Field of Real Numbers

October 18, 2024

Proof the real numbers are a complete field.

Remembering the real numbers construction, we can define the set of real numbers as the set of all equivalence classes of Cauchy sequences of rational numbers:

R≡CQ/R. \mathbb{R} \equiv \mathcal{C}_\mathbb{Q} / R.

Then, we can define the field of real numbers as the set of all equivalence classes of Cauchy sequences of rational numbers with the operations of addition and multiplication defined as:

[(an)]+[(bn)]=[(an+bn)],[(an)]⋅[(bn)]=[(an⋅bn)].\begin{aligned} [(a_n)] + [(b_n)] &= [(a_n + b_n)], \\ [(a_n)] \cdot [(b_n)] &= [(a_n \cdot b_n)]. \end{aligned}

These operations are well-defined, as the sum and product of two Cauchy sequences are also Cauchy sequences, and the equivalence classes are preserved under these operations. Thus, the set of real numbers forms a field under these operations, providing a complete model of the real numbers from the rational numbers.

Be (an)∈CQ(a_n) \in \mathcal{C}_\mathbb{Q} a Cauchy sequence of real numbers, then for each nn, we can choose a sequence (bk)∈Q(b_k) \in \mathbb{Q} such that an=[(bk)]a_n = [(b_k)], and an element within this sequence bn∈Qb_n \in \mathbb{Q} that verifies bn<an<bn+1/nb_n < a_n < b_n + 1/n.

For each ϵ′∈R,ϵ′>0\epsilon' \in \mathbb{R}, \epsilon' > 0, we can choose n0∈Nn_0 \in \mathbb{N} such that for any n,m∈Nn,m \in \mathbb{N} with n,m≥n0n,m \geq n_0, we have:

∣an−am∣<ϵ′3. |a_n - a_m| < \frac{\epsilon'}{3}.

Also we can choose n0∈Nn_0 \in \mathbb{N} such that 1/n0<ϵ′/31/n_0 < \epsilon'/3. Then:

∣bn−bm∣ =∣bn−an+an−am+am−bm∣≤∣bn−an∣+∣an−am∣+∣am−bm∣<1n+ϵ′3+1m<1n0+ϵ′3+1n0<ϵ′3+ϵ′3+ϵ′3=ϵ′\begin{aligned} |b_n - b_m| &\ = |b_n - a_n + a_n - a_m + a_m - b_m| \\ &\leq |b_n - a_n| + |a_n - a_m| + |a_m - b_m| \\ &< \frac{1}{n} + \frac{\epsilon'}{3} + \frac{1}{m} \\ &< \frac{1}{n_0} + \frac{\epsilon'}{3} + \frac{1}{n_0} \\ &< \frac{\epsilon'}{3} + \frac{\epsilon'}{3} + \frac{\epsilon'}{3} \\ &= \epsilon' \end{aligned}

So, (bn)(b_n) is a Cauchy sequence of rational numbers, so, exists a∈Ra \in \mathbb{R} such that a=[(bn)]a = [(b_n)]. Now, let’s consider ϵ∈R,ϵ>0\epsilon \in \mathbb{R}, \epsilon > 0, then exists n1∈Nn_1 \in \mathbb{N} such that for any n∈Nn \in \mathbb{N} with n≥n1n \geq n_1, we have:

∣bn−a∣<ϵ2. |b_n - a| < \frac{\epsilon}{2}.

As we did before, we can choose n1∈Nn_1 \in \mathbb{N} such that 1/n1<ϵ/21/n_1 < \epsilon/2. Then:

∣an−a∣≤∣an−bn∣+∣bn−a∣<1n+ϵ2<1n1+ϵ2<ϵ2+ϵ2=ϵ\begin{aligned} |a_n - a| &\leq |a_n - b_n| + |b_n - a| \\ &< \frac{1}{n} + \frac{\epsilon}{2} \\ &< \frac{1}{n_1} + \frac{\epsilon}{2} \\ &< \frac{\epsilon}{2} + \frac{\epsilon}{2} \\ &= \epsilon \end{aligned}

So, (an)(a_n) is a Cauchy that converges to aa, and we can conclude that the set of real numbers is complete.