author: niplav, created: 2019-03-20, modified: 2022-07-23, language: english, status: in progress, importance: 2, confidence: likely

“Naive Set Theory” by Paul Halmos is a short introduction to set theory. Here, I present solutions to the explicitely stated exercises and problems in that book.

Solutions to “Naïve Set Theory”

Section 3

Exercise 1

It seems like there is no way one could use either insetting (putting a given set into another set) and pairing or pairing on two different inputs to obtain the same set. However, if one sees pairing the same set, then pairing $\emptyset$ with $\emptyset$ would result in $\{\emptyset\}$, which is also the result of insetting $\emptyset$.


Insetting and pairing must have different results because insetting will always result in a set with 1 element, and pairing will always result in a set with 2 elements. Therefore, they can't be the same set.

Pairing the sets a, b and c, d can't result in the same set unless $a=c$ and $b=d$ or $a=d$ and $b=c$. Otherwise, $\{a,b\}$ would contain at least one element not in $\{c,d\}$.

Section 4

Exercise 1

I am not exactly sure what I'm supposed to do here. I guess "observe" means "prove" here, so "prove that the condition has nothing to do with the set B".


$$(A \cap B) \cup C = A \cap (B \cup C) \Leftrightarrow C \subset A\\ (A \cup C) \cap (B \cup C) = A \cap (B \cup C) \Leftrightarrow C \subset A\\ A \cup C = A \Leftrightarrow C \subset A$$

The last statement is trivially true.

Section 5

Some Easy Exercises

$A-B=A \cap B'$


$A-B=\{a | a \in A \land a \not \in B\}=\{a | a \in A \land a \in B'\}=A \cap B'$

$A \subset B \hbox{ if and only if } A-B=\emptyset$


$$A \subset B \Leftrightarrow \forall a \in A:\\ a \in B \Leftrightarrow \exists C:\\ B=A \cup C \Leftrightarrow A-(A \cup C)=\emptyset \Leftrightarrow A-B=\emptyset$$

$A-(A-B)=A \cap B$


$$ A-(A-B)=\\ A-(A \cap B')=\\ A \cap (A \cap B')'=\\ A \cap (A' \cup B)=\\ A \cap A' \cup A \cap B=\\ \emptyset \cup A \cap B=\\ A\cap B$$

$A \cap (B-C)=(A \cap B)-(A \cap C)$


$$(A \cap B)-(A \cap C)=\\ (A \cap B) \cap (A \cap C)'=\\ (A \cap B) \cap (A' \cup C')=\\ (A \cap B \cap A') \cup (A \cap B \cap C')=\\ A \cap B \cap C'=\\ A \cap (B-C)$$

$A \cap B \subset (A \cap C) \cup (B \cap C')$


$$A \cap B \subset (A \cap C) \cup (B \cap C')\\ =((A \cap C) \cup B) \cap ((A \cap C) \cup C')\\ =((A \cap C) \cup B) \cap ((A \cup C') \cap (C \cup C'))\\ =((A \cap C) \cup B) \cap (A \cup C')\\ =(A \cup B) \cap (C \cup B) \cap (A \cup C')$$

$A \cap B \subset (A \cup B) \cap (C \cup B) \cap (A \cup C')$ is true because $A \subset (A \cup B)$ and $B \subset (C \cup B)$ and $A \subset (A \cup C')$.

$ (A \cup C) \cap (B \cup C') \subset A \cup B $


$$ (A \cup C) \cap (B \cup C')\\ = ((A \cup C) \cap B) \cup ((A \cup C) \cap C')\\ = ((A \cup C) \cap B) \cup A\\ = (A \cap B) \cup (C \cap B) \cup A \subset A \cup B $$

This is the case because $(A \cap B) \cup (C \cap B) \subset B$ (since intersections with $B$ are subsets of $B$), and the union with $A$ doesn't change the equation.

Exercise 1

To be shown: The power set of a set with n elements has $2^n$ elements. Proof by induction.


Induction base: The power set of the empty set contains 1 element:


Induction assumption:


Induction step:

To be shown: $|P(A \cup \{a\}|=2*2^{|A|}=2^{|A|+1}$.

$P(A \cup \{a\})$ contains two disjunct subsets: $P(A)$ and $N=\{\{a\}\cup S | S \in P(A)\}$. Those are disjunct because every element in $N$ contains $a$ ($\forall n \in N: a \in n$), but there is no element of $P(A)$ that contains $a$. Also, it holds that $P(A) \cup N=P(A \cup\{a\})$, because elements in the power set can either contain $a$ or not, there is no middle ground. It is clear that $|N|=|P(A)|$, therefore $|P(A \cup \{a\})|=|P(A)|+|N|=2*|P(A)|=2*2^{|A|}=2^{|A|+1}$.

Exercise 2

To be shown:

${\cal{P}}(E) \cap {\cal{P}}(F)={\cal{P}}(E \cap F)$


If $S \in {\cal{P}}(E \cap F)$, then $\forall s \in S: s \in E \cap F$. Therefore, $S \subset E$ and $S \subset F$ and thereby $S \in {\cal{P}}(E)$ and $S \in {\cal{P}}(F)$. This means that $S \in {\cal{P}}(E) \cap {\cal{P}}(F)$.

If $S \in {\cal{P}}(E) \cap {\cal{P}}(F)$, then a very similar proof can be written: $S \subset E$ and $S \subset F$, so $\forall s \in S:s \in E$ and $\forall s \in S: s \in F$. Then $S \subset E \cap F$ and therefore $S \in {\cal{P}}(E \cap F)$.

To be shown:

${\cal{P}}(E) \cup {\cal{P}}(F)\subset{\cal{P}}(E \cup F)$


If $S \in {\cal{P}}(E) \cup {\cal{P}}(F)$, then $S \in {\cal{P}}(E) \Leftrightarrow S \subset E$ or $S \in {\cal{P}}(F) \Leftrightarrow S \subset F$. Since it is true for any set $X$ that $S \subset E \Rightarrow S \in {\cal{P}}(E \cup X)$, it is true that $S \in {\cal{P}}(E \cup F)$ (similar argumentation if $S \subset F$).

A reasonable interpretation for the introduced notation: If ${\cal{C}}={X_1, X_2, \dots, X_n}$, then

$\bigcap_{X \in \cal{C}} X=X_1 \cap X_2 \cap \dots X_n$

Similarly, if ${\cal{C}}={X_1, X_2, \dots, X_n}$, then

$\bigcup_{X \in \cal{C}} X=X_1 \cup X_2 \cup \dots X_n$

The symbol ${\cal{P}}$ still stands for the power set.

To be shown:

$\bigcap_{X \in \cal{C}} {\cal{P}}(X)={\cal{P}}(\bigcap_{X \in \cal{C}} X)$

Proof by induction.

Induction base:

${\cal{P}}(E) \cap {\cal{P}}(F)={\cal{P}}(E \cap F)$

Induction assumption:

$\bigcap_{X \in \cal{C}} {\cal{P}}(X)={\cal{P}}(\bigcap_{X \in \cal{C}} X)$

Induction step:

$${\cal{P}}(Y) \cap \bigcap_{X \in \cal{C}} {\cal{P}}(X)\\ ={\cal{P}}(Y) \cap {\cal{P}}(\bigcap_{X \in \cal{C}} X)\\ ={\cal{P}}(Y \cap \bigcap_{X \in \cal{C}} X)$$

The last step uses ${\cal{P}}(E) \cap {\cal{P}}(F)={\cal{P}}(E \cap F)$, since $\bigcap_{X \in \cal{C}} X$ is also just a set.

To be shown:

$\bigcup_{X \in \cal{C}} {\cal{P}}(X) \subset{\cal{P}}(\bigcup_{X \in \cal{C}} X)$

Proof by induction.

Induction base:

${\cal{P}}(E) \cup {\cal{P}}(F) \subset {\cal{P}}(E \cup F)$

Induction assumption:

$\bigcup_{X \in \cal{C}} {\cal{P}}(X) \subset {\cal{P}}(\bigcup_{X \in \cal{C}} X)$

Induction step:

$${\cal{P}}(Y) \cup \bigcup_{X \in \cal{C}} {\cal{P}}(X)\\ \subset {\cal{P}}(Y) \cup {\cal{P}}(\bigcup_{X \in \cal{C}} X)\\ \subset {\cal{P}}(Y \cup \bigcup_{X \in \cal{C}} X)$$

The last step uses ${\cal{P}}(E) \cup {\cal{P}}(F) \subset {\cal{P}}(E\cup F)$, since $\bigcup_{X \in \cal{C}} X$ is also just a set.

To be shown:

$\bigcup {\cal{P}}(E)=E$


$\forall X \in {\cal{P}}(E): X \subset E$. Furthermore, $E \in {\cal{P}}(E)$. Since $A \subset E \Rightarrow A \cup E=E$, it holds that $E=\bigcup_{X \in {\cal{P}}(E)}=\bigcup {\cal{P}}(E)$.

And "E is always equal to $\bigcup_{X \in {\cal{P}}(E)}$ (that is $\bigcup {\cal{P}}(E)=E$), but that the result of applying ${\cal{P}}$ and $\bigcup$ to $E$ in the other order is a set that includes E as a subset, typically a proper subset" (p. 21).

I am not entirely sure what this is supposed to mean. If it means that we treat ${\cal{E}}$ as a collection, then $\forall X \in {\cal{E}}:\bigcup_{E \in {\cal{E}}} E \subset X$. But that doesn't mean that ${\cal{E}} \subset {\cal{P}}(\bigcup_{E \in {\cal{E}}}E)$: If ${\cal{E}}=\{\{a,b\},\{b,c\}\}$, then $\bigcup_{E \in {\cal{E}}} E=\{b\}$, and ${\cal{E}}=\{\{a,b\},\{b,c\}\} \not\subset {\cal{P}}(\{b\})=\{\{b\},\emptyset\}$.

If we treat $E$ simply as a set, then $\bigcup E=E$, and it is of course clear that $E \subset {\cal{P}}(E)$, as for all other subsets of $E$.

Section 6

A Non-Trivial Exercise

"find an intrinsic characterization of those sets of subsets of A that correspond to some order in A"

Let ${\cal{M}} \subset {\cal{P}}({\cal{P}}(A))$ be the set of all possible orderings of $A$. In the case of $A=\{a, b\}$, $\cal{M}$ would be $\{\{\{a\}, \{a, b\}\}, \{\{b\}, \{a, b\}\}\}$.

Some facts about every element $M \in \cal{M}$:

$\bigcap M=\{min\}$, where $\{min\}$ is the smallest element in the ordering $M$, the element in $M$ for which there is no other element $a \in M$ so that $a \subset \{min\}$ (which means that $\emptyset \not \in M$).

$\bigcup M=A$. $A$ must therefore be in $M$ and be the biggest element (no element $a \in M$ so that $a \supset A$).

For all elements $m \in M$ except $\{min\}$ there exists at least one element $n \in M$ so that $n \subset m$.

Similarly, for all elements $m \in M$ except $A$ there exists at least one element $n \in M$ so that $n \supset m$.

For every $m \in M$ except $A$ and $\{min\}$, there exist two unique elements $x,y \in M$ so that $m$ is the only set in $M$ for which it is true that $x \subset m \subset y$.

For every $a \in A$, there must exist two sets $m,n \in M$ so that $n=m \cup \{a\}$ (except for $min$). This means that the $|A|=|M|$, the size of $A$ is the size of $M$.

These conditions characterise $\cal{M}$ intrinsically and are the solution to the question.

Exercise 1

(i) To be shown: $(A \cup B) \times X=(A \times X) \cup (B \times X)$


$$(A \cup B) \times X=\\ \{(e, x): e \in A \lor e \in B, x \in X\}=\\ \{(e, x): e \in A, x \in X\} \cup \{(e, x): e \in B, x \in X\}=\\ (A \times X) \cup (B \times X)$$

(ii) Te be shown: $(A \cap B) \times (X \cap Y)=(A \times X) \cap (B \times Y)$

$$(A \times X) \cap (B \times Y)=\\ \{(x,y), x \in A, y \in X\} \cap \{(v,w), v \in B, w \in Y\}=\\ \{(x,y), x \in A \land x \in B, y \in X \land y \in Y\}=\\ \{(x,y), x \in A \cap B, y \in X \cap Y\}=\\ (A \cap B) \times (X \cap Y)$$

(iii) To be shown: $(A-B) \times X = (A \times X)-(B \times X)$

Two-sided proof by contradiction:

1. $(A-B)\times X \subset (A \times X)-(B \times X)$

Let $(u,v) \in (A-B) \times X$. Then $u \in (A-B)$, and $v \in X$. Suppose $(u, v) \not \in (A \times X)-(B\times X)$. Then $(u,v) \in (A \times X) \cap (B \times X)$. Then $(u,v) \in (A \cap B) \times (X \cap X)$. Then $u \in A \cap B$ and $v \in X$. But if $u \in A \cap B$, then $u \not \in A-B$! Contradiction.

2. $ (A \times X)-(B \times X) \subset (A-B)\times X$

Let $(u,v) \in (A \times X)-(B\times X)$. Then $(u,v)\in(A \times X)$ and $(u,v)\not \in (B \times X)$. Because $v$ must be in $X$, and there is no flexibility there, $u \not \in B$. Suppose $(u,v)\not \in (A-B) \times X$. Since necessariliy $v \in X$, $u \not \in A-B$. But if $u \not \in A-B$, $u$ must be an element of $A\cap B$. Then $u \in B$, and there is a contradiction.

Section 7

Exercise 1

Reflexive, but neither symmetric nor transitive (symmetry violation: $(b,a)\not\in$, transitivity violation: $(a,c)\not\in$): $\{(a,a),(a,b),(b,b),(b,c),(c,c)\}$

Symmetric, but neither reflexive nor transitive (reflexivity violation: $(a,a)\not\in$, transitivity violation: $(a,c)\not\in$): $\{(a,b),(b,a),(b,c),(c,b)\}$

Transitive, but neither reflexive nor symmetric (reflexivity violation: $(a,a)\not\in$, symmetry violation: $(b,a)\not\in$): $\{(a,b),(b,c),(a,c)\}$

Exercise 2

We shall write $X/R$ for the set of all equivalence classe. (Pronounce $X/R$ as “X modulo R,“ or, in abbreviated form, “X mod R.“ Exercise: show that $X/R$ is indeed a set by exhibiting a condition that specifies exactly the subset $X/R$ of the power set ${\cal{P}}(X)$).

Paul Halmos, “Naïve Set Theory“ p. 38, 1960

To be honest, I'm not quite sure what exactly I am supposed to do here. $X/R$ has been defined as being a set, how can I prove a definition?

But I can try and construct $X/R$ from ${\cal{P}}(X)$:

$X/R=\{E: (\forall x, y \in E: x R y) \land E \in {\cal{P}}(X) \} \subset {\cal{P}}(X)$

□, I guess?

Section 8

Exercise 1

Basically, the question is "Which projections are one-to-one", or, "Which projections are injective"?

The answer is: A projection $p: X\times Y \mapsto X$ is injective iff $\forall (x,y)\in X \times Y: \nexists (x,z) \in X \times Y: z \neq y$. Or, simpler: Every element of $X$ occurs at most once in the relation. This can be extended easily to relations composed of more than 2 sets.

Exercise 2

(i) To be shown: $Y^{\emptyset}=\{\emptyset\}$

1. $\emptyset:\emptyset \rightarrow Y$ (the empty set is a function from $\emptyset$ to $Y$).

This is true because $\emptyset$ is a relation so that $\emptyset \subset \emptyset \times Y$, and $\forall x \in \emptyset: \exists (x,y) \in \emptyset$. Or: $\emptyset$ is a set of pairs that maps all elements in $\emptyset$ to $X$, and therefore a function from $\emptyset$ to $X$.

2. Assume $\exists x \in Y^{\emptyset}: x \neq \emptyset$

Then $x: \emptyset \rightarrow Y$, and $x \subset \emptyset \times Y$. But $\emptyset \times X$ can only be $\emptyset$, but it was assumed that $x \neq \emptyset$. Therefore, no such $x$ can exist.

(ii) To be shown: $X \neq \emptyset \Rightarrow \emptyset^{X}=\emptyset$

Assume $\exists f \in \emptyset^{X}$. Then $f \subset X \times \emptyset \land \forall x \in X: \exists (x,y) \in f: y \in \emptyset$ (or: $f$ maps all elements of $X$ to an element in $\emptyset$). However there are no elements in the empty set (that I know of), so $f$ can't exist.

However, if $X=\emptyset$, then (i) applies. So $\emptyset^{\emptyset}=\{\emptyset\}$.

Section 9

Exercise 1

Here, the reader is asked to formulate and prove the commutative law for unions of families of sets.

For context, the associative law for unions of families of sets is formulated as follows:

Suppose, for instance, that $\{I_{j}\}$ is a family of sets with domain $J$, say; write $K=\bigcup_{j} I_{j}$, and let $\{A_{k}\}$ be a family of sets with domain $K$. It is then not difficult to prove that

$$\bigcup_{k \in K} A_{k}=\bigcup_{j \in J}(\bigcup_{i \in I_{j}} A_{i}) $$

Okay, this is all fine and dandy, but where am I going with this? Well, I have probably misunderstood this, because the way I understand it, the correct way to formulate the commutative law for unions of families does not hold.

Let's say $X$ is a set indexed by $I$, or, in other words, $x_{i}$ is a family. Let's then define a new operator index-union to make these expressions easier to read: $I໔X$ as $\bigcup_{i \in I} X_{i}$.

The associative law then expanded reads as

$$\bigcup_{k \in \bigcup_{j \in J} I_{j}} A_{k}=\bigcup_{j \in J}(\bigcup_{i \in I_{j}} A_{i})$$

or, simpler, as $(J໔I)໔A=J໔(I໔A)$.

Then, simply, the commutative version of the law would be $A໔B=B໔A$.

Then there is a very simple counterexample:

$A=\{1,2\}$, $B=\{a,b\}$. Then a family from A to B could be $\{(1,\{a\}),(2,\{b\})\}$ and a family from B to a could be $\{(a,\{1\}),(b,\{2\})\}$. Then $A໔B=\bigcup_{a \in A} B_{a}=\{1,2\}$ and $B໔A=\bigcup_{b \in B} A_{b}=\{a,b\}$, and those two are different sets.

Despite my obvious love for unnecessarily inventing new notation, I'm not a very good mathematician, and believe (credence $\ge 99\%$) that I have misunderstood something here (the rest is taken up by this being an editing/printing mistake). I am not sure what, but would be glad about some pointers where I'm wrong.

Other Failing Ways of Interpreting the Exercise

Other ways of interpreting the exercise also have obvious counter-examples.

If $f, g$ are two families in $X$ (as functions: $f: X \rightarrow X, g: X \rightarrow X$), then the counterexample is

$$X=\{a,b\}\\ f_{a}=b, f_{b}=b\\ g_{a}=a, g_{b}=a\\ f໔g=\{b\}\\ g໔f=\{a\}$$

If $f, g$ are two families of sets in $X$ (as functions: $f: X \rightarrow {\cal{P}}(X), g: X \rightarrow {\cal{P}}(X)$), then the counterexample is

$$X=\{a,b,c\}\\ f_{a}=\{b\}, f_{b}=\{b\}, f_{c}=\{b\}\\ g_{a}=\{a\}, g_{b}=\{c\}, g_{c}=\{c\}\\ f໔g=\{b\}\\ g໔f=\{c\}$$

Exercise 2

To be shown:

(i): $(\bigcup_{i \in I} A_{i}) \cap (\bigcup_{j \in J} B_{j})=\bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$

1. $(\bigcup_{i \in I} A_{i}) \cap (\bigcup_{j \in J} B_{j}) \subset \bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$

Let $e \in (\bigcup_{i \in I} A_{i}) \cap (\bigcup_{j \in J} B_{j})$. Then there exists an $i_{e} \in I$ so that $e \in A_{i_{e}}$ and a $j_{e} \in J$ so that $e \in B_{j_{e}}$. Then $(i_{e}, j_{e}) \in I \times J$, and furthermore $e \in A_{i_{e}} \cap B_{j_{e}}$. Since $A_{i_{e}} \cap B_{j_{e}} \subset \bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$, $e$ is contained in there as well.

Only in this case I will attempt to write out the the proof more formally:

$$e \in (\bigcup_{i \in I} A_{i}) \cap (\bigcup_{j \in J} B_{j}) \Rightarrow \\ e \in (\bigcup_{i \in I} A_{i}) \land e \in (\bigcup_{j \in J} B_{j}) \Rightarrow \\ (\exists i_{e} \in I: e \in A_{i_{e}}) \land (\exists j_{e} \in J: e \in B_{j_{e}}) \Rightarrow \\ \exists i_{e} \in I: \exists j_{e} \in J: e \in A_{i_{e}} \land e \in B_{j_{e}} \Rightarrow \\ \exists (i_{e}, j_{e}) \in I \times J: e \in (A_{i_{e}} \cap B_{j_{e}}) \Rightarrow \\ e \in \bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$$

2. $(\bigcup_{i \in I} A_{i}) \cap (\bigcup_{j \in J} B_{j}) \supset \bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$

Let $e \in \bigcup_{(i,j) \in I \times J} (A_{i} \cap B_{j})$. Then there exists $(i_{e}, j_{e}) \in I \times J$ so that $e \in (A_{i_{e}} \cap B_{j_{e}})$. But then $e \in \bigcup_{i \in I} A_{i}$ and $e \in \bigcup_{j \in J} B_{j}$, which means that $e$ is also in their intersection.

(ii): $(\bigcap_{i \in I} A_{i}) \cup (\bigcap_{j \in J} B_{j})=\bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$

1. $(\bigcap_{i \in I} A_{i}) \cup (\bigcap_{j \in J} B_{j}) \subset \bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$

Let $e \in (\bigcap_{i \in I} A_{i}) \cup (\bigcap_{j \in J}B_{j})$. Then $e$ is an element of all of $A_{i}$ or an element of all of $B_{j}$ (or both). Since that is the case, $e$ is always an element of $A_{i} \cup B_{j}$. Then $e$ is also an element of the intersection of all of these unions $\bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$.

This can be written down more formally as well:

$$e \in (\bigcap_{i \in I} A_{i}) \cup (\bigcap_{j \in J} B_{j}) \Rightarrow \\ e \in (\bigcap_{i \in I} A_{i}) \lor e \in (\bigcap_{j \in J} B_{j}) \Rightarrow \\ \forall i \in I: e \in A_{i} \lor \forall j \in J: e \in B_{j} \Rightarrow \\ \forall i \in I: \forall j \in J: e \in A_{i} \lor e \in B_{j} \Rightarrow \\ \forall i \in I: \forall j \in J: e \in A_{i} \cup B_{j} \Rightarrow \\ e \in \bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$$

2. $(\bigcap_{i \in I} A_{i}) \cup (\bigcap_{j \in J} B_{j}) \supset \bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$

Let $e \in \bigcap_{(i,j) \in I \times J} (A_{i} \cup B_{j})$. Then for every $i$ and $j$, $e \in A_{i} \cup B_{j}$. That means that for every $i$, $e \in A_{i}$: if there is just one $j$ so that $e \not \in B_{j}$, for the last sentence to be true, $A_{i}$ must compensate for that. Or, if not an $e$ in every $A_{i}$, then this must be true for every $B_{j}$ (with a similar reasoning as for $A_{i}$). But if $e$ in every $A_{i}$ or every $B_{j}$, it surely must be in their union.

Exercise 3

To be proven:

$(\bigcup_{i} A_{i}) \times (\bigcup_{j} B_{j})=\bigcup_{i,j}(A_{i} \times B_{j})$

$$ e \in \bigcup_{i,j}(A_{i} \times B_{j}) \\ (e_{1},e_{2}) \in \bigcup_{i,j}(A_{i} \times B_{j}) \Leftrightarrow \\ \exists i \in I, j \in J: (e_{1},e_{2}) \in A_{i} \times B_{j} \Leftrightarrow \\ \exists i \in I, j \in J: (e_{1} A_{i}) \land (e_{2} \in B_{j}) \Leftrightarrow \\ \exists i \in I: (e_{1} A_{i}) \land \exists j \in J: (e_{2} \in B_{j}) \Leftrightarrow \\ (e_{1} \in \bigcup_{i} A_{i}) \land (e_{2} \in \bigcup_{j} B_{j}) \Leftrightarrow \\ (e_{1}, e_{2}) \in (\bigcup_{i} A_{i}) \times (\bigcup_{j} B_{j}) \Leftrightarrow \\ e \in (\bigcup_{i} A_{i}) \times (\bigcup_{j} B_{j}) $$

One can say that $(e_{1}, e_{2}) \in (\bigcup_{i} A_{i}) \times (\bigcup_{j} B_{j}) = \exists i \in I, j \in J: e_{1} \in A_{i} \land e_{2} \in B_{j}$ because there must be at least one $i$ so that $e \in A_{i}$, and because all combinations of elements are chosen from $A$ and $B$.

To be proven:

$(\bigcap_{i} A_{i}) \times (\bigcap_{j} B_{j})=\bigcap_{i,j}(A_{i} \times B_{j})$

$$ e \in \bigcap_{i,j}(A_{i} \times B_{j}) \\ (e_{1},e_{2}) \in \bigcap_{i,j}(A_{i} \times B_{j}) \Leftrightarrow \\ \forall i \in I, j \in J: (e_{1},e_{2}) \in A_{i} \times B_{j} \Leftrightarrow \\ \forall i \in I, j \in J: (e_{1} A_{i}) \land (e_{2} \in B_{j}) \Leftrightarrow \\ \forall i \in I: (e_{1} A_{i}) \land \forall j \in J: (e_{2} \in B_{j}) \Leftrightarrow \\ (e_{1} \in \bigcap_{i} A_{i}) \land (e_{2} \in \bigcap_{j} B_{j}) \Leftrightarrow \\ (e_{1}, e_{2}) \in (\bigcap_{i} A_{i}) \times (\bigcap_{j} B_{j}) \Leftrightarrow \\ e \in (\bigcap_{i} A_{i}) \times (\bigcap_{j} B_{j}) $$

To b proven:

$\bigcap_{i} X_{i} \subset X_{j} \subset \bigcup_{i} X_{i}$


$$ e \in \bigcap_{i} X_{i} \Rightarrow \\ \forall i: e \in X_{i} \Rightarrow \\ \forall i: \exists j: {i,j} \subset I \land i=j \land e \in X_{i} \Rightarrow \\ \forall i: \exists j: i \in I \land j \in J \land i=j \land e \in X_{j} \Rightarrow \\ e \in X_{j} \Rightarrow \\ \exists j \in I: e \in X_{j} \Rightarrow \\ \exists i \in I: e \in X_{i} \Rightarrow \\ e \in \bigcup_{i} X_{i} $$

Section 10

Exercise 1

(i) To be shown: $f(\bigcup_{i} A_{i}) = \bigcup_{i} f(A_{i}$

If $e \in f(\bigcup_{i} A_{i})$, then there exists an $i$ for which $e \in f(A_{i})$. Then also $e \in \bigcup_{i}f(A_{i})$ (since the same set of indices is iterated over).

If otherwise $e \in \bigcup_{i} f(A_{i})$, then there is also an $i$ for which $e \in f(A_{i}$. Then $e$ is at least once an element of $f(\bigcup_{i} A_{i})$.

(ii) To be shown: $f(\bigcap_{i} A_{i}) \neq \bigcap_{i} f(A_{i}$

Example: $I=\{1, 2\}, A_1=\{1\}, A_2=\{2\}, f(\{1\})=\{a\}, f(\{2\})=\{a\}, f(\emptyset)=\emptyset, f(\{1,2\})=\{a\}$

Then $f(\bigcap_{i} A_{i})=f(\{1\} \cap \{2\})=f(\emptyset)=\emptyset$, but $\bigcap_{i} f(A_{i})=\{a\} \cap \{a\}=\{a\}$.

Exercise 2

A necessary and sufficient condition that $f$ map $X$ onto $Y$ is that the inverse image under $f$ of each non-empty subset of $Y$ be a non-empty subset of $X$. (Proof?)

But that's—false? Unless I understand something different by "map" (which I just take as relates from one set to another, possibly in a many-to-one relation).


$X=\{1,2\}, Y=\{a,b\}$. $f(\{1\})=\{a\}, f(\{2\})=\{a\}$, etc for all subsets of $X$. Then $f^{-1}(\{b\})=\emptyset$.

Exercise 3

To be proven: $h(gf)=(hg)f$



Query 1

what do $R^{-1}$, $S^{-1}$, $RS$, and $R^{-1}S^{-1}$ mean?"

$yR^{-1}x$: $y$ father of $x$, $z S^{-1}y$: $z$ is brother of $y$. $$xRSy$: $x$ nephew of $y$. $xR^{-1}S^{-1}$.

Exercise 4

To be proven: $(SR)^{-1}=R^{-1}S^{-1}$


Let $(z,x) \in (SR)^{-1}$. Then $xSRz$. Then $\exists y: xSy \land yRs$. Then $yS^{-1}x$ and $zR^{-1}y$, and then $zR^{-1}S^{-1}x$.

Query 2

is there a connection among I, $RR^{-1}$, and $R^{-1}R$?

Shouldn't $I=RR^{-1}$? No. Because if $xRy$ and $zRy$, then $xRR^{-1}z$. But $I \subset RR^{-1}$, and $I \subset R^{-1}R$. I think alsso that $RR^{-1}=(R^{-1}R)^{-1}$, and $R^{-1}R=(RR^{-1})^{-1}$ (which is true by the previously proven theorem). I don't think there's any further connections here.

Exercise 5

(i) Given $g: Y \rightarrow X$ and $gf(x)=x$, then $f$ is "one-to-one" and "$g$ maps $Y$ onto $X$".

Judging from what I understand, "$f$ is one-to-one" would mean that $f$ is injective, and "$g$ maps $X$ onto $X$" just means that $g$ is surjective?

Wikipedia agrees with these hunches.

$\forall x_1, x_2 \in X: x_1 \neq x_2 \Leftrightarrow f(x_1) \neq f(x_2)$, because if $f(x_1)=f(x_2)=y$ for $x_1 \neq x_2$, then $f(y)=x_1$ and $f(y)=x_2$, which is not possible.

$g$ is surjective, since every element in $\hbox{dom } f \subset Y$ has exactly one corresponding element in $X$ through $g$, but elements in $Y - \hbox{dom } f$ must also be mapped to $X$ (it could be that $Y - \hbox{dom } f=\emptyset$, but that is not guaranteed).

(ii) To be proven: $f(A \cap B)=f(A) \cap f(B)$ iff for all subsets of $A$ and $B$, $f$ injective.

First case: $f$ injective $\Rightarrow f(A \cap B) = f(A) \cap f(B)$.

Let $y \in f(A \cap B)$, then there exists exactly one $x \in X$ so that $f(x)=y$. That means $x \in A \land x \in B$. Then $e \in f(A) \land e \in f(B)$.

Second case: $f(A \cap B)=f(A) \cap f(B) \Rightarrow f$ injective.

More formal:

$f(A \cap B)=f(A) \cap f(B) \Rightarrow \forall x_1 \neq x_2 \in X, f(x_1) \neq f(x_2)$.

Let then $x_1 \neq x_2 \in X$, so that $f(x_1)=f(x_2)=y$. Let then $A, B$ so that $x_1 \in A \backslash B$ and $x_2 \in B \backslash A$.

Then $y \in f(A) \cap f(B)$, but $y \not \in f(A \cap B)$. So those $x_1, x_2$ can't exist, therefore f is injective.

(iii) To be proven: f injective $\Leftrightarrow f(X-A) \subset Y-f(A)$ for all subsets of A.

First case: f inj. $\Rightarrow f(X-A) \subset Y-f(A)$

Let $e \in f(X-A)$ and $e \not \in Y-f(A)$. Then $\exists x \in X-A$ so that $f(x)=e$, and there is no $x_2 \in A: f(x_2)=e$ (since f is injective). But since $e \in Y$ (by definition) and $e \not \in f(A)$, e must be in $Y-f(A)$.

Second case: $f(X-A) \subset Y-f(A) \Rightarrow f$ inj.

Assume $x_1, x_2 (x_1 \not = x_2)$ so that $f(x_1)=f(x_2)=y$. Let then $x_1 \not \in A$ and $x_2 \in A$. Then $y \in f(X-A)$ (since $x_1 \in X-A$). But since $y \in f(A)$, $y \not \in Y-f(A)$. Contradiction, f must be injective.

(iv) To be shown: $\forall y : \exists x: f(x)=y \Leftrightarrow Y-f(A) \subset f(X-A)$

First case: f surj. $\Rightarrow \forall A \subset X: Y-f(A) \subset f(X-A)$

Assume $y \in Y$ and $y \not \in fNA)$. If $y \not \in f(X-A)$, then $\lnot \exists x \in X-A$ so that $f(x)=y$. But since f is surjective, there must be an x so that $f(x)=y$ and $x \not \in A$! Contradiction.

Second case: $\forall A \subset X: Y-f(A) \subset f(X-A) \Rightarrow \forall y: \exists x: f(x)=y$.

Assume $ \exists y \in Y: \lnot \exists x: f(x)=y$. Then $y \in Y-f(A)$ for all A, but $y \not \in f(X-A)$ (since y is not in the range). Contradiction of the assumption, such a y can't exist.

Section 11

Exercise 1

Since the intersection of every (non-empty) family of successor sets is a successor set itself (proof?)

Let $\{S_i\}$ be a family of successor sets.

Every $s_i \in \{S_i\}$ is a successor set, then it is guaranteed that $0 \in s_i$. One can assume that $\forall s_i \in \{S_i\}: x \in s_i$, which implies that $x \in \bigcap_{i \in I} s_i$. But since all $s_i$ are successor sets, $x^+$ must also be an element of all $s_i$ Then $x^+ \in \bigcap_{i \in I} s_i$, and the intersection of the family is a successor set.

However, remember the definition of a family: a family of sets is a function that maps from an index set into the powerset of a set, so this function is $f: I \rightarrow \mathcal{P}(A)$.

In this case, we could set $I=\{\emptyset\}$ and $\hbox{ran} f=\{\{\{\emptyset, \{\emptyset\}\}\}\}$, with $f(\emptyset)=\{\{\emptyset, \{\emptyset\}\}\}$. That would make f a non-empty family, but $\bigcap_{i \in I} s_i=\{\{\{\emptyset, \{\emptyset\}\}\}\}$, which is definitely not a successor set. Since the solution I presented above seems exactly like the thing Halmos would want me to do, I guess I have misunderstood the definition for "family of sets".

Section 12

Exercise 1

Prove that if $n$ is a natural number, then $n \not = n^{+}$

For $n=0=\emptyset$, $n \not = n^+$:

$$n^+=\\ n \cup \{n\}=\\ \emptyset \cup \{\emptyset\}=\\ \{\emptyset\}\not =\\ \emptyset=\\ n$$

Let $n^-$ be the number such that ${n^-}^+=n$, and assume that $n^- \not =n$.

Then for $n^+=n \cup \{n\}=n^- \cup \{n^-\} \cup \{n\}$ to be equal to $n$, $\{n\}$ would need to be equal to $n^-$ or $\{n^-\}$. That would contradict the assumption, so $n^+ \not =n$.

This would be much easier to prove if the Axiom of Regularity had been introduced: if we know that it can't be that $a \in a$, then $n^+=n \cup \{n\}$ would need to be $\{n\}$ for $n^+$ to be equal to $n$, which could only be the case if $n \in n$, violating the axiom.

Exercise 2

Prove that if $n$ is a natural number, […] if $n \not = 0$, then $n=m^+$ for some natural number $m$.

This feels true almost by definition?

On pg. 44 a natural number is defined as "an element of the minimal successor set $ω$". Then, if $n \in ω$, there must be an element $n^-$ of $ω$ so that ${n^-}^+=n$ (by the second Peano axiom and the condition that $ω$ can contain no superfluous elements, which would be the third Peano axiom).

Exercise 3

Prove that $ω$ is transitive.

I don't understand what is being asked from me here. $ω$ is not a relation since it's not a set of ordered sets with two elements (tuples), so this question is underspecified.

Exercise 4

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 $ whenever $m$ is an element of $E$ distinct from $k$.

Trying to decode this first into first-order logic and then natural language again, I get that if $E \subset N \in ℕ$, then $∃k\in E: ∀ m \not = k \in E: k \in m$. Decoding into natural language, this roughly means that for every subset of the natural numbers, that subset has a minimum.

Lemma: Any Successor of $n$ Contains $n$

Statement: $n \in n^{+_k}$ for all $k$.

Basis: For $k=1: n \in \{n\} \subset n^+$.

Assume this holds for $k-1$.

Step $n^{+_k}=n^{+_{k-1}} \cup \{n^{+_{k-1}}\}$. $n \in \{n^{+_{k-1}}\}$, so $n \in n^{+_k}$.

Assume such a $k$ did not exist. Then for every $k_i$, it would hold that there is some $m_i \not= k$ so that $k_i \not \in m_i$.

But then it must hold that $m_i \in k_i$ (it can't be that $m_i \not \in k_i$ and $k_i \not \in m_i$ for natural numbers). If that isn't the case (so $m_i \not \in k_i$), then there must exists some $r_i$ that $m_i \not \in r_i$. But since $E$ contains only finitely many elements, this leads to a cycle (which is not possible, since that would mean that $k_i \not \in k_j$ and $k_j \not \in k_i$). So such a $k$ must exist.

Section 13

Stray Exercise 1

The discovery and establishment of the properties of powers, as well as the detailed proofs of the statements about products, can safely be left as exercises to the readers.

To be shown: $k \cdot (m+n)=k \cdot m+k \cdot n$.

Induction over $n$.

Induction base: $k \cdot (m+0)=k \cdot m=k \cdot m+k \cdot 0$.

Induction assumption: $k \cdot (m+n)=k \cdot m+k \cdot n$.

Induction step: $k \cdot (m+n^+)=k \cdot (m+n)^+=k \cdot (m+n)+k=(k \cdot m+k\cdot n)+k=k \cdot m+(k \cdot n + k)=k \cdot m + k \cdot n^+$

To be shown: $m \cdot n=n \cdot m$.

First, $1 \cdot n=n$ because $1 \cdot 0=0$, and if $1 \cdot n=n$, then $1 \cdot n^+=(1 \cdot n)+1=n+1=n^+$.

Second, $m^+ \cdot n=(m \cdot n)+m$ for $n \ge 1$, because $m^+ \cdot n=(m+1) \cdot n=(m \cdot n)+m$ (as per distributivity).

Induction over $m$.

Induction base: If $m=0$, then $m \cdot n=0 \cdot n=0=n \cdot 0=n \cdot m$.

Induction assumption: $m \cdot n=n \cdot m$.

Induction step: $m^+ \cdot n=(m \cdot n)+n=(n \cdot m)+n=n \cdot m^+$ (as per definition of the product).

To be shown: $(k \cdot m) \cdot n=k \cdot (m \cdot n)$.

Induction over $n$.

Induction base: If $n=0$, then $(k \cdot m) \cdot 0=0=k \cdot (m \cdot 0)$ (since $p_{k \cdot m}(0)=0$).

Induction assumption: $(k \cdot m) \cdot n=k \cdot (m \cdot n)$

Induction step: $k \cdot (m \cdot n^+)=k \cdot ((m \cdot n)+m)=k \cdot (m \cdot n)+k \cdot m=(k \cdot m) \cdot n+k \cdot m=(k \cdot m) \cdot n^+$ (using distributivity).

As for the properties of the power, I don't really want to spend much time and energy proving them. Let it suffice to say that the powers are neither associative (counterexample: $2^{(3^2)}=512 \not =64=(2^3)^2$) nor commutative (counterexample: $2^3=8\not =9=3^2$) nor distributive over either addition (counterexample: $2^{3+4}=128 \not =24=2^3+2^4$) or multiplication (counterexample: $2^{3 \cdot 4}=4068 \not=128=2^3 \cdot 2^4$).

They have two interesting properties: $a^{b\cdot c}=(a^b)^c$, and $a^{b+c}=a^b \cdot a^c$.

For commutative hyperoperators, see Ghalimi 2019.

Exercise 1

To be shown: if $m<n$, then $m+k < n+k$

Induction over $k$

Induction base: If $k=0$, then $m+k=m<n=n+k$

Induction assumption: $m+k<n+k$

Induction step:

$$m+k^+ < n+k^+ \Leftrightarrow \\ (m+k)^+ < (n+k)^+ \Leftrightarrow \\ \{m+k\} \cup (m+k) \subset \{n+k\} \cup (n+k)$$

We know that $m+k \subset n+k$.

We can prove that $\{m+k\} \subset n+k$:

In the base case, $m+k^+=(m+k)^+=\{m+k\} \cup (m+k)=n+k$, in which $\{m+k\} \subset n+k$ clearly holds. Assume that $m+k \subset n+k$. Then $n+k^+=(n+k)^+=\{n+k\} \cup (n+k) \supset n+k \supset m+k$.

Therefore, we know that $\{m+k\} \subset n+k$ and therefore $m+k \subset n+k$ and therefore $m+k < n+k$.

To be shown: if $m<n$ and $k \not =0$, then $m \cdot k<n \cdot k$.

Induction over $k$. Base case: $m \cdot 1=m<n=n \cdot 1$ (I don't think the text proves that $p_1(n)=n$, but I also don't think it would be that useful for me to do that here).

Induction assumption: $m \cdot k<n \cdot k$.

Induction step: It must hold that $m \cdot k^+=p_m(k)+m=(m \cdot k)+m<(n \cdot k)+n=p_n(k)+n=n \cdot k^+$. In general, if $m<n$ and $k<l$, then $m+k<n+l$ because $m+k<n+k=k+n<l+n$ (two applications of the theorem above, and using the commutativity of addition), so $(m \cdot k)+m<(n \cdot k)+n$.

Therefore, $m \cdot k<n \cdot k$.

To be shown: Ever non-empty set $E \subset ℕ$ has a minimum.

(I tried to prove this constructively, but the result was some weird amalgam of a constructive and and an inductive proof. Oh well.)

Let $e \in N$ be any element in $E$. Then if $e=\emptyset$, then there can be no element of $E$ smaller than $e$, and $e$ is smaller than any other natural number.

If $e \not=\emptyset$, then for any $e \in N$, if we know that $e \not \in E$ and $e^+ \in E$, then $e^+$ is the number so that no natural number smaller than $e^+ \in E$, since we know that all numbers $< e^+$ are not in $E$.

Exercise 2

Assume that there exists a bijection $f$ between $ω$ and some finite natural number $n$. Then $n \subset ω$, but $ω \not \subseteq n$, since $n \subset n^+=\{n\} \cup n \subseteq ω$. Then, by the pigeonhole principle, $f$ can't exist: there is at least one element too many in $ω$ to be matched to $n$.

Therefore, such an $n$ can't exist, $ω$ is infinite.

Exercise 3

The proof here is the same as the one in exercise 2.

Exercise 4

To be shown: the union of a finite set of finite sets is finite (wouldn't this better be if it was a finite family of finite sets? Whatever.).

Proof: Let $\mathcal{S}$ be our set, and $\bigcup_{S \in \mathcal{S}} S$ be the union of the finitely many elements of $\mathcal{S}$.

If $\#(\mathcal{S})=0$, then the union is the empty set $\emptyset$, which is clearly finite (equivalent to $0$.

If $\#(\mathcal{S})=1$, then the resulting set is just the only element of $\mathcal{S}$, which is per definition finite.

Assume that $\#(\mathcal{S})=n$, and that $\bigcup_{S \in \mathcal{S}} S$ is finite. Let then $\mathcal{S}'=\mathcal{S} \cup \{S_+\}$, where $S_+$ is a finite set. Then $\bigcup_{S \in \mathcal{S}'} S=S_+ \cup \bigcup_{S \in \mathcal{S}} S$. Since we know that both $\bigcup_{S \in \mathcal{S}} S$ and $S_+$ are finite, we know that $\bigcup_{S \in \mathcal{S}'} S$ is therefore also finite, per the statement in the text that the union of two finite sets is also finite.

But we can also prove that $E \cup F$ is finite if $E$ and $F$ are finite: If $f \in F$, then $E \cup \{f\}$ is still finite (either $\#(E \cup \{f\})=\#(E)$ or $\#(E \cup \{f\})=(\#(E))^+$). We can then set $F:=F \backslash \{f\}$ and $E:=E \cup \{f\}$. Since we can repeat this only finitely many times, the resulting $E$ must be still finite when $F=\emptyset$ (this is one of my very few constructive proofs, be gentle please).

To be shown: if $E$ is finite, then $\mathcal{P}(E)$ is finite, and $\#(\mathcal{P}(E))=2^{\#(E)}$.

Proof: This was shown in Exercise 2 in Section 5 (I use the notation of $|E|$ for the size of a set there). Since we can construct the size of $\mathcal{P}(E)$, it is equivalent to some natural number, and therefore finite.

To be shown: 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 $E$.

We start and set $k:=0$. We then choose an element from $m \in E$, and set $E:=E \backslash \{m\}$. If $k \le m$, we set $k:=m$. Else we do nothing. We know this process is repeated finitely many times, and once we end up with $E=\emptyset$, we have created a $k$ such that $m \le k$ for all $m \in E$.

(Is this a constructive proof? It feels like one, but also like writing code.)

Section 14

Exercise 1

Totality: $R \cup R^{-1}=X \times X$

Antisymmetry: $R \cap R^{-1}=\{(x,x)| x \in X\}$

Stray Exercise 1

A set $E$ may have no lower bounds or upper bounds at all, or it may have many; in the latter case it could happen that none of them belongs to $E$. (Examples?)

Example for set without lower & upper bounds is $X=∅$ (where always $E=∅$), if $X=\{a,b,c,d\}$ and the partial order on $X$ is $a<b<d$ and $a<c<d$ and $E=\{d\}$ then the lower bounds are $\{a,b,c\}$, if $E=∅$, then every element of $X$ is a lower bound of $E$ (since the empty all-quantification is true), but none of them are in $E$.

Section 15

Stray Exercise 1

To be proven: The Cartesian product of a finite family of sets $\{X_i\}$, at least one of which is empty, is necessarily and sufficiently also empty.

Base case: As suggested by the book, I start the induction at 1: In this case $\{X_i\}=\{\{\}\}$, the set containing the empty set. The Cartesian product of $\{\}$ with no other set is also the empty set.

Induction assumption: If $\{X_i\}$ is a family of sets with $n$ elements, at least one of which is empty, then $X_1 \times X_2 \times \dots \times X_n=\emptyset$.

Induction step: If $\{X_i\}$ is a family of sets with $n+1$ elements, at least one of which is the empty set, then either $X_{n+1}$ is the empty set, or the empty set is is the first $n$ elements.

Let $X_{\star}=X_1 \times X_2 \times \dots \times X_n$. Then in the former case we know that $X_{\star} \times X_{n+1}=X_{\star} \times \emptyset=\emptyset$ (as per the text). In the latter case we know that $X_{\star} \times X_{n+1}=\emptyset \times X_{n+1}=\emptyset$, so in both cases we get the empty set as a result.

Stray Exercise 2

To prove:


there exists a function $f$ with domain $\mathcal{C}$ such that if $A \in \mathcal{C}$, then $f(A) \in A$



if $\mathcal{C}$ is a collection of pairwise disjoint non-empty sets, then there exists a set $A$ such that $A \cap C$ is a singleton for each $C$ in $\mathcal{C}$

are equivalent to the axiom of choice:

Axiom of choice. The Cartesian product of a non-empty family of non-empty sets is non-empty.


We need to prove two directions: $\text{I} \Rightarrow \text{AOC}$ and $\text{AOC} \Rightarrow \text{I}$.

For $\text{AOC} \Rightarrow \text{I}$, let $\mathcal{C}^{\times}=C_1 \times C_2 \times \dots$. Let $C^{\times}$ be any element of $\mathcal{C}^{\times}$ (Can we do that? Or does this then circularly depend on a choice function on $\mathcal{C}^{\times}$?). Then we can define $f$ by $f(C_i)=C^{\times}(i)$ (since $C^{\times}$ is a tuple).

For $\text{I} \Rightarrow \text{AOC}$, we can construct at least one element of the Cartesian product of $\mathcal{C}$ by constructing the element $(f(C_1), f(C_2), f(C_3), \dots)$. This tuple must be an element of the Cartesian product of the elements in $\mathcal{C}$, therefore that Cartesian product can't be empty.


This is actually a special case of I: we just define $f(C)=C \cap A$. This gives us the results from above.