55 lines
2.9 KiB
TeX
55 lines
2.9 KiB
TeX
\documentclass[../main_text.tex]{subfiles}
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\begin{document}
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\section{Week 2}
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\exercise*[1.1.1]
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\begin{tcolorbox}
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Let $F$ be an ordered field and $x,y,z \in F$. If $x < 0$ and $y < z$, then $xy > xz$.
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\end{tcolorbox}
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So let's assume the premise. $F$ is an ordered field and $x,y,z \in F$,
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and we choose $x,y$ and $z$ such that $x < 0$ and $y < z$.
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From $x < 0$ it follows that $(-x) > 0$. From $y < z$ it follows that $0 < z - y$.
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From both of these, we can conclude that $0 < (-x)(z-y)$.
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Working out the right side with the distributive law, gives $0 < (-x*z)-(-x*y)$.
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Using $-1*-1 = 1$, gives $0 < (-xz)-(-xy)$, thus $0 < xy - xz$. The right part can be split again: $xz < xy$.
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Then, the < can be flipped, which gives $xy > xz$. \qed
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\exercise[1.1.2]
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\begin{tcolorbox}
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Let $S$ be an ordered set. Let $A \subset S$ be a non-empty finite subset. Then A is bounded. Furthermore,
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$\text{inf} A$ exists and in in $A$ and $\text{sup} A$ exists and is in $A$.
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\end{tcolorbox}
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In order to prove that $A$ is bounded, we have to prove that it has an upper and a lower bound. Let us prove
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that $A$ is bounded above first.
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In particular, we have to prove that $\exists a \in A$ such that $x \leq b$ for all $x \in E$.
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Since $A$ is non-empty and finite, we can use induction on the cardinality of $A$, since that will always
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be some natural number $n$. So, we have to prove two cases: the base case, where $|A|=1$, and the inductive step,
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where we will assume that when $A$ has an upper bound when it has cardinality $m$,
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then it also has an upper bound when its cardinality is equal to $m+1$.
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The base case is quite simple; if $A = \{x\}$, then $x$ is the greatest element and $A$ has an upper bound.
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Now for the inductive step. We assume that for some set $B \subset S$ with cardinality $m$,
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$B$ is bounded above. Thus, there is some $b \in B$ such that $b$ is greater than all other
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elements in $B$. Now, let's add a new element $h \in S$ to $B$, such that $h$ is distinct from all elements already in
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$B$ and the cardinality of $B$ is now $m+1$. Then, since $S$ is well ordered, we can compare $h$ also to $b$.
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Either $h$ is greater than this $b$, in which case $h$ is the new greatest element, or it is less than $b$, in which
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case $b$ stays the greatest element of $B$. In both cases however, $B$ remains bounded above. \qed
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A similar argument can be made to prove the existence of the lower bound, the supremum of $A$ in $A$ and the infimum
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of $A$ in $A$ \footnote{It might even be that I have already proven that $A$ has a supremum present in $A$.
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Then that's also good enough to show that $A$ is bounded, since in order for $A$ to have a supremum,
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it must also be bounded.}. This will be left to the reader.
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\exercise[1.1.5]
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\begin{tcolorbox}
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Let $S$ be an ordered set. Let $A \subset S$ and suppose $b$ is an upper bound for $A$. Suppose $b \in A$.
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Show that $b = \text{sup} A$.
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\end{tcolorbox}
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\end{document}
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