Home | | Latest | About | Random
# Dedekind completeness and intermediate value theorem.
The real numbers $\mathbb{R}$ as we know is an ordered field that satisfies the **least upperbound property** (LUB), or also known as **Dedekind completeness**. This "completeness axiom" of the reals give the reals a rich structure that enables calculus to work. For instance, we have the following familiar theorems in calculus:
(1) **Intermediate value theorem**: If $f:\mathbb{R}\to\mathbb{R}$ is continuous, and that $a < b$ is such that $f(a)\neq f(b)$, then for any $y$ strictly between $f(a)$ and $f(b)$, there exists some $c$ such that $f(c)=y$.
(2) **Mean value theorem**: If $f:\mathbb{R}\to\mathbb{R}$ is differentiable, then for any $a < b$, there exists some $c\in (a,b)$ such that $f'(c)= \frac{f(b)-f(a)}{b-a}$.
(3) **Ratio test**: If a real sequence $(a_n)$ is such that $|\frac{a_{n+1}}{a_n}|\to L$ with $L < 1$, then $\sum a_n$ converges.
(4) **Alternating series test**: If $a_n \to 0$ with $a_n > 0$, then the alternating series $\sum (-1)^n a_n$ converges.
(5) **Connectedness**: $\mathbb{R}$ is not the union of nonempty proper open sets.
(6) **Cauchy sequences in $\mathbb{R}$ converges**.
... etc.
As it turns out, some of these calculus theorems are in fact **equivalent** to Dedekind completeness in an ordered field (and **some are not**). In some sense, these Dedekind equivalent ones can be thought a a core theorem that one desires for our theory of calculus, and strong enough to define the reals (and consequently the rest of calculus)!
Let us focus on the intermediate value theorem. This is an "intuitive" statement: If you have the graph of a continuous function $f(x)$ through two points $(a,f(a))$ and $(b,f(b))$, then the graph ought to go through all the values between $f(a)$ and $f(b)$. As it turns out, we indeed have the equivalence:
> **PROPOSITION.**
> In an ordered field $(X,\le)$ with the natural ordered topology on $X$, $X$ is Dedekind complete if and only if $X$ has the intermediate value property.
Here an ordered field $X$ having the **intermediate value property** we mean: For any continuous function $f:X\to X$, for any $a < b$ in $X$ such that $f(a) \neq f(b)$, if $y$ is any value in between $f(a)$ and $f(b)$, then there exists some $c \in (a,b)$ such that $f(c) = y$.
And since we are working with in an arbitrary ordered field $(X,\le)$, we ought to define what we mean by its order topology and continuity on $X$.
## Continuity in an order topology.
Let $(X,\le)$ be a **totally ordered set** (often also known as **linearly ordered set**), then we can define an a natural **order topology** on $X$ by specifying every open sets in $X$ as some union (possibly infinite) of the following:
(1) Open intervals : $(a,b)=\{x \in X : a < x < b\}$, and
(2) Open rays: $(a,\rightarrow) = \{x\in X : x > a\}$ and $(\leftarrow,a) = \{x \in X : x < a\}$.
(In the usual reals, this agrees with the usual real metric on $\mathbb{R}$.)
Okay, now we have open sets, we can speak of continuity in $X$ with respect to this order topology.
For $(X,\le)$ a totally ordered set with the natural order topology on $X$, we say a function $f:X\to X$ is **continuous at $x \in X$** if for any open interval $J$ containing $f(x)$, one can find an open interval $I$ containing $x$ such that the image $f(I)\subset J$.
In other words, all near-by (enough) points of $x$ will all map to near-by points of $f(x)$.
![[1 teaching/CSU summer 2023/problems/---files/dedekind-completeness-and-intermediate-value-theorem 2023-06-05 11.08.03.excalidraw.svg]]
%%[[1 teaching/CSU summer 2023/problems/---files/dedekind-completeness-and-intermediate-value-theorem 2023-06-05 11.08.03.excalidraw|🖋 Edit in Excalidraw]], and the [[CSU summer 2023/problems/---files/dedekind-completeness-and-intermediate-value-theorem 2023-06-05 11.08.03.excalidraw.dark.svg|dark exported image]]%%
We further say $f:X\to X$ is **continuous** if it is continuous at each point $x \in X$.
![[---images/---assets/---icons/question-icon.svg]] When $\mathbb{R},\mathbb{Q}, \mathbb{Z}$ are endowed with their usual ordered topology, it is equivalent two their usual metric topology. Consider the piecewise function
$$
f(x)=\begin{cases}
1 & \text{if } x \ge 0\\ 0 & \text{if } x < 0
\end{cases}
$$as functions from $\mathbb{R}\to \mathbb{R}$, $\mathbb{Q}\to \mathbb{Q}$, and $\mathbb{Z}\to \mathbb{Z}$. Which of these is continuous?
What about the function
$$f(x)=\begin{cases}
1 & \text{if } x^2 \ge 2 \\ 0 & \text{else}
\end{cases}$$as a functions on $\mathbb{R}$, $\mathbb{Q}$, and $\mathbb{Z}$?
![[---images/---assets/---icons/question-icon.svg]] Show that $\mathbb{Q}$ and $\mathbb{Z}$ do not satisfy intermediate value property.
## The reals satisfy the intermediate value property.
![[---images/---assets/---icons/question-icon.svg]] Using the fact that the reals are Dedekind complete (having the least upperbound property), the reals have the intermediate value property.
## Intermediate value property implies Dekekind completeness.
![[---images/---assets/---icons/question-icon.svg]] Let $(X,\le)$ be an ordered field, show that if $X$ has the intermediate value property, then every nonempty bounded above subset $A \subset X$ has a least upperbound in $X$, namely, $\sup(A)$ exists in $X$.
Hint. Show by contradiction, suppose there exists a nonempty subset $A$ that is bounded above, but $A$ has no least upperbound, consider the function
$$
f(x)=\begin{cases}
1 & \text{if } x \text{ is an upperbound of } A \\
0 & \text{else}
\end{cases}
$$
Use this function to help you.
## Other notions in an ordered topology.
Let us define some other notions in an ordered topology, which are all compatible with the usual metric topology on $\mathbb{R}$.
Given a sequence $(a_n)$ in an ordered field $(X,\le)$, what does it mean for it to converge?
Let us define it this way: We say $(a_n)$ **converges** in $X$ if there exists some $a \in X$ such that for every open interval $I \ni a$, there exists a natural $N$ such that $n > N$ implies $a_n \in I$. In this case, we write $a_n \to a$, and say $a$ is the **limit** of the sequence.
Given a function $f : X \to X$, what does it mean to say the limit of $f$ exists at $x=a$?
We say the limit of $f$ exists at $x=a$ if there exists some $L \in X$ such that for every open interval $J$ containing $L$, there exists some open interval $I$ containing $a$ such that the image of the punctured interval $I-\{a\}$ lies in $J$. In this case we write $\lim_{x\to a} f(x) = L$.
Given a function $f:X\to X$, what does it mean to say $f$ is differentiable at $x=a$?
Here we need to carry out arithmetics, so $X$ is in particular an ordered field (note previously we need not $X$ to be a field). We say $f:X\to X$ is differentiable at $x=a$ if the limit of the difference quotient function $g(x) = \begin{cases}\frac{f(x)-f(a)}{x-a} & \text{if }x\neq a \\ \ast & \text{else}\end{cases}$ exists at $x=a$. In this case, the value of the limit of $g$ at $x=a$ is denoted as $f'(a)$, and called the derivative of $f$ at $x=a$.
Given an series $\sum a_n$, what do we mean by the series converge in $X$?
We define a series $\sum a_n$ converges if the partial sum sequence $s_n = a_1 + \cdots + a_n$ converges in $X$.
![[---images/---assets/---icons/question-icon.svg]] Show that in an ordered field $(X,\le)$, if $X$ satisfies the mean value theorem, then $X$ is Dedekind complete.