Real Numbers

See also the Dedekind-cut construction of $ℝ$ .

Rational Sequence

A rational sequence

(a_{n})

is a sequence such that

a_{n} \in ℚ

Cauchy Sequence

Let

(a_{n})

be a rational sequence. We say that

(a_{n})

is a cauchy sequence if for any

ϵ \in ℝ^{+}

there exists some

N \in ℕ_{0}

such that for every

n, m \geq N

| a_{n} - a_{m} | \leq ϵ

A Convergent Rational Sequence is Cauchy

Suppose that

a_{n} \to L

for some

L \in ℝ

, then

(a_{n})

is cauchy

Let

ϵ \in ℝ^{+}

since

(a_{n})

converges to

q

then we know that there exists some

M \in ℕ_{0}

such that for any

n \geq N

we have that

| a_{n} - L | < \frac{ϵ}{2}

, take

N = M

and let

n, m \geq M

then we have

| a_{n} - a_{m} | = | (a_{n} - L) - (a_{m} - L) | \leq | a_{n} - L | + | a_{m} - q | < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

where we've used the triangle inequality

A Cauchy Sequence is Bounded

(a_{n})

is a cauchy sequence, then there exists some

M \in ℚ

such that

| a_{n} | \leq M

Since $(a_{n})$ is cauchy then with $ϵ = 1$ we get an $N \in ℕ_{0}$ such that for any $m, n \geq N$ we have that $| a_{n} - a_{m} | < 1$ . Specificially since $N + 1 \geq N$ then we would know that $| a_{N + 1} - a_{m} | \leq 1$ which is the same as $a_{N + 1} - 1 < a_{n} < a_{N + 1} + 1$ so that $| a_{n} | \leq \max (| a_{N + 1} - 1 |, | a_{N + 1} + 1 |)$

Therefore set $M = \max (| a_{0} |, | a_{1} |, \dots, | a_{N} |, | a_{N + 1} - 1 |, | a_{N + 1} + 1 |)$ With this, let $k \in ℕ_{0}$ then we know that if $k \leq N$ then $| a_{k} | \leq M$ as they are directly included in our maximum, on the other hand if $k > N$ then $| a_{k} | < \max (| a_{N + 1} - 1 |, | a_{N + 1} + 1 |) \leq M$ and therefore $(a_{n})$ is bounded.

The Collection of All Cauchy Sequence

We use the notation

𝒞_{ℚ}

to denote the set of all cauchy sequences of rational numbers

Cauchy Subtraction Relation

Let

(a_{n}), (b_{n}) \in 𝒞_{ℚ}

, and we define the relation such that they are related if

a_{n} - b_{n} \to 0

The Cauchy Subtraction Relation is an Equivalence Relation

The cauchy subtraction relation is an equivalence relation

Let $(a_{n}), (b_{n}), (c_{n}) \in 𝒞_{ℚ}$ , let's show that the relation is reflexive, first we have to show that $a_{n} - a_{n} \to 0$ the sequence $a_{n} - a_{n}$ is zero for every index so it trivially tends to 0.

To show that the relation is symmetric we assume that $a_{n} - b_{n} \to 0$ , and we must show that $b_{n} - a_{n} \to 0$ , so let $ϵ \in ℝ^{+}$ , so we get some $N \in ℕ_{0}$ such that for all $n \geq ℕ_{0}$ we have that $| a_{n} - b_{n} | < ϵ$ but since we can move things around inside of the absolute value bars we get $| b_{n} - a_{n} | < ϵ$ as needed.