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Real Analysis Proofs #1

Avishek Sen Gupta on 18 May 2021

Since I’m currently self-studying Real Analysis, I’ll be listing down proofs I either initially had trouble understanding, or enjoyed proving, here. These are very mathematical posts, and are for personal documentation, mostly.

Recursive Definitions

Source: Analysis 1 by Terence Tao

Definitions

Peano Axioms Used

  1. \(0\) is a natural number.
  2. If \(n\) is a natural number, \(\mathbf{n++}\) is also a natural number.
  3. \(0\) is not the successor to any natural number, i.e., \(n++ \neq 0, \forall n\in\mathbb{N}\).
  4. Different natural numbers must have different successors. If \(m\neq n\), then \(m++ \neq n++\). Conversely, if \(m++ \neq n++\), then \(m=n\).

Proposition

Suppose there exists a function \(f_n:\mathbb{R}\rightarrow\mathbb{R}\). Let \(c\in\mathbb{N}\). Then we can assign a unique natural number \(a_n\) for each natural number \(n\), such that \(a_0=c\), and \(a_{n++}=f_n(a_n) \forall n\in\mathbb{N}\).

Proof by Induction

For zero

Let \(0\) be assigned \(a_0=c\). Then, \(a_{0++}=f_0(a_0)\). Since \(0\) is never a successor to any natural number by Axiom (3), \(a_0\) will not recur as for \(a_{0++}\).

For \(n\)

From Axiom (4), we can infer that:

\[n++\neq n,n-1,n-2,...,1,0 \\ \Rightarrow a_{n++}\neq a_n,a_{n-1},a_{n-2},...,a_1,a_0\]

Thus, \(a_{n++}\) is unique in the set \(\{a_0,a_1,a_2,...,a_n,a_{n++}\}\).

For \(n++\)

By extension, for \((n++)++\), we can write:

\[(n++)++\neq n++,n,n-1,n-2,...,1,0 \\ \Rightarrow a_{(n++)++}\neq a_{n++},a_n,a_{n-1},a_{n-2},...,a_1,a_0\]

Thus, \(a_{(n++)++}\) is unique in the set \(\{a_0,a_1,a_2,...,a_n,a_{n++},a_{(n++)++}\}\). Thus, we can assign a unique natural number \(a_{(n++)++}\) such that \(a_{(n++)++}=f_{n++}(a_{n++})\).

\[\blacksquare\]

Proof of Existence of Real Cube Roots

Let \(r\in\mathbb{N}\) For the case of \(r=0\), the cube root is \(0\).

Consider the set \(\mathbb{S}=\{x:x^3\leq r, x\in \mathbb{R}, r\in \mathbb{R}\}\).

This set is non-empty because $0\in\mathbb{S}$. It is also bounded by \(r+1\) because \((r+1)^3=r^3+3r^2+3r+1>r\).

Therefore, by the Completeness Axiom, \(\mathbb{S}\) has a least upper bound. Denote this least upper bound by \(x\).

By the Trichotomy property, these are the possible cases:

Case 1: Assume that: \(\mathbb{x^3<r}\)

Then, by our definition of \(\mathbb{S}\), \(x\in\mathbb{S}\) and is its least upper bound, i.e., there are no elements in \(\mathbb{S}\) which are greater than \(x\).

If the cube of the least upper bound \(x\) is less than \(r\), then it is enough to show that there exists a \(x+\delta:\delta>0\) whose cube is also less than \(r\).

Assume that \(0<\delta<1\). There can exist \(\delta>1\), but that would restrict the choice of upper bounds we have to play about with:

Then, we’d like to find a \(0<\delta<1\) such that \((x+\delta)^3<r\). This gives us:

\[(x+\delta)^3<r \\ x^3+\delta^3+3x^2\delta+3x\delta^2<r \\ (x^3-r)+\delta^3+3x^2\delta+3x\delta^2<0\]

We know that \(3x\delta^2<3x\delta\) is a positive quantity, and note that \(\delta^3<\delta\), thus we can say:

\[(x^3-r)+\delta^3+3x^2\delta+3x\delta^2<(x^3-r)+\delta+3x^2\delta+3x\delta\]

Then, it is enough to prove that:

\[(x^3-r)+\delta+3x^2\delta+3x\delta<0\]

With some algebraic manipulation, we get:

\[(x^3-r)+\delta+3x^2\delta+3x\delta<0 \\ \Rightarrow \delta(1+3x^2+3x)<r-x^3 \\ \Rightarrow \delta<\frac{r-x^3}{1+3x^2+3x}\]

If we assume \(\delta=\frac{1}{k}:k\in\mathbb{N}\), then we can say:

\[k>\frac{1+3x^2+3x}{r-x^3}: k\in\mathbb{N}\]

Since the Archimedean property states that natural numbers have no upper bound, \(k\) must exist. This means, we have proven that there is a \(k\in\mathbb{N}\), for which there exists a cube root \((x+\frac{1}{k})\) which is larger than \(x\), such that \((x+\frac{1}{k})^3<r\). This implies that \((x+\frac{1}{k})\) exists in \(\mathbb{S}\). However, this contradicts our assumption that no element greater than \(x\) exists in \(\mathbb{S}\).

Thus, the statement \(x^3<r\) is false.

Case 2: Assume that: \(\mathbb{x^3>r}\)

If the cube of the least upper bound \(x\) is greater than \(r\), then it is enough to show that there exists a \(x-\delta:\delta>0\) whose cube is also greater than \(r\).

Assume that \(0<\delta<1\). There can exist \(\delta>1\), but that would restrict the choice of upper bounds we have to play about with:

Then, we’d like to find a \(0<\delta<1\) such that \((x+\delta)^3>r\). This gives us:

\[(x-\delta)^3>r \\ x^3-\delta^3-3x^2\delta+3x\delta^2>r \\ (x^3-r)-\delta^3-3x^2\delta+3x\delta^2>0 \\\]

Again note that since \(\delta^3<\delta\), and \(3x\delta^2\) is positive, we can write:

\[(x^3-r)-\delta^3-3x^2\delta+3x\delta^2>(x^3-r)-\delta-3x^2\delta \\\]

Thus it is enough to prove that:

\[(x^3-r)-\delta-3x^2\delta>0\]

Some algebraic manipulation gives us:

\[(1+3x^2)\delta<x^3-r \\ \Rightarrow \delta<\frac{x^3-r}{1+3x^2}\]

If we assume \(\delta=\frac{1}{k}:k\in\mathbb{N}\), then we can say:

\[k>\frac{1+3x^2}{x^3-r}: k\in\mathbb{N}\]

Since the Archimedean property states that natural numbers have no upper bound, \(k\) must exist. This means, we have proven that there is a \(k\in\mathbb{N}\), for which there exists a cube root \((x-\frac{1}{k})\) which is smaller than \(x\), such that \((x-\frac{1}{k})^3<r\). Thus, \((x+\frac{1}{k})\) is a least upper bound for \(\mathbb{S}\); however, this contradicts our assumtion that \(x\) is the least upper bound.

Thus, the statement \(x^3>r\) is false.

Thus, the only possibility is that Case 3 is true, i.e., \(x^3=r\), thus implying the existence of real cube roots of real numbers.

\[\blacksquare\]
tags: Real Analysis - Mathematics - Proof