gamma\), and
- it has no largest member since for all \(q \in \gamma\), there exists \(\alpha \in A\) such that \(\gamma \in \alpha\) but \(\alpha\) has no largest member so there exists \(r \in \alpha\) such that $q < r hence \(r \in \gamma\).

\(\gamma\) is an upper bound because \(\gamma\) contains all \(\alpha \in A\) and order on cuts is defined by inclusion.

\(\gamma\) is the least upper bound because for all \(\delta < \gamma\), there exists \(x \in \gamma\) such that \(x \notin \delta\), so there exists \(\alpha \in A\) such that \(x \in \alpha\), hence \(\delta\) is not an upper bound for \(\alpha\) and thus \(\delta\) is not an upper bound for \(A\). QED.

\(\mathbb{R}\) is the *only* ordered field with the least upper bound property.

Consequence: the length “\(\sqrt{2}\)” can be thought of as \(\sup\{1, 1.4, 1.41, 1.414, 1.4142, \dots\}\), whose shorthand is \(1.4142135\dots\).

More generally, \(a^{\frac{1}{n}} := \sup\{r \in \mathbb{Q} : r^n < a\}\), i.e. roots exist.

### Consequences

*Archimedean property:* if \(x, y \in \mathbb{R}\) and \(x > 0\), then there exists \(n \in \mathbb{N}\) such that \(nx > y\).

*Proof (by contradiction).* Consider \(A = \{nx : n \in \mathbb{N}\}\). If \(A\) is bounded by \(y\) (i.e. \(nx < y\) for all \(n\)) then \(A\) has a least upper bound \(\alpha\) (by the least upper bound property). Then \(\alpha - x\) is not an upper bound for \(A\). Hence, \(\alpha - x < mx\) for some \(m \in \mathbb{N}\). Then \(\alpha < (m + 1)x\). Thus \(\alpha\) is not an upper bound for \(A\), a contradiction! Therefore, if \(x, y \in \mathbb{R}\) and \(x > 0\), then there exists \(n \in \mathbb{N}\) such that \(nx > y\). QED.

*Corollary.* If \(x > 0\) then there exists \(n \in \mathbb{N}\) such that \(\frac{1}{n} < x\).

*Theorem* (\(\mathbb{Q}\) is dense in \(\mathbb{R}\))*.* Between \(x, y \in \mathbb{R}\), and \(x < y\), there is a \(q \in \mathbb{Q}\) such that \(x < q < y\).

*Proof.* Choose \(n\) such that \(\frac{1}{n} < y - x\) (by Archimedean property). Consider multiples of \(\frac{1}{n}\). These are unbounded (by the Archimedean property). Choose the first multiple such that \(\frac{m}{n} > x\). If \(\frac{m}{n}\) is not less than \(y\) then \(\frac{m - 1}{n} < x\) and \(\frac{m}{n} > y\), which implies \(\frac{1}{n} > y - x\), a constradiction. So, \(\frac{m}{n} < y\). QED.