Ian Charlesworth | Teaching | Math 104

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Due: September 6th, 2019

Math 104 Assignment 1

Properties of ordered sets
Suppose $S$ is an ordered set with ordering $\preceq$.
1. Prove that if $E \subseteq S$, then $\sup E$ is unique if it exists.
2. Prove that if $e \in S$ is a lower bound for $E$ and $e \in E$, then $e = \inf E$.
3. Suppose that $E \subseteq S$ is non-empty, that $x\in S$ is a lower bound for $E$, and that $y \in S$ is an upper bound for $E$. Prove that $x \preceq y$. Must it be true that $x \prec y$?
  (A word on notation: the statement "$E \subseteq S$ is non-empty" here means "$E$ is non-empty and a subset of $S$"; it does not mean "$E$ is a subset of $S$ and $S$ is non-empty".)
4. Prove that if $E \subseteq S$ is finite and non-empty, then $\sup E$ exists in $S$ (hint: use induction). As a result, show that if $S$ is finite then it has the Least Upper Bound Property.
5. Show that $\emptyset$ has a least upper bound if and only if $S$ has a minimum element. (An element $y \in S$ is the minimum of $S$ if $y \preceq x$ for every $x \in S$.)
Suprema depend on the ordered set
1. Give an example of sets $E \subseteq S_1 \subseteq S_2 \subseteq S_3 \subseteq \Q$ such that $E$ has a least upper bound in $S_1$ and in $S_3$, but not in $S_2$.
2. Prove that for any example with the properties above (not only the one you happened to write down), the least upper bound of $E$ in $S_1$ must be different from the least upper bound of $E$ in $S_3$.
3. Does there exist an example with the above properties such that $E = S_1$? Provide an example or prove that it is impossible.
Properties of ordered fields
Let $\F$ be an ordered field.
1. Show that $\F$ does not have a minimum element.
2. Show that $\F$ does not have a minimum positive element: that is, there is no minimum element of the subset $\F_+ = \set{x \in \F \mid x \gt 0}$.
3. Suppose that $E \subseteq \F$ has a least upper bound, and let $x \in \F$. Show that $\sup\set{x + e \mid e \in E} = x + \sup E$.
4. Suppose again that $E \subseteq \F$ has a least upper bound, and let $x \in \F$. Denote by $xE$ the set $\set{xe \mid e \in E}$. Show that if $0 \preceq x$ then $\sup(xE) = x\sup E$, while if $x \preceq 0$ then $\inf(xE) = x\sup E$.
Some explicit extrema
1. Prove that $\sup\set{x + y + z \mid x, y, z \in \Q, 0 \gt x \geq y \gt z} = 0$.
2. Determine which of the following extrema exist in $\R$, and find their values. You need not supply proofs, but you should convince yourself that you could produce a proof if you were asked to do so.
  1. $\inf\set{x + y + z \mid x, y, z \in \Q, 1 \lt x \lt y \lt z}$
  2. $\inf\set{x - y + z \mid x, y, z \in \Q, 1 \lt x \lt y \lt z}$
  3. $\inf\set{x + y - z \mid x, y, z \in \Q, 1 \lt x \lt y \lt z}$
  4. $\sup\set{x + y - z \mid x, y, z \in \Q, 1 \lt x \lt y \lt z}$
  5. $\sup\set{x + y - 2z \mid x, y, z \in \Q, 1 \lt x \lt y \lt z}$
Properties of irrational numbers
1. Prove that if $q \in \Q$ and $t \in \R\setminus \Q$, then $q + t \in \R\setminus\Q$.
2. Prove that if $q \in \Q$ and $t \in \R\setminus \Q$, then $qt \in \set{0}\cup(\R\setminus\Q)$.
3. Prove that if $x, y \in \R$ with $x \lt y$, there is some $t \in \R\setminus\Q$ with $x \lt t \lt y$.

Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.

Further musings on Problem 2
1. Is there an infinite chain of sets as in 2.A.? That is, is there a sequence of sets $S_n$ so that $E \subseteq S_n \subseteq S_{n+1} \subseteq \Q$ holds for every $n \in \N$, and $E$ has a supremum in $S_n$ if and only if $n$ is odd?
2. Is there a set $E \subseteq \Q$ so that $E$ has a supremum in $\Q$ which is different from its supremum in $\R$?
Partial orders can be extended to orders
A partial order on a set $S$ is a relation $\preceq$ on $S$ which is transitive and antisymmetric (that is, if $x, y \in S$ with $x \preceq y$ and $y \preceq x$ then $x = y$, and if $x, y, z \in S$ with $x \preceq y$ and $y \preceq z$ then $x \preceq z$). In particular, note that $S$ may have incomparable elements: $x, y \in S$ so that neither $x \preceq y$ nor $y \preceq x$.

Prove that if $\preceq$ is a partial order on $S$, there is an order on $S$ which extends $\preceq$: that is, an order $\preceq'$ on $S$ so that if $x, y \in S$ with $x \preceq y$, then $x \preceq' y$.

To prove this, you may assume the following:

Zorn's Lemma. Suppose that $Z$ is a set and $\preceq$ is a partial order on $Z$ with the property that every totally-ordered subset of $Z$ is bounded above in $Z$. Then $Z$ contains a maximal element. (A subset $W \subseteq Z$ is totally ordered if $\preceq$ restricted to $W$ is an order, i.e., if whenever $x, y \in W$ we have either $x \preceq y$ or $y \preceq x$. An element $x \in Z$ is maximal if whenever $y \in Z$ with $x \preceq y$, we have $y = x$.)

Math 104 Assignment 1

Properties of ordered sets

Suprema depend on the ordered set

Properties of ordered fields

Some explicit extrema

Properties of irrational numbers

Further musings on Problem 2

Partial orders can be extended to orders