Paul R. Halmos. Naive Set Theory. Van Nostrand Reinhold Company. 1960.

# Section 1, “The Axiom of Extension”

p. 2:

Axiom of extension. Two sets are equal if and only if they have the same elements.

# Section 2, “The Axiom of Specification”

p. 6:

Axiom of specification. To every set $$A$$ and to every condition $$S(x)$$ there corresponds a set $$B$$ whose elements are exactly those elements $$x$$ of $$A$$ for which $$S(x)$$ holds.

A “condition” here is just a sentence. The symbolism is intended to indicate that the letter $$x$$ is free in the sentence $$S(x)$$; that means that $$x$$ occurs in $$S(x)$$ at least once without being introduced by one of the phrases “for some $$x$$” or “for all $$x$$.” It is an immediate consequence of the axiom of extension that the axiom of specification determines the set $$B$$ uniquely. To indicate the way $$B$$ is obtained from $$A$$ and from $$S(x)$$ it is customary to write

$B = \{x \in A : S(x) \}.$

In other words, in the above $$S(x)$$ is a predicate.

# Section 3, “Unordered Pairs”

p. 9:

Axiom of pairing. For any two sets there exists a set that they both belong to.

p. 10:

If a is a set, we may form the unordered pair $$\{a, a\}$$. That unordered pair is denoted by $$\{a\}$$ and is called the singleton of $$a$$; it is uniquely characterized by the statement that it has $$a$$ as its only element. Thus, for instance, $$\emptyset$$ and $$\{\emptyset\}$$ are very different sets; the former has no elements, whereas the latter has the unique element $$\emptyset$$. To say that $$a \in A$$ is equivalent to saying that $$\{a\} \subset A$$.

# Section 4, “Unions and Intersections”

p. 12:

Axiom of unions. For every collection of sets there exists a set that contains all the elements that belong to at least one set of the given collection.

# Section 5, “Complements and Powers”

p. 19:

Axiom of powers. For each set there exists a collection of sets that contains among its elements all the subsets of the given set.

For a set $$A$$, denote by $$\mathscr{P}(A)$$ the power set of $$A$$, the set whose elements are the subsets of $$A$$.

# Section 6, “Ordered Pairs”

p. 23:

The ordered pair of $$a$$ and $$b$$, with first coordinate $$a$$ and second coordinate $$b$$, is the set $$(a,b)$$ defined by

$(a,b) = \{\{a\},\{a,b\}\}.$

However convincing the motivation of this definition may be, we must still prove that the result has the main property that an ordered pair must have to deserve its name. We must show that if $$(a, b)$$ and $$(x, y)$$ are ordered pairs and if $$(a,b)=(x,y)$$, then $$a = x$$ and $$b = y$$. To prove this, we note first that if $$a$$ and $$b$$ happen to be equal, then the ordered pair $$(a,b)$$ is the same as the singleton $$\{\{a\}\}$$. If, conversely, $$(a, b)$$ is a singleton, then $$\{a\}=\{a,b\}$$, so that $$b \in \{a\}$$, and therefore $$a=b$$. Suppose now that $$(a,b)=(x,y)$$. If $$a = b$$, then both $$(a, b)$$ and $$(x, y)$$ are singletons, so that $$x = y$$; since $$\{x\} \in (a,b)$$ and $$\{a\} \in (x,y)$$, it follows that $$a$$, $$b$$, $$x$$, and $$y$$ are all equal. If $$a \neq b$$, then both $$(a, b)$$ and $$(x, y)$$ contain exactly one singleton, namely $$\{a\}$$ and $$\{x\}$$ respectively, so that $$a = x$$. Since in this case it is also true that both $$(a, b)$$ and $$(x, y)$$ contain exactly one unordered pair that is not a singleton, namely $$\{a, b\}$$ and $$\{x, y\}$$ respectively, it follows that $$\{a,b\}=\{x,y\}$$, and therefore, in particular, $$b \in \{x,y\}$$. Since $$b$$ cannot be $$x$$ (for then we should have $$a = x$$ and $$b = x$$, and, therefore, $$a = b$$), we must have $$b = y$$, and the proof is complete.

pp. 23-24:

If $$A$$ and $$B$$ are sets, does there exist a set that contains all the ordered pairs $$(a,b)$$ with $$a$$ in $$A$$ and $$b$$ in $$B$$? It is quite easy to see that the answer is yes. Indeed, if $$a \in A$$ and $$b \in B$$, then $$\{a\} \subset A$$ and $$\{b\} \subset B$$, and therefore $$\{a,b\} \subset A \cup B$$. Since also $$\{a\} \subset A \cup B$$, it follows that both $$\{a\}$$ and $$\{a,b\}$$ are elements of $$\mathscr{P}(A \cup B)$$. This implies that $$\{\{a\},\{a,b\}\}$$ is a subset of $$\mathscr{P}(A \cup B)$$, and hence that it is an element of $$\mathscr{P}(\mathscr{P}(A \cup B))$$; in other words $$(a,b) \in \mathscr{P}(\mathscr{P}(A \cup B))$$ whenever $$a \in A$$ and $$b \in B$$. Once this is known, it is a routine matter to apply the axiom of specification and the axiom of extension to produce the unique set $$A \times B$$ that consists exactly of the ordered pairs $$(a,b)$$ with $$a$$ in $$A$$ and $$b$$ in $$B$$. This set is called the Cartesian product of $$A$$ and $$B$$; it is characterized by the fact that

$A \times B = \{x : x=(a,b) \; \textrm{for some} \; a \; \textrm{in} \; A \; \textrm{and for some} \; b \; \textrm{in} \; B\}.$

p. 25:

The charge of artificiality is true, but it is not too high a price to pay for conceptual economy. The concept of an ordered pair could have been introduced as an additional primitive, axiomatically endowed with just the right properties, no more and no less. In some theories this is done. The mathematician’s choice is between having to remember a few more axioms and having to forget a few accidental facts; the choice is pretty clearly a matter of taste. Similar choices occur frequently in mathematics; in this book, for instance, we shall encounter them again in connection with the definitions of numbers of various kinds.

p. 25:

Exercise. If either $$A=\emptyset$$ or $$B=\emptyset$$, then $$A \times B = \emptyset$$, and conversely. If $$A \subset X$$ and $$B \subset Y$$, then $$A \times B \subset X \times Y$$, and (provided $$A \times B \neq \emptyset$$) conversely.

# Section 7, “Relations”

pp. 26-27:

We may not know what a relation is, but we do know what a set is, and the preceding considerations establish a close connection between relations and sets. The precise set-theoretic treatment of relations takes advantage of that heuristic connection; the simplest thing to do is to define a relation to be the corresponding set. This is what we do; we hereby define a relation as a set of ordered pairs. Explicitly: a set $$R$$ is a relation if each element of $$R$$ is an ordered pair; this means, of course, that if $$z \in R$$, then there exist $$x$$ and $$y$$ so that $$z=(x,y)$$. If $$R$$ is a relation, it is sometimes convenient to express the fact that $$(x,y) \in R$$ by writing

$xRy$

and saying, as in everyday language, that $$x$$ stands in the relation $$R$$ to $$y$$.

p.27:

In the preceding section we saw that associated with every set $$R$$ of ordered pairs there are two sets called the projections of $$R$$ onto the first and second coordinates. In the theory of relations these sets are known as the domain and the range of $$R$$ (abbreviated dom $$R$$ and ran $$R$$); we recall that they are defined by

$\textrm{dom} R = \{x : \textrm{for some} \; y \; (x R y)\}$

and

$\textrm{ran} R = \{y: \textrm{for some} \; x \; (x R y)\}.$

p. 27:

If $$R$$ is a relation included in a Cartesian product $$X \times Y$$ (so that $$\textrm{dom} R \subset X$$ and $$\textrm{ran} R \subset Y$$), it is sometimes convenient to say that $$R$$ is a relation from $$X$$ to $$Y$$; instead of a relation from $$X$$ to $$X$$ we may speak of a relation in $$X$$.

# Section 8, “Functions”

p. 30:

If $$X$$ and $$Y$$ are sets, a function from (or on) $$X$$ to (or into) $$Y$$ is a relation $$f$$ such that $$\textrm{dom} f = X$$ and such that for each $$x$$ in $$X$$ there is a unique element $$y$$ in $$Y$$ with $$(x,y) \in f$$. The uniqueness condition can be formulated explicitly as follows: if $$(x,y) \in f$$ and $$(x,z) \in f$$, then $$y=z$$. For each $$x$$ in $$X$$, the unique $$y$$ in $$Y$$ such that $$(x,y) \in f$$ is denoted by $$f(x)$$. For functions this notation and its minor variants supersede the others used for more general relations; from now on, if $$f$$ is a function, we shall write $$f(x)=y$$ instead of $$(x,y) \in f$$ or $$x f y$$. The element $$y$$ is called the value that the function sends or maps or transforms $$x$$ onto $$y$$. The words map or mapping, transformation, correspondence, and operator are among some of the many that are sometimes used as synonyms for function. The symbol

$f:X \to Y$

is sometimes used as an abbreviation for “$$f$$ is a function from $$X$$ to $$Y$$.” The set of all functions from $$X$$ to $$Y$$ is a subset of the power set $$\mathscr{P}(X \times Y)$$; it will be denoted by $$Y^X$$.

p. 33:

Exercise. (i) $$Y^\emptyset$$ has exactly one element, namely $$\emptyset$$, whether $$Y$$ is empty or not, and (ii) if $$X$$ is not empty, then $$\emptyset^X$$ is empty.

# Section 10, “Inverses and Composites”

pp. 38-39:

A correspondence between the elements of $$X$$ and the elements of $$Y$$ does always induce a well-behaved correspondence between the subsets of $$X$$ and the subsets of $$Y$$, not forward, by the formation of images, but backward, by the formation of inverse images. Given a function $$f$$ from $$X$$ to $$Y$$, let $$f^{-1}$$, the inverse of $$f$$, be the function from $$\mathscr{P}(Y)$$ to $$\mathscr{P}(X)$$ such that if $$B \subset Y$$, then

$f^{-1}(B) = \{x \in X: f(x) \in B\}.$

In words: $$f^{-1}(B)$$ consists of exactly those elements of X that $$f$$ maps into $$Y$$; the set $$f^{-1}(B)$$ is called the inverse image of $$B$$ under $$f$$. 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?) A necessary and sufficient condition that $$f$$ be one-to-one is that the inverse image under $$f$$ of each singleton in the range of $$f$$ be a singleton in $$X$$.

If the last condition is satisfied, then the symbol $$f^{-1}$$ is frequently assigned a second interpretation, namely as the function whose domain is the range of $$f$$, and whose value for each $$y$$ in the range of $$f$$ is the unique $$x$$ in $$X$$ for which $$f(x) = y$$. In other words, for one-to-one functions $$f$$ we may write $$f^{-1}(y)=x$$ if and only if $$f(x) = y$$. This use of the notation is mildly inconsistent with our first interpretation of $$f^{-1}$$, but the double meaning is not likely to lead to any confusion.

pp. 39-40:

The discussion of inverses shows that what a function does can in a certain sense be undone; the next thing we shall see is that what two functions do can sometimes be done in one step. If, to be explicit, $$f$$ is a function from $$X$$ to $$Y$$ and $$g$$ is a function from $$Y$$ to $$Z$$, then every element in the range of $$f$$ belongs to the domain of $$g$$, and, consequently, $$g(f(x))$$ makes sense for each $$x$$ in $$X$$. The function $$h$$ from $$X$$ to $$Z$$, defined by $$h(x) = g(f(x))$$ is called the composite of the functions $$f$$ and $$g$$; it is denoted by $$g \circ f$$ or, more simply, by $$gf$$. (Since we shall not have occasion to consider any other kind of multiplication for functions, in this book we shall use the latter, simpler notation only.)

Observe that the order of events is important in the theory of functional composition. In order that $$gf$$ be defined, the range of $$f$$ must be included in the domain of $$g$$, and this can happen without it necessarily happening in the other direction at the same time. Even if both $$fg$$ and $$gf$$ are defined, which happens if, for instance, $$f$$ maps $$X$$ into $$Y$$ and $$g$$ maps $$Y$$ into $$X$$, the functions $$fg$$ and $$gf$$ need not be the same; in other words, functional composition is not necessarily commutative.

Functional composition may not be commutative, but it is always associative. If $$f$$ maps $$X$$ into $$Y$$, if $$g$$ maps $$Y$$ into $$Z$$, and if $$h$$ maps $$Z$$ into $$U$$, then we can form the composite of $$h$$ with $$gf$$ and the composite of $$hg$$ with $$f$$; it is a simple exercise to show that the result is the same in either case.

The connection between inversion and composition is important; something like it crops up all over mathematics. If $$f$$ maps $$X$$ into $$Y$$ and $$g$$ maps $$Y$$ into $$Z$$, then $$f^{-1}$$ maps $$\mathscr{P}(Y)$$ into $$\mathscr{P}(X)$$ and $$g^{-1}$$ maps $$\mathscr{P}(Z)$$ into $$\mathscr{P}(Y)$$. In this situation, the composites that are formable are $$gf$$ and $$f^{-1}g^{-1}$$; the assertion is that the latter is the inverse of the former. Proof: if $$x \in (gf)^{-1}(C)$$, where $$x \in X$$ and $$C \subset Z$$, then $$g(f(x)) \in C$$, so that $$f(x) \in g^{-1}(C)$$, and therefore $$x \in f^{-1}(g^{-1}(C))$$; the steps of the argument are reversible.

# Categories

A category C comprises the following data: 1 2

1. A class of objects $$\textrm{Obj}(\mathbf{C})$$.

2. For each pair of objects $$X,Y$$, a class of morphisms $$\textrm{hom}_{\mathbf{C}}(X,Y)$$. A morphism $$f \in \textrm{hom}_{\mathbf{C}}(X,Y)$$ has domain $$\textrm{dom}(f)=X$$ and codomain $$\textrm{cod}(f)=Y$$.

3. For each triple of objects $$X,Y,Z$$, a map $$\begin{equation} \circ_{X,Y,Z}:\textrm{hom}_{\mathbf{C}}(X,Y) \times \textrm{hom}_{\mathbf{C}}(Y,Z) \to \textrm{hom}_{\mathbf{C}}(X,Z) \end{equation}$$ called composition and denoted by $$\circ_{X,Y,Z}:(f,g) \mapsto g \circ f$$. 3

4. For each object $$X$$, a morphism $$\textrm{id}_X \in \textrm{hom}_{\mathbf{C}}(X,X)$$, called the identity morphism.

The data has to satisfy the following rules:

1. For each morphism $$f$$, it holds that $$f \circ \textrm{id}_{\textrm{dom}(f)} = f$$, called the right identity law. 4

2. For each morphism $$f$$, it holds that $$\textrm{id}_{\textrm{cod}(f)} \circ f = f$$, called the left identity law.

3. For each triple of morphisms $$f,g,h$$, if $$\textrm{cod}(f)=\textrm{dom}(g)$$ and $$\textrm{cod}(g)=\textrm{dom}(h)$$ then
$$h \circ (g \circ f) = (h \circ g) \circ f,$$ called the associative law. 5

# The category structure of Set

$$\textrm{Obj}(\mathbf{Set})$$ is the class of sets.

For each pair of sets $$X,Y$$, the class of morphisms $$\textrm{hom}_{\mathbf{Set}}(X,Y)$$ is the set of functions $$Y^X$$. 6

$$Y^\emptyset$$ has exactly one element, namely $$\emptyset$$, whether $$Y$$ is empty or not, and (ii) if $$X$$ is not empty, then $$\emptyset^X$$ is empty.

For each triple of objects $$X,Y,Z$$, the composition

$\circ_{X,Y,Z}:Y^X \times Z^Z \to Z^X$

## $$X \neq \emptyset, Y \neq \emptyset, Z \neq \emptyset$$

$\circ_{X,Y,Z}(f,g)(x) = g(f(x)),\quad f \in Y^X, g \in Z^Y, \quad x \in X.$

## $$X = \emptyset, Y \neq \emptyset, Z \neq \emptyset$$

$Y^X=\{\emptyset\}$ $\circ_{X,Y,Z}(\emptyset,g) = \emptyset, \quad g \in Z^Y.$

## $$X = \emptyset, Y = \emptyset, Z = \emptyset$$

For each set $$X$$ that is not empty, the identity morphism $$\textrm{id}_X \in X^X$$ is

$\textrm{id}_X(x) = x, \quad x \in X.$

When $$X$$ is empty,

$\mathrm{id}_\emptyset = \emptyset \in \emptyset^\emptyset.$
1. When for each pair of objects $$X,Y$$ it holds that the class of morphisms $$\textrm{hom}_{\mathbf{C}}(X,Y)$$ is a set, the category C is called locally small. See locally small category in nLab

When the class of objects and the class of morphisms are both sets, the category is called small. For doing algebraic geometry my impression is that there one works with small categories. See The Stacks project