Announcement: This is reading notes excerpted from Walter Rudin’s Principles of Mathematical Analysis, from Chapter one, with special attention on how to construct \(\mathbb{R}\) from \(\mathbb{Q}\). This question arised with a theorem’s proof in the book Probability Essentials presented as below.
Theorem 2.1. The Borel \(\sigma\)-algebra of \(\mathbb{R}\) is generated by intervals of the form \((-\infty, a]\), where \(a \in \mathbb{Q}\) (\(\mathbb{Q}\) = rationals).
So first of all, what is a rational number. From wikipedia, a rational number is any number that could be represented as the quotient or fraction of two integers, \(\frac{p}{q}\). Although, as a causalist, this leads me to question ‘What is an integer?’, we here not care about its mathematical rigor and assume the axiomatic result.
So to prove \(\sqrt{2}\) is not a rational number, means to prove that \(\sqrt{2}\) can not be expressed as the form \(\frac{p}{q}\). We use proof by contradiction here:
Proof.
Suppose \(\sqrt{2}\) can be expressed as \(\frac{p}{q}\). More formally, the assumption can be written as: \(\exists p, q \in \mathbb{Z}, \sqrt{2} = \frac{p}{q}\). Then, \(\sqrt{2} q = p \Rightarrow 2q^2 = p^2\). So there exits p, q such that \(2p^2 = q^2\) holds. When problem reduces to equation of integers, the use of parity is always turned to. If q is odd, so does \(q^2\), the equality will not hold, since the left-hand side of the equal should be even. So q is even. Let \(p = 2^{i}k\), \(q=2^{j}l\), where \(k, l\) are odd and \(j > 0\). According to the equality, we have: \(2 * 2^{2i}k^2 = 2^{2j}l^2\). That is: \(2^{2i+1-2j}k^2 = l^2\). It is impossible, since \(k, l\) are odd, and \(2i+1-2j \ne 0\). So the assumption is not true. \(\square\)
Def. Order. an order is a relation \(R=\{<,=,>\}\) over a set, such that only three specific relationships hold between any x and y in the set and each of them is transitive, denoted as \(xRy\).
So an Ordered Set \(S\) is a set where an order is defined. Since \(S\) is an ordered set, we could define bounds for its subset \(E \subset S\). If there exists \(\beta \in S\) such that \(x \le \beta\) for every \(x \in E\), we say that \(E\) is bounded above and \(\beta\) is an Upper Bound of \(E\). Lower can be defined similarly.
Moreover, if for any \(\gamma < \beta\) and \(\gamma \in S\), \(\gamma\) is not an upper bound of \(E\), then \(\beta\) is the Least Upper Bound of set \(E\), which is denoted as \(\beta\) = sup\(E\). Greatest lower bound can be defined in the similar way, and is denoted by \(\beta\) = inf\(E\)
An ordered set \(S\) is said to have least-upper-bound property if for any subset \(E \in S\) and \(E \ne \phi\), if \(E\) is bounded above, then \(E\) has its least upper bound \(\beta\) = sup\(E \in S\). Then we could have a theorem that connects the relationship between least upper bound and greatest lower bound.
Theorem.
Suppose \(S\) is a set that has least-upper-bound property. If \(E \in S\) is bounded below and we denote all possible lower bounds of \(E\) as a set \(L\). According to the least-upper-bound property of \(S\), suppose the least uppper bound of \(L\) is \(\alpha\) = sup\(L\). Then we have: \(\alpha\) = inf\(E\).
Def. Field. A field is a set \(F\) with two operations, i.e. addition, multiplication, which satisfies the following field axioms:
(A) Axioms for addition
- (A1) If \(x, y \in F\), then their sum \(x + y \in F\).
- (A2) Addition is commutative, i.e. \(x + y = y + x\).
- (A3) Addition is associative, i.e. \(x + (y + z) = (x + y) + z\).
- (A4) \(F\) contains an element 0 such that \(0 + x = x\) for every \(x \in F\).
- (A5) To every \(x \in F\) corresponds an element \(-x \in F\) such that \(x + -x = 0\).
(M) Axioms for multiplication
- (M1) If \(x, y \in F\), then their sum \(x y \in F\).
- (M2) Addition is commutative, i.e. \(x y = y x\).
- (M3) Addition is associative, i.e. \(x (y z) = (x y) z\).
- (M4) \(F\) contains an element 1 such that \(1 x = x\) for every \(x \in F\).
- (M5) To every \(x \in F\) corresponds an element \(1/x \in F\) such that \(x (1/x) = 1\).
(D) Distributive Law
- \(x(y + z) = xy + xz\) holds for all \(x, y, z \in F\).