adding docx format

Author: luifrancgom

_quarto.yml

@@ -16,7 +16,7 @@ book:
       affiliations:
         - name: University of Michigan
   cover-image: images/cover_halmos_naive_set_theory.png
-  downloads: [pdf]
+  downloads: [pdf, docx]
   favicon: images/favicon_halmos_naive_set_theory.ico
   repo-url: https://github.com/luifrancgom/halmos_naive_set_theory_quarto/
   issue-url: https://github.com/luifrancgom/halmos_naive_set_theory_quarto/issues
@@ -101,10 +101,12 @@ format:
       - partials/latex/before-body.tex
     # keep-md: true
     # keep-tex: true
-  # docx:
-  #   # table of contents  
-  #   toc: true
-  #   number-sections: true
+  docx:
+    # format options
+    reference-doc: word/custom-reference-doc.docx
+    # table of contents  
+    toc: true
+    number-sections: true
 
 # citations
 bibliography: references/halmos_naive_set_theory_quarto.bib

chapter_1.qmd

@@ -59,4 +59,8 @@ The wording of the definition implies that each set must be considered to be inc
 
 If $A$ and $B$ are sets such that $A \subset B$ and $B \subset A$, then $A$ and $B$ have the same elements and therefore, by the axiom of extension, $A = B$. This fact is described by saying that set inclusion is *antisymmetric*`\index{antisymmetric}`{=latex}. (In this respect set inclusion behaves differently from equality. Equality is *symmetric*`\index{symmetric}`{=latex}, in the sense that if $A = B$, then necessarily $B = A$.) The axiom of extension can, in fact, be reformulated in these terms: if $A$ and $B$ are sets, then a necessary and sufficient condition that $A = B$ is that both $A \subset B$ and $B \subset A$. Correspondingly , almost all proofs of equalities between two sets $A$ and $B$ are split into two parts; first show that $A \subset B$, and then show that $B \subset A$.
 
-Observe that belonging ($\in$) and inclusion ($\subset$) are conceptually very different indeed. One important difference has already manifested itself above: inclusion is always reflexive, whereas it is not at all clear that belonging is ever reflexive. That is: $A \subset A$ is always true; is $A \in A$ ever true? It is certainly not true of any reasonable set that anyone has ever seen. Observe, along the same lines, that inclusion is transitive, whereas belonging is not. Everyday examples, involving, for instance, super-organizations whose members are organizations, will readily occur to the interested reader.
+Observe that belonging ($\in$) and inclusion ($\subset$) are conceptually very different indeed. One important difference has already manifested itself above: inclusion is always reflexive, whereas it is not at all clear that belonging is ever reflexive. That is: $A \subset A$ is always true; is $A \in A$ ever true? It is certainly not true of any reasonable set that anyone has ever seen. Observe, along the same lines, that inclusion is transitive, whereas belonging is not. Everyday examples, involving, for instance, super-organizations whose members are organizations, will readily occur to the interested reader.
+
+::: {.content-visible when-format="docx"}
+{{< pagebreak >}}
+:::

chapter_10.qmd

@@ -21,9 +21,7 @@ f(f^{-1}(B)) \subset B.
 $$
 
 ::: {.proof}
-
 If $y \in f(f^{-1}(B))$, then $y = f(x)$ for some $x$ in $f^{-1}(B)$; this means that $y = f(x)$ and $f(x) \in B$, and therefore $y \in B$.
-
 :::
 
 If $f$ maps $X$ onto $Y$, then 
@@ -33,9 +31,7 @@ f(f^{-1}(B)) = B.
 $$
 
 ::: {.proof}
-
 If $y \in B$, then $y = f(x)$ for some $x$ in $X$, and therefore for some $x$ in $f^{-1}(B)$; this means that $y \in f(f^{-1}(B))$.
-
 :::
 
 If $A \subset X$, then 
@@ -45,9 +41,7 @@ A \subset f^{-1}(f(A))
 $$
 
 ::: {.proof}
-
 If $x \in A$, then $f(x) \in f(A)$; this means that $x \in f^{-1}(f(A))$.
-
 :::
 
 If $f$ is one-to-one, then 
@@ -57,9 +51,7 @@ A = f^{-1}(f(A))
 $$
 
 ::: {.proof}
-
 If $x \in f^{-1}(f(A))$, then $f(x) \in f(A)$, and therefore $f(x) = f(u)$ for some $u$ in $A$; this implies that $x = u$ and hence that $x \in A$. 
-
 :::
 
 The algebraic behavior of $f^{-1}$ is unexceptionable. If $\{ B_{i} \}$ is a family of subsets of $Y$, then 
@@ -99,7 +91,9 @@ The algebraic properties of inversion and composition are the same for relations
 The algebra of relations provides some amusing formulas. Suppose that, temporarily, we consider relations in one set $X$ only, and, in particular, let the relation of equality in $X$ (which is the same as the identity mapping on $X$). The relation $I$ acts as a multiplicative unit; this means that $IR = RI = R$ for every relation $R$ in $X$. Query: is there a connection among $I$, $RR^{-1}$, and $R^{-1}R$? The three defining properties of an equivalence relation can be formulated in algebraic terms as follows: reflexivity means $I \subset R$, symmetry means $R \subset R^{-1}$, and transitivity means $RR \subset R$.
 
 ::: {#exr-10-1}
-
 (Assume in each case that $f$ is a function from $X$ to $Y$) $(i)$ If $g$ is a function from $Y$ to $X$ such that $gf$ is the identity on $X$, then $f$ is one-to-one and $g$ maps $Y$ onto $X$. $(ii)$ A necessary and sufficient condition that $f(A \cap B) = f(A) \cap f(B)$ for all subsets $A$ and $B$ of $X$ is that $f$ be one-to-one. $(iii)$ A necessary and sufficient condition that $f(X - A) \subset Y - f(A)$ for all subsets $A$ of $X$ is that $f$ be one-to-one. $(iv)$ A necessary and sufficient condition that $Y - f(A) \subset f(X - A)$ for all subsets $A$ of $X$ is that $f$ map $X$ onto $Y$.
+:::
 
+::: {.content-visible when-format="docx"}
+{{< pagebreak >}}
 :::

chapter_11.qmd

@@ -41,11 +41,9 @@ etc. The "etc." means that we hereby adopt the usual notation, and, in what foll
 From what has been said so far it does not follow that the construction of successors can be carried out ad infinitum within one and the same set. What we need is a new set-theoretic principle.
 
 ::: {#cnj-infinity}
-
 ## Axiom of infinity`\index{axiom of infinity}`{=latex}
 
 There exists a set containing $0$ and containing the successor of each of its elements.
-
 :::
 
 The reason for the name of the axiom should be clear. We have not yet given a precise definition of infinity, but it seems reasonable that sets such as the ones that the axiom of infinity describes deserve to be called infinite.
@@ -84,4 +82,8 @@ $$
 \end{gathered}
 $$
 
-The word "sequence" is used in a few different ways in the mathematical literature, but the differences among them are more notational than conceptual. The most common alternative starts at $1$ instead of $0$; in other words, it refers to a family whose index set is $\omega - \{ 0 \}$ instead of $\omega$.
+The word "sequence" is used in a few different ways in the mathematical literature, but the differences among them are more notational than conceptual. The most common alternative starts at $1$ instead of $0$; in other words, it refers to a family whose index set is $\omega - \{ 0 \}$ instead of $\omega$.
+
+::: {.content-visible when-format="docx"}
+{{< pagebreak >}}
+:::

chapter_12.qmd

@@ -46,27 +46,25 @@ The assertions @eq-12-1 — @eq-12-5 are known as the Peano`\index{Peano}`{=late
 Induction is often used not only to prove things but also to define things. Suppose, to be specific, that $f$ is a function from a set $X$ into the same set $X$, and suppose that $a$ is an element of $X$. It seems natural to try to define an infinite sequence $\{ u(n) \}$ of elements of $X$ (that is, a function $u$ from $\omega$ to $X$) in some such way as this: write $u(0) = a$, $u(1) = f(u(0))$, $u(2) = f(u(1))$, and so on. If the would-be definer were pressed to explain the "and so on," he might lean on induction. What it all means, he might say, is that we define $u(0)$ as $a$, and then, inductively, we define $u(n^{+})$ as $f(u(n))$ for every $n$. This may sound plausible, but, as justification for an existential assertion, it is insufficient. The principle of mathematical induction does indeed prove, easily, that there can be at most one function satisfying all the stated conditions, but it does not establish the existence of such a function. What is needed is the following result.
 
 ::: {#thm-recursion}
-
 ## Recursion theorem`\index{recursion}`{=latex}
 
 If $a$ is an element of a set $X$, and if $f$ is a function from $X$ into $X$, then there exists a function $u$ from $\omega$ into $X$ such that $u(0) = a$ and such that $u(n^{+}) = f(u(n))$ for all $n$ in $\omega$. 
-
 :::
 
 ::: {.proof}
-
 Recall that a function from $\omega$ to $X$ is a certain kind of subset of $\omega \times X$; we shall construct $u$ explicitly as a set of ordered  pairs. Consider, for this purpose, the collection $\mathcal{C}$ of all those subsets $A$ of $\omega \times X$ for which $(0, a) \in A$ and for which $(n^{+}, f(x)) \in A$ whenever $(n, x) \in A$. Since $\omega \times X$ has these properties, the collection $\mathcal{C}$ is not empty. We may, therefore, form the intersection of all the sets of the collection $\mathcal{C}$. Since it is easy to see that $u$ itself belongs to $\mathcal{C}$, it remains only to prove that $u$ is a function. We are to prove, in other words, that for each natural number $n$ there exists at most one element $x$ of $X$ such that $(n, x) \in u$. (Explicitly: if both $(n, x)$ and $(n, y)$ belong to $u$, then $x = y$.) The proof is inductive. Let $S$ be the set of all those natural numbers $n$ for which it is indeed true that $(n, x) \in u$ for at most one $x$. We shall prove that $0 \in S$ and that if $n \in S$, then $n^{+} \in S$.
 
 Does $0$ belong to $S$? If not, then $(0,b) \in u$ for some $b$ distinct from $a$. Consider, in this case, the set $u - \{ (0, b) \}$. Observe that this diminished set still contains $(0, a)$ (since $a \neq b$), and that if the diminished set contains $(n, x)$, then it contains $(n^{+}, f(x))$ also. The reason for the second assertion is that since $n^{+} \neq 0$, the discarded element is not equal to $(n^{+}, f(x))$. In other words, $u - \{ (0, b) \} \in \mathcal{C}$. This contradicts the fact that $u$ is the smallest set in $\mathcal{C}$, and we may conclude that $0 \in S$. 
 
 Suppose now that $n \in S$; this means that there exists a unique element $x$ in $X$ such that $(n, x) \in u$. Since $(n, x) \in u$, it follows that $(n^{+}, f(x)) \in u$. If $n^{+}$ does not belong to $S$, then $(n^{+}, y) \in u$ for some $y$ different from $f(x)$. Consider, in this case, the set $u - \{ (n^{+}, y) \}$. Observe that this diminished set contains $(0, a)$ (since $n^{+} \neq 0$), and that if the diminished set contains $(m, t)$, say, then it contains $(m^{+}, f(t))$ also. Indeed, if $m = n$, then $t$ must be $x$, and the reason the diminished set contains $(n^{+}, f(x))$ is that $f(x) \neq y$; if, on the other hand, $m \neq n$, then the reason the diminished set contains $(m^{+}, f(t))$ is that $m^{+} \neq n^{+}$. In other words, $u - \{ ( n^{+}, y) \} \in \mathcal{C}$. This again contradicts the fact that $u$ is the smallest set in $\mathcal{C}$, and we may conclude that $n^{+} \in S$.
-
 :::
 
 The proof of the recursion theorem is complete. An application of the recursion theorem is called *definition by induction*`\index{definition by induction}`{=latex}.
 
 ::: {#exr-12-1}
-
 Prove that if $n$ is a natural number, then $n \neq n^{+}$; if $n \neq 0$, then $n = m^{+}$ for some natural number $m$. Prove that $\omega$ is transitive. Prove that if $E$ is a non-empty subset of some natural number, then there exists an element $k$ in $E$ such that $k \in m$ whenever $m$ is an element of $E$ distinct from $k$.
+:::
 
+::: {.content-visible when-format="docx"}
+{{< pagebreak >}}
 :::

chapter_13.qmd

@@ -15,9 +15,7 @@ The next topic that deserves some attention is the theory of order in the the se
 The preceding paragraph implies that if $m$ and $n$ are in $\omega$, then at least one of the three possibilities ($m \in n$, $m = n$, $n \in m$) must hold; it is easy to see that, in fact, always exactly one of them holds. (The reason is another application of the fact that a natural number is not a subset of one of its elements.) Another consequence of the preceding paragraph is that if $n$ and $m$ are distinct natural numbers, then a necessary and sufficient condition that $m \in n$ is that $m \subset n$. Indeed, the implication from $m \in n$ to $m \subset n$ is just the transitivity of $n$. If, conversely, $m \subset n$ and $m \neq n$, then $n \in m$ cannot happen (for then $m$ would be a subset of one of its elements), and therefore $m \in n$. If $m \in n$, or if, equivalently, $m$ is a proper subset of $n$, we shall write $m < n$ and we shall say that $m$ is *less* than $n$. If $m$ is known to be either less than $n$ or else equal to $n$, we write $m \le n$. Note that $\le$ and $<$ are relations in $\omega$. The former is reflexive, but the latter is not; neither is symmetric; both are transitive. If $m \le n$, and $n \le m$, then $m = n$.
 
 ::: {#exr-13-1}
-
-Prove that if $m < n$, then $m + k < n + k$, and prove that if $m < n$ and $k \neq 0$, then $m \cdot k <n \cdot k$. Prove that if $E$ is a non-empty set of natural numbers, then there exists an element $k$ in $E$ such that $k \le m$ for all $m$ in $E$. 
-
+Prove that if $m < n$, then $m + k < n + k$, and prove that if $m < n$ and $k \neq 0$, then $m \cdot k <n \cdot k$. Prove that if $E$ is a non-empty set of natural numbers, then there exists an element $k$ in $E$ such that $k \le m$ for all $m$ in $E$.
 :::
 
 Two sets $E$ and $F$ (not necessarily subsets of $\omega$) are called *equivalent*`\index{equivalent}`{=latex}, in symbols $E \sim F$, if there exists a one-to-one correspondence between them. It is easy to verify that equivalence in this sense, for subsets of some particular set $X$, is an equivalence relation in the power set $\mathcal{P}(X)$.
@@ -29,25 +27,23 @@ It is a mildly shocking fact that a set can be equivalent to a proper subset of
 A set $E$ is called *finite*`\index{finite}`{=latex} if it is equivalent to some natural number; otherwise $E$ is *infinite*`\index{infinite}`{=latex}.
 
 ::: {#exr-13-2}
-
 Use this definition to prove that $\omega$ is infinite.
-
 :::
 
 A set can be equivalent to at most one natural number. (Proof: we know that for any two distinct natural numbers one must be an element and therefore proper subset of the other; it follows from the preceding paragraph that they cannot be equivalent.) We may infer that a finite set is never equivalent to a proper subset; in other words, as long as we stick to finite sets, the whole is always greater than any of its parts. 
 
 ::: {#exr-13-3}
-
 Use this consequence of the definition of finiteness to prove that $\omega$ is infinite.
-
 :::
 
 Since every subset of a natural number is equivalent to a natural number, it follows also that every subset of a finite set is finite. 
 
 The *number of elements*`\index{number of elements}`{=latex} in a finite set $E$ is, by definition, the unique natural number equivalent to $E$; we shall denote it by $\# (E)$. It is clear that if the correspondence between $E$ and $\# (E)$ is restricted to the finite subsets of some set $X$, the result is a function from a subset of the power set $\mathcal{P}(X)$ to $\omega$. This function is pleasantly related to the familiar set-theoretic relations and operations. Thus, for example, if $E$ and $F$ are finite sets such that $E \subset F$, then $\# (E) \le \# (F)$. (The reason is that since $E \sim \# (E)$ and $F \sim \# (F)$, it follows that $\# (E)$ is equivalent to a subset of $\# (F)$.) Another example is the assertion that if $E$ and $F$ are finite sets then $E \cup F$ is finite, and, moreover, if $E$ and $F$ are disjoint, then $\# (E \cup F) = \# (E) + \# (F)$. The crucial step in the proof is the fact that if $m$ and $n$ are natural numbers, then the complement of $m$ in the sum $m + n$ is equivalent to $n$; the proof of this auxiliary fact is achieved by induction on $n$. Similar techniques prove that if $E$ and $F$ are finite sets, then so also are $E \times F$ and $E^{F}$, and, moreover, $\# (E \times F) = \# (E) \cdot \# (F)$ and  $\# (E^{F}) = \# (E)^{\# (F)}$.
 
 ::: {#exr-13-4}
-
 The union of a finite set of finite sets is finite. If $E$ is finite, then $\mathcal{P}(E)$ is finite and, moreover, $\# (\mathcal{P}(E)) = 2^{\# (E)}$. If $E$ is a non-empty finite set of natural numbers, then there exists an element $k$ in $E$ such that $m \le k$ for all $m$ in $E$.
+:::
 
+::: {.content-visible when-format="docx"}
+{{< pagebreak >}}
 :::