Finish formally proving Apostol chapter I.3.
parent
5256c4e81a
commit
acd8b3edff
|
@ -131,7 +131,7 @@ lemma leq_nat_abs_ceil_self (x : ℝ) : x ≤ Int.natAbs ⌈x⌉ := by
|
|||
unfold Int.natAbs
|
||||
have k' : k = ⌈x⌉ := rfl
|
||||
rw [←k']
|
||||
have _ : k ≥ 0 := by -- Hint for match below
|
||||
have : k ≥ 0 := by -- Hint for match below
|
||||
rw [k', ge_iff_le]
|
||||
exact Int.ceil_nonneg (ge_iff_le.mp h)
|
||||
match k with
|
||||
|
@ -150,12 +150,12 @@ lemma leq_nat_abs_ceil_self (x : ℝ) : x ≤ Int.natAbs ⌈x⌉ := by
|
|||
For every real `x` there exists a positive integer `n` such that `n > x`.
|
||||
-/
|
||||
theorem exists_pnat_geq_self (x : ℝ) : ∃ n : ℕ+, ↑n > x := by
|
||||
let x' : ℕ+ := ⟨Int.natAbs ⌈x⌉ + 1, by simp⟩
|
||||
have h : x < x' := calc x
|
||||
let n : ℕ+ := ⟨Int.natAbs ⌈x⌉ + 1, by simp⟩
|
||||
have : x < n := calc x
|
||||
_ ≤ Int.natAbs ⌈x⌉ := leq_nat_abs_ceil_self x
|
||||
_ < ↑↑(Int.natAbs ⌈x⌉ + 1) := by simp
|
||||
_ = x' := rfl
|
||||
exact ⟨x', h⟩
|
||||
_ = n := rfl
|
||||
exact ⟨n, this⟩
|
||||
|
||||
/-- #### Theorem I.30
|
||||
|
||||
|
@ -451,7 +451,7 @@ theorem forall_mem_le_forall_mem_imp_sup_le_inf (S T : Set ℝ)
|
|||
(hS : S.Nonempty) (hT : T.Nonempty)
|
||||
(p : ∀ s ∈ S, ∀ t ∈ T, s ≤ t)
|
||||
: ∃ (s : ℝ), IsLUB S s ∧ ∃ (t : ℝ), IsGLB T t ∧ s ≤ t := by
|
||||
-- Sshow a supremum of `S` and an infimum of `T` exists (since each set bounds
|
||||
-- Show a supremum of `S` and an infimum of `T` exists (since each set bounds
|
||||
-- above and below the other, respectively).
|
||||
let ⟨s, hs⟩ := hS
|
||||
let ⟨t, ht⟩ := hT
|
||||
|
|
|
@ -12,7 +12,27 @@
|
|||
|
||||
\header{A Set of Axioms for the Real-Number System}{Tom M. Apostol}
|
||||
|
||||
\section*{\proceeding{Theorem I.27}}%
|
||||
\section*{\verified{Lemma 1}}%
|
||||
\hyperlabel{sec:lemma-1}%
|
||||
|
||||
Nonempty set $S$ has supremum $L$ if and only if set $-S$ has infimum $-L$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{is\_lub\_neg\_set\_iff\_is\_glb\_set\_neg}
|
||||
|
||||
\divider
|
||||
|
||||
Suppose $L = \sup{S}$ and fix $x \in S$.
|
||||
By definition of the supremum, $x \leq L$ and $L$ is the smallest value
|
||||
satisfying this inequality.
|
||||
Negating both sides of the inequality yields $-x \geq -L$.
|
||||
Furthermore, $-L$ must be the largest value satisfying this inequality.
|
||||
Therefore $-L = \inf{-S}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.27}}%
|
||||
\hyperlabel{sec:theorem-i.27}%
|
||||
|
||||
Every nonempty set $S$ that is bounded below has a greatest lower bound; that
|
||||
|
@ -22,9 +42,17 @@ is, there is a real number $L$ such that $L = \inf{S}$.
|
|||
|
||||
\link{exists\_isGLB}
|
||||
|
||||
\divider
|
||||
|
||||
Let $S$ be a nonempty set bounded below by $x$.
|
||||
Then $-S$ is nonempty and bounded above by $x$.
|
||||
By the completeness axiom, there exists a supremum $L$ of $-S$.
|
||||
By \nameref{sec:lemma-1}, $L$ is a supremum of $-S$ if and only if $-L$ is an
|
||||
infimum of $S$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.29}}%
|
||||
\section*{\verified{Theorem I.29}}%
|
||||
\hyperlabel{sec:theorem-i.29}
|
||||
|
||||
For every real $x$ there exists a positive integer $n$ such that $n > x$.
|
||||
|
@ -33,9 +61,18 @@ For every real $x$ there exists a positive integer $n$ such that $n > x$.
|
|||
|
||||
\link{exists\_pnat\_geq\_self}
|
||||
|
||||
\divider
|
||||
|
||||
Let $n = \abs{\ceil{x}} + 1$.
|
||||
It is trivial to see $n$ is a positive integer satisfying $n \geq 1$.
|
||||
Thus all that remains to be shown is that $n > x$.
|
||||
If $x$ is nonpositive, $n > x$ immediately follows from $n \geq 1$.
|
||||
If $x$ is positive,
|
||||
$$x = \abs{x} \leq \abs{\ceil{x}} < \abs{\ceil{x}} + 1 = n.$$
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.30}}%
|
||||
\section*{\verified{Theorem I.30}}%
|
||||
\hyperlabel{sec:theorem-i.30}%
|
||||
|
||||
If $x > 0$ and if $y$ is an arbitrary real number, there exists a positive
|
||||
|
@ -47,82 +84,324 @@ integer $n$ such that $nx > y$.
|
|||
|
||||
\link{exists\_pnat\_mul\_self\_geq\_of\_pos}
|
||||
|
||||
\divider
|
||||
|
||||
Let $x > 0$ and $y$ be an arbitrary real number.
|
||||
By \nameref{sec:theorem-i.29}, there exists a positive integer $n$ such that
|
||||
$n > y / x$.
|
||||
Multiplying both sides of the inequality yields $nx > y$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.31}}%
|
||||
\section*{\verified{Theorem I.31}}%
|
||||
\hyperlabel{sec:theorem-i.31}%
|
||||
|
||||
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
|
||||
$$a \leq x \leq a + \frac{y}{n}$$
|
||||
for every integer $n \geq 1$, then $x = a$.
|
||||
$$a \leq x \leq a + \frac{y}{n}$$ for every integer $n \geq 1$, then $x = a$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{forall\_pnat\_leq\_self\_leq\_frac\_imp\_eq}
|
||||
|
||||
\divider
|
||||
|
||||
By the trichotomy of the reals, there are three cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $x = a$.
|
||||
Then we are immediately finished.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $x < a$.
|
||||
But by hypothesis, $a \leq x$.
|
||||
Thus $a < a$, a contradiction.
|
||||
|
||||
\paragraph{Case 3}%
|
||||
|
||||
Suppose $x > a$.
|
||||
Then there exists some $c > 0$ such that $a + c = x$.
|
||||
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
|
||||
$nc > y$.
|
||||
Rearranging terms, we see $y / n < c$.
|
||||
Therefore $a + y / n < a + c = x$.
|
||||
But by hypothesis, $x \leq a + y / n$.
|
||||
Thus $a + y / n < a + y / n$, a contradiction.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since these cases are exhaustive and both case 2 and 3 lead to
|
||||
contradictions, $x = a$ is the only possibility.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.32}}%
|
||||
\hyperlabel{sec:theorem-i.32}%
|
||||
\section*{\verified{Lemma 2}}%
|
||||
\hyperlabel{sec:lemma-2}%
|
||||
|
||||
Let $h$ be a given positive number and let $S$ be a set of real numbers.
|
||||
\begin{enumerate}[(a)]
|
||||
\item If $S$ has a supremum, then for some $x$ in $S$ we have
|
||||
$$x > \sup{S} - h.$$
|
||||
\item If $S$ has an infimum, then for some $x$ in $S$ we have
|
||||
$$x < \inf{S} + h.$$
|
||||
\end{enumerate}
|
||||
If three real numbers $a$, $x$, and $y$ satisfy the inequalities
|
||||
$$a - y / n \leq x \leq a$$ for every integer $n \geq 1$, then $x = a$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\ % Force space prior to *Proof.*
|
||||
\link{forall\_pnat\_frac\_leq\_self\_leq\_imp\_eq}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item \link{sup\_imp\_exists\_gt\_sup\_sub\_delta}
|
||||
\item \link{inf\_imp\_exists\_lt\_inf\_add\_delta}
|
||||
\end{enumerate}
|
||||
\divider
|
||||
|
||||
By the trichotomy of the reals, there are three cases to consider:
|
||||
|
||||
\paragraph{Case 1}%
|
||||
|
||||
Suppose $x = a$.
|
||||
Then we are immediately finished.
|
||||
|
||||
\paragraph{Case 2}%
|
||||
|
||||
Suppose $x < a$.
|
||||
Then there exists some $c > 0$ such that $x = a - c$.
|
||||
By \nameref{sec:theorem-i.30}, there exists an integer $n > 0$ such that
|
||||
$nc > y$.
|
||||
Rearranging terms, we see that $y / n < c$.
|
||||
Therefore $a - y / n > a - c = x$.
|
||||
But by hypothesis, $x \geq a - y / n$.
|
||||
Thus $a - y / n < a - y / n$, a contradiction.
|
||||
|
||||
\paragraph{Case 3}%
|
||||
|
||||
Suppose $x > a$.
|
||||
But by hypothesis $x \leq a$.
|
||||
Thus $a < a$, a contradiction.
|
||||
|
||||
\paragraph{Conclusion}%
|
||||
|
||||
Since these cases are exhaustive and both case 2 and 3 lead to
|
||||
contradictions, $x = a$ is the only possibility.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.33}}%
|
||||
\section*{Theorem I.32}%
|
||||
\hyperlabel{sec:theorem-i.32}%
|
||||
|
||||
Let $h$ be a given positive number and let $S$ be a set of real numbers.
|
||||
|
||||
\subsection*{\verified{Theorem I.32a}}%
|
||||
\hyperlabel{sub:theorem-i.32a}%
|
||||
|
||||
If $S$ has a supremum, then for some $x$ in $S$ we have $x > \sup{S} - h$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{sup\_imp\_exists\_gt\_sup\_sub\_delta}
|
||||
|
||||
\divider
|
||||
|
||||
By definition of a supremum, $\sup{S}$ is the least upper bound of $S$.
|
||||
For the sake of contradiction, suppose for all $x \in S$,
|
||||
$x \leq \sup{S} - h$.
|
||||
This immediately implies $\sup{S} - h$ is an upper bound of $S$.
|
||||
But $\sup{S} - h < \sup{S}$, contradicting $\sup{S}$ being the \textit{least}
|
||||
upper bound.
|
||||
Therefore our original hypothesis was wrong.
|
||||
That is, there exists some $x \in S$ such that $x > \sup{S} - h$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\subsection*{\verified{Theorem I.32b}}%
|
||||
\hyperlabel{sub:theorem-i.32b}%
|
||||
|
||||
If $S$ has an infimum, then for some $x$ in $S$ we have $x < \inf{S} + h$.
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{inf\_imp\_exists\_lt\_inf\_add\_delta}
|
||||
|
||||
\divider
|
||||
|
||||
By definition of an infimum, $\inf{S}$ is the greatest lower bound of $S$.
|
||||
For the sake of contradiction, suppose for all $x \in S$,
|
||||
$x \geq \inf{S} + h$.
|
||||
This immediately implies $\inf{S} + h$ is a lower bound of $S$.
|
||||
But $\inf{S} + h > \inf{S}$, contradicting $\inf{S}$ being the
|
||||
\textit{greatest} lower bound.
|
||||
Therefore our original hypothesis was wrong.
|
||||
That is, there exists some $x \in S$ such that $x < \inf{S} + h$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{Theorem I.33}%
|
||||
\hyperlabel{sec:theorem-i.33}%
|
||||
|
||||
Given nonempty subsets $A$ and $B$ of $\mathbb{R}$, let $C$ denote the set
|
||||
$$C = \{a + b : a \in A, b \in B\}.$$
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
|
||||
$$\sup{C} = \sup{A} + \sup{B}.$$
|
||||
\item If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
|
||||
$$\inf{C} = \inf{A} + \inf{B}.$$
|
||||
\end{enumerate}
|
||||
|
||||
\note{This is known as the "Additive Property."}
|
||||
|
||||
\subsection*{\verified{Theorem I.33a}}%
|
||||
\hyperlabel{sub:theorem-i.33a}%
|
||||
|
||||
If each of $A$ and $B$ has a supremum, then $C$ has a supremum, and
|
||||
$$\sup{C} = \sup{A} + \sup{B}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\ % Force space prior to *Proof.*
|
||||
\link{sup\_minkowski\_sum\_eq\_sup\_add\_sup}
|
||||
|
||||
\begin{enumerate}[(a)]
|
||||
\item \link{sup\_minkowski\_sum\_eq\_sup\_add\_sup}
|
||||
\item \link{inf\_minkowski\_sum\_eq\_inf\_add\_inf}
|
||||
\end{enumerate}
|
||||
\divider
|
||||
|
||||
We prove (i) $\sup{A} + \sup{B}$ is an upper bound of $C$ and (ii)
|
||||
$\sup{A} + \sup{B}$ is the \textit{least} upper bound of $C$.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\hyperlabel{par:theorem-i.33a-i}%
|
||||
|
||||
Let $x \in C$.
|
||||
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
||||
that $x = a' + b'$.
|
||||
By definition of a supremum, $a' \leq \sup{A}$.
|
||||
Likewise, $b' \leq \sup{B}$.
|
||||
Therefore $a' + b' \leq \sup{A} + \sup{B}$.
|
||||
Since $x = a' + b'$ was arbitrarily chosen, it follows $\sup{A} + \sup{B}$
|
||||
is an upper bound of $C$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Since $A$ and $B$ have supremums, $C$ is nonempty.
|
||||
By \nameref{par:theorem-i.33a-i}, $C$ is bounded above.
|
||||
Therefore the completeness axiom tells us $C$ has a supremum.
|
||||
Let $n > 0$ be an integer.
|
||||
We now prove that
|
||||
\begin{equation}
|
||||
\label{par:theorem-i.33a-ii-eq1}
|
||||
\sup{C} \leq \sup{A} + \sup{B} \leq \sup{C} + 1 / n.
|
||||
\end{equation}
|
||||
|
||||
\subparagraph{Left-Hand Side}%
|
||||
|
||||
First consider the left-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
||||
By \nameref{par:theorem-i.33a-i}, $\sup{A} + \sup{B}$ is an upper bound of
|
||||
$C$.
|
||||
Since $\sup{C}$ is the \textit{least} upper bound of $C$, it follows
|
||||
$\sup{C} \leq \sup{A} + \sup{B}$.
|
||||
|
||||
\subparagraph{Right-Hand Side}%
|
||||
|
||||
Next consider the right-hand side of \eqref{par:theorem-i.33a-ii-eq1}.
|
||||
By \nameref{sub:theorem-i.32a}, there exists some $a' \in A$ such that
|
||||
$\sup{A} < a' + 1 / (2n)$.
|
||||
Likewise, there exists some $b' \in B$ such that
|
||||
$\sup{B} < b' + 1 / (2n)$.
|
||||
Adding these two inequalities together shows
|
||||
\begin{align*}
|
||||
\sup{A} + \sup{B}
|
||||
& < a' + b' + 1 / n \\
|
||||
& \leq \sup{C} + 1 / n.
|
||||
\end{align*}
|
||||
|
||||
\subparagraph{Conclusion}%
|
||||
|
||||
Applying \nameref{sec:theorem-i.31} to \eqref{par:theorem-i.33a-ii-eq1}
|
||||
proves $\sup{C} = \sup{A} + \sup{B}$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\proceeding{Theorem I.34}}%
|
||||
\subsection*{\verified{Theorem I.33b}}%
|
||||
\hyperlabel{sub:theorem-i.33b}%
|
||||
|
||||
If each of $A$ and $B$ has an infimum, then $C$ has an infimum, and
|
||||
$$\inf{C} = \inf{A} + \inf{B}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{inf\_minkowski\_sum\_eq\_inf\_add\_inf}
|
||||
|
||||
\divider
|
||||
|
||||
We prove (i) $\inf{A} + \inf{B}$ is a lower bound of $C$ and (ii)
|
||||
$\inf{A} + \inf{B}$ is the \textit{greatest} lower bound of $C$.
|
||||
|
||||
\paragraph{(i)}%
|
||||
\hyperlabel{par:theorem-i.33b-i}%
|
||||
|
||||
Let $x \in C$.
|
||||
By definition of $C$, there exist elements $a' \in A$ and $b' \in B$ such
|
||||
that $x = a' + b'$.
|
||||
By definition of an infimum, $a' \geq \inf{A}$.
|
||||
Likewise, $b' \geq \inf{B}$.
|
||||
Therefore $a' + b' \geq \inf{A} + \inf{B}$.
|
||||
Since $x = a' + b'$ was arbitrarily chosen, it follows $\inf{A} + \inf{B}$
|
||||
is a lower bound of $C$.
|
||||
|
||||
\paragraph{(ii)}%
|
||||
|
||||
Since $A$ and $B$ have infimums, $C$ is nonempty.
|
||||
By \nameref{par:theorem-i.33b-i}, $C$ is bounded below.
|
||||
Therefore \nameref{sec:theorem-i.27} tells us $C$ has an infimum.
|
||||
Let $n > 0$ be an integer.
|
||||
We now prove that
|
||||
\begin{equation}
|
||||
\label{par:theorem-i.33b-ii-eq1}
|
||||
\inf{C} - 1 / n \leq \inf{A} + \inf{B} \leq \inf{C}.
|
||||
\end{equation}
|
||||
|
||||
\subparagraph{Right-Hand Side}%
|
||||
|
||||
First consider the right-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
||||
By \nameref{par:theorem-i.33b-i}, $\inf{A} + \inf{B}$ is a lower bound of
|
||||
$C$.
|
||||
Since $\inf{C}$ is the \textit{greatest} upper bound of $C$, it follows
|
||||
$\inf{C} \geq \inf{A} + \inf{B}$.
|
||||
|
||||
\subparagraph{Left-Hand Side}%
|
||||
|
||||
Next consider the left-hand side of \eqref{par:theorem-i.33b-ii-eq1}.
|
||||
By \nameref{sub:theorem-i.32b}, there exists some $a' \in A$ such that
|
||||
$\inf{A} > a' - 1 / (2n)$.
|
||||
Likewise, there exists some $b' \in B$ such that
|
||||
$\inf{B} > b' - 1 / (2n)$.
|
||||
Adding these two inequalities together shows
|
||||
\begin{align*}
|
||||
\inf{A} + \inf{B}
|
||||
& > a' + b' - 1 / n \\
|
||||
& \geq \inf{C} - 1 / n.
|
||||
\end{align*}
|
||||
|
||||
\subparagraph{Conclusion}%
|
||||
|
||||
Applying \nameref{sec:lemma-2} to \eqref{par:theorem-i.33b-ii-eq1}
|
||||
proves $\inf{C} = \inf{A} + \inf{B}$ as expected.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\section*{\verified{Theorem I.34}}%
|
||||
\hyperlabel{sec:theorem-i.34}%
|
||||
|
||||
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that
|
||||
$$s \leq t$$
|
||||
Given two nonempty subsets $S$ and $T$ of $\mathbb{R}$ such that $$s \leq t$$
|
||||
for every $s$ in $S$ and every $t$ in $T$. Then $S$ has a supremum, and $T$
|
||||
has an infimum, and they satisfy the inequality
|
||||
$$\sup{S} \leq \inf{T}.$$
|
||||
has an infimum, and they satisfy the inequality $$\sup{S} \leq \inf{T}.$$
|
||||
|
||||
\begin{proof}
|
||||
|
||||
\link{forall\_mem\_le\_forall\_mem\_imp\_sup\_le\_inf}
|
||||
|
||||
\divider
|
||||
|
||||
By hypothesis, $S$ and $T$ are nonempty sets.
|
||||
Let $s \in S$ and $t \in T$.
|
||||
Then $t$ is an upper bound of $S$ and $s$ is a lower bound of $T$.
|
||||
By the completeness axiom, $S$ has a supremum.
|
||||
By \nameref{sec:theorem-i.27}, $T$ has an infimum.
|
||||
All that remains is showing $\sup{S} \leq \inf{T}$.
|
||||
|
||||
For the sake of contradiction, suppose $\sup{S} > \inf{T}$.
|
||||
Then there exists some $c > 0$ such that $\sup{S} = \inf{T} + c$.
|
||||
Therefore $\inf{T} < \sup{S} - c / 2$.
|
||||
By \nameref{sub:theorem-i.32a}, there exists some $x \in S$ such that
|
||||
$\sup{S} - c / 2 < x$.
|
||||
Thus $$\inf{T} < \sup{S} - c / 2 < x.$$
|
||||
But by hypothesis, $x \in S$ is a lower bound of $T$ meaning $x \leq \inf{T}$.
|
||||
Therefore $x < x$, a contradiction.
|
||||
Out original assumption is incorrect; that is, $\sup{S} \leq \inf{T}$.
|
||||
|
||||
\end{proof}
|
||||
|
||||
\end{document}
|
||||
|
|
|
@ -42,7 +42,7 @@
|
|||
\doublebox{
|
||||
\begin{minipage}{0.95\textwidth}
|
||||
\vspace{2pt}
|
||||
\hl{Note}: #1
|
||||
\hl{Note:} #1
|
||||
\vspace{2pt}
|
||||
\end{minipage}}
|
||||
\end{center}}
|
||||
|
|
Loading…
Reference in New Issue