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

  1. (nontriviality) i.e. \(\alpha \neq \emptyset, \mathbb{Q}\),
  2. (closed downwards or closed to the left) if \(p \in \alpha\), \(q \in \mathbb{Q}\), and \(q < p\), then \(q \in \alpha\), and
  3. (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

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

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

Proof of well-definedness.

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

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

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

\(\gamma\) 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.

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