is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Completeness Axiom of the Real Numbers

A Few Definitions


What is completeness? Intuitively, it just means the real numbers have no “gap”. For example, the rational numbers have gaps at every irrational number.

Bounded Above

This means that a given set has an upper bound. This just means the set doesn’t have infinitely large values.

Least Upper Bound

A least upper bound is the smallest upper bound for a given set.

Least Upper Bound Property

There are a few ways of expressing completeness. One is called the least upper bound property.

MathJax TeX Test Page Every non-empty set in $\mathbb{R}$ with an upper bound has a least upper bound in $\mathbb{R}$.

Theorem: If X is non-empty and bounded above, there exists a sequence in X that converges to the LUB(x).

MathJax TeX Test Page Let $p = LUB(X)$. $$S = \{p - \frac{1}{i}\}_{i = 1}^{\infty}$$ Claim: $\lim S \to p$ We want to prove this statement. We need to do a discovery phase though. $$\frac{1}{i} < \epsilon \to i > \frac{1}{\epsilon}$$ Because we need an N s.t. $\forall i \ge N$, let's let $$i > \text{ceil}\left(\frac{1}{\epsilon}\right) + 1$$ $$\forall \epsilon>0, \exists N \text{ s.t. } \forall i \ge N, |S_i - p| \lt \epsilon$$ Now, we start our proof $$\text{Let } \epsilon > 0$$ $$\text{Let N = } \text{ceil}\left(\frac{1}{\epsilon}\right) + 1$$ $$\text{Let i } \ge N$$ $$i \ge \text{ceil}\left(\frac{1}{\epsilon}\right) + 1 \gt \frac{1}{\epsilon}$$ $$\frac{1}{i} \lt \epsilon$$ $$|p - \frac{1}{i} - p| \lt \epsilon$$ Therefore, the limit approaches p, or the LUB of X.

Convergent Sequences are Bounded

conv (1).png

This is true. Why is this non-obvious? If you have a sequence with five numbers, the lower bound is the minimum of those five, and the upper bound is the maximum of those five. However, an infinite sequence may not necessarily have a finite max and minimum.

MathJax TeX Test Page We want to prove that there exists an M such that $\forall i \in \mathbb{N}, |x_i| < M $. How do we do this? We know that the definition of the limit of a sequence is $\forall \epsilon > 0, \exists N s.t. \forall i \ge N, |x_i - L| < \epsilon$ Because we know that values get closer and closer to the limit, we can split this sequence up into two parts. Values that are within 1 of L and values that aren't. $$\text{Let } \epsilon = 1$$ $$\forall i \ge N, |x_i - L| < 1$$ $$\forall i \ge N, |x_i| - |L| < 1$$ Now, from $x_N ... x_{\infty}$, |L| + 1 is greater than every element. We now can look at all elements from $x_1 \dots x_{N-1}$. Because there are a finite number of elements, we can take the maximum of $\{|x_1|, \dots, |x_{N-1}|\}$ as the bound. Now, we have the bound for the whole sequence: either it's the max after $x_N$ or before $x_N$.

Bounded Sequences are Convergent

Not true. Consider the sequence [0,1,0,1,0,1,0,1 … ] This does not converge. However, this leads us to one of the most important theorems in real analysis.

Bounded Sequences have a Convergent Subsequence

This is also known as the Bolzano-Weierstrass Theorem.

bolzano (1).png
MathJax TeX Test Page Let's say you have an infinite sequence from $a_1$ to $b_1$. We create the following subsequence. Within each interval, there are guaranteed to be infinitely many points in one half of the interval. Take a point in that interval. Next, there are guaranteed to be infinitely many points in one half of that interval. Take a point in that interval. Because we continue to choose points with infinitely many points, we can divide each region in half. Eventually, the sequence will approach a single point.

Extreme Value Theorem

MathJax TeX Test Page The extreme value states that given a compact region (closed and bounded), then there exists a maximum value of $f(x)$ such that x is in the region. Written formally, you write: $$\text{If } f: X \to \mathbb{R} \text{ is real-valued and continuous, then } \exists x \in X \text{ s.t. } \forall y \in X, f(x) \geq f(y)$$

Bolzano-Weierstrass is required to prove this theorem. With this theorem, we are able to optimization, meaning we maximize a function given a constraint.

David Witten

Multivariable Second Derivative Test

Multivariable Differentials