0.1  (§3) Relations

• Def. A relation on $A$ is a subset $C\subset A\times A$. Write $xCy$ for $(x,y)\in C$; read “$x$ is in the relation $C$ to $y$”.

  Note. Contrast with §2. The graph of a function $A\to A$ is also a subset of $A\times A$, but a very special one: each $x$ occurs exactly once as a first coordinate. Delete that condition and what remains is a relation. Nothing else is required — any subset will do.

• Rmk. Two families of relations carry the weight in this course:

  • equivalence relations — used from §22 onwards (quotients);

  • order relations — used from §14 onwards (the order topology).

Equivalence relations and partitions

• Def. $C$ is an equivalence relation on $A$ if for all $x,y,z\in A$:

  1. (reflexivity) $xCx$;

  2. (symmetry) $xCy\Rightarrow yCx$;

  3. (transitivity) $xCy$ and $yCz$ $\Rightarrow$ $xCz$.

• Notation. Write $x\sim y$. The three conditions become $x\sim x$; $x\sim y\Rightarrow y\sim x$; $x\sim y$ and $y\sim z\Rightarrow x\sim z$.

• Caution. Symmetry and transitivity do not give reflexivity. The empty relation on a nonempty set satisfies (2) and (3) and fails (1).

  Note. The tempting bogus argument: “$x\sim y$ gives $y\sim x$, and transitivity then gives $x\sim x$.” It assumes there is some $y$ with $x\sim y$. Worth putting on the board and letting them find the hole; it is Munkres Exercise 3.3.

• Def. For $x\in A$ the equivalence class of $x$ is

  $$E_x=\{y\in A\mid y\sim x\}.$$

  Nonempty, since $x\in E_x$ by reflexivity.

• $\mathrm{★}$ Lem. Two equivalence classes are either disjoint or equal.

  Proof. Suppose $z\in E_x\cap E_{x^{\prime }}$, so $z\sim x$ and $z\sim x^{\prime }$; by symmetry and transitivity $x\sim x^{\prime }$. For any $w\in E_x$ we get $w\sim x\sim x^{\prime }$, hence $w\in E_{x^{\prime }}$; so $E_x\subset E_{x^{\prime }}$. The situation is symmetric in $x$ and $x^{\prime }$, so the reverse inclusion holds too.

• Def. A partition of $A$ is a collection of disjoint nonempty subsets whose union is $A$.

• $\mathrm{★}$ Thm. Equivalence relations on $A$ and partitions of $A$ determine one another:

  • $\sim$ gives the partition $𝒜=\{E_x\mid x\in A\}$ (disjoint by the lemma, nonempty by reflexivity, covering $A$ since $x\in E_x$);

  • a partition $𝒟$ gives $x\sim y$ iff $x,y$ lie in the same element of $𝒟$;

  • the two constructions are mutually inverse.

  Proof. For the second bullet: symmetry is immediate; reflexivity holds because $𝒟$ covers $A$; transitivity holds because distinct elements of $𝒟$ are disjoint, so the element containing $x$ is determined by $x$. The equivalence classes of this relation are exactly the members of $𝒟$. For uniqueness, if ${\sim }_1$ and ${\sim }_2$ give the same partition then for each $x$ the classes $E_x^1$ and $E_x^2$ are both the unique member of $𝒟$ containing $x$, hence equal; so ${\sim }_1={\sim }_2$.

• Ex. On ${ℝ}^2$, declare $p\sim q$ when $|p|=|q|$. Classes: circles centred at the origin, together with $\{0\}$.

• Ex. On ${ℝ}^2$, declare $(x_0,y_0)\sim (x_1,y_1)$ when $y_0=y_1$. Classes: the horizontal lines. More generally $y_0-x_0^2=y_1-x_1^2$ gives the vertical translates of a parabola.

• Caution. The collection of all lines in the plane is not a partition: two distinct lines can meet. Disjointness is the condition that fails.

  Note. Do one example where the classes are visibly a partition and one where the proposed pieces overlap. The picture does more than the axioms here. Point forward: in §22 the set of equivalence classes becomes a space, and this bookkeeping becomes the quotient topology.

Order relations

• Def. $C$ is an order relation (simple, or linear, order) on $A$ if:

  1. (comparability) $x\ne y\Rightarrow xCy$ or $yCx$;

  2. (nonreflexivity) $xCx$ holds for no $x$;

  3. (transitivity) $xCy$ and $yCz$ $\Rightarrow$ $xCz$.

• Notation. Write . Then $x\le y$ means or $x=y$; $y>x$ means .

• Rmk. (2) and (3) together forbid and at once: transitivity would force . So exactly one of , $x=y$, holds.

• Ex. The usual order on $ℝ$. A less familiar one: $x\prec y$ iff , or $x^2=y^2$ and .

• Ex. Relations satisfying (2) and (3) but not (1) — strict partial orders — are common; inclusion of subsets is one. [Mk §11]

• Def. For the open interval . If it is empty, $a$ is the immediate predecessor of $b$ and $b$ the immediate successor of $a$.

• Def.  and have the same order type if some bijection $f:A\to B$ satisfies .

• Ex. $(-1,1)$ and $ℝ$ have the same order type, via

  $$f(x)=\frac{x}{1-x^2}.$$

• Def. Dictionary order on $A\times B$:

• Ex. ${\mathbb{Z}}_{+}\times [0,1)$ in the dictionary order has the order type of $[0,\mathrm{\infty })$; $[0,1)\times {\mathbb{Z}}_{+}$ does not — in the latter every element has an immediate successor.

  Note. Same two factors, order of the factors swapped, genuinely different order types. Draw the two pictures: a line broken into intervals versus a stack of copies of ${\mathbb{Z}}_{+}$. This is the example that makes §14 worth doing.

The least upper bound property

• Def. Let $A_0\subset A$. An element $b\in A$ is an upper bound for $A_0$ if $x\le b$ for all $x\in A_0$; $A_0$ is bounded above if one exists. A smallest upper bound is the least upper bound, $sup{A}_0$. Dually: lower bound, bounded below, $inf{A}_0$.

• Caution. $sup{A}_0$ need not lie in $A_0$. If it does, it is the largest element of $A_0$.

• $\mathrm{★}$ Def. An ordered set $A$ has the least upper bound property if every nonempty subset that is bounded above has a least upper bound. Greatest lower bound property: dually.

• Thm. The two properties are equivalent.

• Ex. $(-1,1)$ has the least upper bound property; a subset bounded above in $(-1,1)$ has a real supremum, which must lie in $(-1,1)$.

• Ex. $B=(-1,0)\cup (0,1)$ does not. The set $\{-1/2n\mid n\in {\mathbb{Z}}_{+}\}$ is bounded above by every element of $(0,1)$, and has no least upper bound in $B$: the candidate $0$ has been removed.

  Note. One deleted point destroys the property. Keep this example — an ordered set with this property and no immediate successors is a linear continuum, and that is exactly what makes the intermediate value theorem work in §24.

0.2  (§4) The Integers and the Real Numbers

• Rmk. §§1–3 were the logical foundations: sets, functions, relations. Now the mathematical foundation — $\mathbb{Z}$ and $ℝ$ — stated as axioms rather than constructed.

  Note. Two routes are available: build $ℝ$ from set theory with bare hands, or assume it and list what you assumed. The first is honest logic and costs weeks; the second costs one blackboard. Take the second, say why, and move on.

• Def. A binary operation on $A$ is a function $A\times A\to A$. Written infix: $x+y$, $x\cdot y$, $x\circ y$ rather than $+(x,y)$.

Assumption — there is a set $ℝ$ with $+$, $\cdot$, and an order such that:

• Algebraic.

  1. $+$ and $\cdot$ are associative;

  2. $+$ and $\cdot$ are commutative;

  3. there are $0\ne 1$ with $x+0=x$ and $x\cdot 1=x$ for all $x$;

  4. every $x$ has an additive inverse; every $x\ne 0$ has a multiplicative inverse;

  5. $x\cdot (y+z)=x\cdot y+x\cdot z$.

• Mixed.

  1. $x>y\Rightarrow x+z>y+z$;  $x>y$ and $z>0\Rightarrow x\cdot z>y\cdot z$.

• $\mathrm{★}$ Order.

  1. has the least upper bound property [Mk §3];

  2. there is $z$ with .

• Notation. $-x$, $x-y$, $1/x$, $z/x$ are defined from (1)–(5); the usual laws of signs and of fractions are then theorems, not axioms. Likewise the laws of inequalities follow once (6) is adjoined.

• Def. A set with (1)–(5) is a field; with (6) as well, an ordered field; a set with an order satisfying (7) and (8) is a linear continuum.

  Note. Split the list on the board into the algebraist’s half and the topologist’s half. Only (7) and (8) survive into this course — they involve no arithmetic at all, and they are exactly what §24 will use.

• Rmk. (8) is redundant: given , the element $z=(x+y)/(1+1)$ lies strictly between. It is listed separately only because it, with (7), is what the topology depends on.

• Notation. $x$ is positive if $x>0$, negative if . ${ℝ}_{+}$ = positive reals; ${\overline{ℝ}}_{+}$ = nonnegative reals.

Defining the integers

• Def. $A\subset ℝ$ is inductive if $1\in A$ and $x\in A\Rightarrow x+1\in A$.

• $\mathrm{★}$ Def. ${\mathbb{Z}}_{+}={\bigcap }_{A\in 𝒜}A$, the intersection of all inductive subsets of $ℝ$.

  Note. The smallest inductive set. Nothing is being constructed — the positive integers are being carved out of a set we have already assumed. Say this; otherwise the definition looks like a trick.

• Rmk. ${ℝ}_{+}$ is inductive, so ${\mathbb{Z}}_{+}\subset {ℝ}_{+}$: positive integers really are positive. Also $\{x\mid x\ge 1\}$ is inductive, so $1$ is the smallest element of ${\mathbb{Z}}_{+}$.

• Thm. Basic properties:

  • ${\mathbb{Z}}_{+}$ is inductive;

  • (induction) an inductive $A\subset {\mathbb{Z}}_{+}$ equals ${\mathbb{Z}}_{+}$.

  Note. The principle of induction is not an extra axiom here. It is immediate from the definition: ${\mathbb{Z}}_{+}$ is contained in every inductive set.

• Def. $\mathbb{Z}={\mathbb{Z}}_{+}\cup \{0\}\cup (-{\mathbb{Z}}_{+})$;  $\mathbb{Q}=\{p/q\mid p,q\in \mathbb{Z},q\ne 0\}$. Sums, differences and products of integers are integers; quotients need not be.

• Rmk. No integer lies strictly between $n$ and $n+1$.

• Notation. $S_n=\{1,\mathrm{\dots },n-1\}$, the section of ${\mathbb{Z}}_{+}$ below $n$; so $S_1=\mathrm{\varnothing }$ and $S_{n+1}=\{1,\mathrm{\dots },n\}$.

Two alternative forms of induction

• $\mathrm{★}$ Thm. (Well-ordering) Every nonempty subset of ${\mathbb{Z}}_{+}$ has a smallest element.

  Proof. First show by induction on $n$ that every nonempty subset of $S_{n+1}=\{1,\mathrm{\dots },n\}$ has a smallest element. For $n=1$ the only such subset is $\{1\}$. For the step, let $C\subset \{1,\mathrm{\dots },n+1\}$ be nonempty: if $C=\{n+1\}$ we are done; otherwise $C\cap \{1,\mathrm{\dots },n\}$ is nonempty and its smallest element is smallest in $C$. Now take any nonempty $D\subset {\mathbb{Z}}_{+}$, pick $n\in D$, and apply this to $D\cap \{1,\mathrm{\dots },n\}$, which is nonempty; its smallest element is smallest in $D$.

• Thm. (Strong induction) If $A\subset {\mathbb{Z}}_{+}$ and, for every $n$, $S_n\subset A$ implies $n\in A$, then $A={\mathbb{Z}}_{+}$.

  Proof. If not, well-ordering gives a smallest $n\notin A$. Every positive integer below $n$ lies in $A$, i.e. $S_n\subset A$, so the hypothesis forces $n\in A$ — contradiction.

  Note. Note the direction of use: strong induction is proved from well-ordering, which is proved from ordinary induction. Worth drawing the arrow; students often assume the three are unrelated.

Where the least upper bound axiom is actually needed

• Rmk. Everything above used only (1)–(6). Axiom (7) has not yet been touched.

• $\mathrm{★}$ Thm. (Archimedean ordering property) ${\mathbb{Z}}_{+}$ has no upper bound in $ℝ$.

  Proof. Suppose it had one. By (7) it has a least upper bound $b$. Then $b-1$ is not an upper bound, so $n>b-1$ for some $n\in {\mathbb{Z}}_{+}$; hence $n+1>b$ with $n+1\in {\mathbb{Z}}_{+}$, contradicting that $b$ bounds ${\mathbb{Z}}_{+}$.

• Rmk. Other consequences of (7): the greatest lower bound property; existence of $\sqrt{x}$ for every $x>0$; and hence the existence of irrational numbers, e.g. $\sqrt{2}\notin \mathbb{Q}$.

  Note. This is the answer to “why assume something as strange as (7)?” Without it the positive integers could be bounded, square roots need not exist, and $ℝ$ could be $\mathbb{Q}$. Three sentences here buy a lot of goodwill.

• Caution. The familiar decimal symbols $2=1+1$, $3=2+1$, … do name the positive integers uniquely, but that is a fact we never need and will not prove.

0.3  (§5) Cartesian products

• Def. Indexing function for $𝒜$: a surjection $f:J\to 𝒜$; write $A_{\alpha }=f(\alpha )$, family ${\{A_{\alpha }\}}_{\alpha \in J}$. Not assumed injective: $A_{\alpha }=A_{\beta }$ is allowed for $\alpha \ne \beta$.

• Notation. ${\bigcap }_{\alpha \in J}{A}_{\alpha }$, ${\bigcup }_{\alpha \in J}{A}_{\alpha }$; finite case $A_1\cap \mathrm{\cdots }\cap A_n$.

• Def. $n$-tuple $=$ function $x:\{1,\mathrm{\dots },n\}\to X$, written $(x_1,\mathrm{\dots },x_n)$; ${\prod }_{i=1}^n{A}_i$;   $X^n$.

• Def. sequence $=$ function $x:\mathbb{N}\to X$, written ${(x_i)}_{i=1}^{\mathrm{\infty }}$ (also called an $\omega$-tuple);  ${\prod }_{i=1}^{\mathrm{\infty }}{A}_i=A_1\times A_2\times \mathrm{\cdots }$;  $X^{\omega }$.

• $\mathrm{★}$ Def. General product. $J$-tuple $=$ function $x:J\to X$, $x_{\alpha }=x(\alpha )$.

  $$\prod _{\alpha \in J}{A}_{\alpha }=\{x:J\to \bigcup _{\alpha \in J}{A}_{\alpha }|x(\alpha )\in A_{\alpha }\forall \alpha \},X^J=\prod _{\alpha \in J}X.$$

  Note. Stress: a point of an infinite product is a function. Everything about product topologies later reads more easily from this description.

0.4  (§6) Finite sets

• Def. Section of $\mathbb{N}$: $\{1,2,\mathrm{\dots },n\}$; for $n=0$ this is $\mathrm{\varnothing }$.

• Lem. $\{1,\mathrm{\dots },m\}\hookrightarrow \{1,\mathrm{\dots },n\}$ injective $\Rightarrow m\le n$.

  Proof. Induction on $n$: delete $f(m)=k$, use a bijection $\{1,\mathrm{\dots },n\}-\{k\}\to \{1,\mathrm{\dots },n-1\}$.

• Cor. Hence there is no injective $\{1,\mathrm{\dots },m\}\to \{1,\mathrm{\dots },n\}$ when $m>n$. (Pigeonhole.)

• Prop. A bijection $\{1,\mathrm{\dots },m\}\to \{1,\mathrm{\dots },n\}$ forces $m=n$.

  Proof. Apply the lemma to $f$ and to $f^{-1}$.

• Cor. Hence there is no bijection $\{1,\mathrm{\dots },m\}\to \{1,\mathrm{\dots },n\}$ when $m\ne n$.

• Def. $A$ is finite of cardinality $n$ if there is a bijection $A\to \{1,\mathrm{\dots },n\}$. Cardinality is well defined. Ex. $\mathrm{\varnothing }$ has cardinality $0$; singletons have cardinality $1$.

• Lem. $A\subset \{1,\mathrm{\dots },n\}$ is finite, of cardinality $\le n$; if $A⊊\{1,\mathrm{\dots },n\}$ the cardinality is .

• $\mathrm{★}$ Thm. A finite set admits no bijection with a proper subset of itself.

• Cor. $\mathbb{N}$ is not finite: $f(x)=x+1$ is a bijection of $\mathbb{N}$ with the proper subset $\mathbb{N}-\{1\}$.

• Cor. Any subset $B$ of a finite set $A$ is finite; if $B⊊A$ then .

• Prop. TFAE: (1) $A$ finite; (2) some $\{1,\mathrm{\dots },n\}\twoheadrightarrow A$; (3) some $A\hookrightarrow \{1,\mathrm{\dots },n\}$.

  Proof. $(2)\Rightarrow (3)$: send $x$ to $\min g^{-1}(x)$.

• Prop. Finite unions and finite products of finite sets are finite.

0.5  (§7) Countable and uncountable sets

• Def. $A$ is infinite if it is not finite. It is countably infinite if there is a bijection $A\to \mathbb{N}$; countable if it is finite or countably infinite; uncountable otherwise.

  Note. Sections of $\mathbb{N}$ are the model for finite; $\mathbb{N}$ itself is the model for countably infinite. Same sentence, one word changed.

• Ex. $\mathbb{Z}$ is countably infinite: $f(n)=2n$ for $n>0$ and $f(n)=-2n+1$ for $n\le 0$ bijects $\mathbb{Z}\to \mathbb{N}$.

• Ex. $\mathbb{N}\times \mathbb{N}$ is countably infinite — count along the anti-diagonals. A clean proof is below; the picture is not the proof.

The countability criterion

• $\mathrm{★}$ Thm. For $B\ne \mathrm{\varnothing }$, TFAE:

  1. $B$ is countable;

  2. there is a surjection $\mathbb{N}\twoheadrightarrow B$;

  3. there is an injection $B\hookrightarrow \mathbb{N}$.

  Proof. $(1)\Rightarrow (2)$: if $B$ is countably infinite this is the definition; if $B$ is finite, extend a bijection $\{1,\mathrm{\dots },n\}\to B$ to all of $\mathbb{N}$ by sending everything above $n$ to $h(1)$. $(2)\Rightarrow (3)$: $g(b)=\min f^{-1}(\{b\})$, as in §6. $(3)\Rightarrow (1)$: restricting the range, $B$ bijects with a subset of $\mathbb{N}$, so it is enough to know every subset of $\mathbb{N}$ is countable — the lemma below.

  Note. This is the workhorse. After this, nothing is proved by exhibiting a bijection; everything is proved by exhibiting a surjection from $\mathbb{N}$ or an injection into it. Say so explicitly.

• $\mathrm{★}$ Lem. An infinite subset $C\subset \mathbb{N}$ is countably infinite.

  Proof. Define $h:\mathbb{N}\to C$ by $h(1)=\min C$ and

  $$h\left(n\right)=\min \left[C-h\left(\{1,\mathrm{\dots },n-1\}\right)\right].$$

  The set being minimised is nonempty — otherwise $h$ would surject $\{1,\mathrm{\dots },n-1\}$ onto $C$, making $C$ finite — so $h$ is defined, using well-ordering at each step. Injective: for , $h(m)\in h(\{1,\mathrm{\dots },n-1\})$ while $h(n)\notin$ it. Surjective: given $c\in C$, the image $h(\mathbb{N})$ is infinite, hence not contained in $\{1,\mathrm{\dots },c\}$, so $h(n)>c$ for some $n$; take $m$ least with $h(m)>c$. Then for , so $c\notin h(\{1,\mathrm{\dots },m-1\})$, whence $h(m)\le c$. So $h(m)=c$.

• Caution. The definition of $h$ above is not a proof by induction. Induction proves a statement about a function already defined; here the function is being brought into existence. What licenses it is the principle of recursive definition: if a formula gives $h(1)$ uniquely, and for $i>1$ gives $h(i)$ uniquely from the values $h(1),\mathrm{\dots },h(i-1)$, then it determines a unique $h:\mathbb{N}\to A$. [Mk §8]

  Note. Not all recursions are legitimate: “$h(i)=\min [C-h(\{1,\mathrm{\dots },i+1\})]$” asserts that $h(i)$ is not in a set it belongs to. Same shape as the barber who shaves exactly the men who do not shave themselves. Worth one minute — it is the difference between a definition and a wish.

Closure properties

• Cor. A subset of a countable set is countable.

  Proof. Restrict the injection into $\mathbb{N}$.

• $\mathrm{★}$ Cor. $\mathbb{N}\times \mathbb{N}$ is countably infinite.

  Proof. $f(n,m)=2^n{3}^m$ is injective: $2^n{3}^m=2^p{3}^q$ with gives $3^m=2^{p-n}{3}^q$, and the left side is odd, so $n=p$; then $3^m=3^q$ forces $m=q$. Apply the criterion.

  Note. The anti-diagonal bijection is prettier but fiddly to verify. Unique factorisation makes it a one-liner — and it generalises immediately to ${\mathbb{N}}^k$.

• Ex. ${\mathbb{Q}}_{+}$ is countably infinite: $g(n,m)=m/n$ surjects $\mathbb{N}\times \mathbb{N}\to {\mathbb{Q}}_{+}$, and ${\mathbb{Q}}_{+}\supset \mathbb{N}$ is infinite. Same argument for $\mathbb{Q}$.

  Note. The rationals are dense in $ℝ$ and yet no more numerous than $\mathbb{N}$. Students find this the first genuinely surprising statement of the course; let it land before moving on.

• $\mathrm{★}$ Thm. A countable union of countable sets is countable.

  Proof. Index the family by $J\subset \mathbb{N}$ and choose surjections $f_n:\mathbb{N}\to A_n$ and $g:\mathbb{N}\to J$. Then $h(k,m)=f_{g(k)}(m)$ surjects $\mathbb{N}\times \mathbb{N}$ onto ${\bigcup }_{n\in J}{A}_n$, and $\mathbb{N}\times \mathbb{N}\leftrightarrow \mathbb{N}$.

• Caution. The word choose there is doing real work: infinitely many $f_n$ are selected at once. This is an appeal to the axiom of choice. [Mk §9]

• Thm. A finite product of countable sets is countable.

  Proof. For two factors, $h(n,m)=(f(n),g(m))$ surjects $\mathbb{N}\times \mathbb{N}\to A\times B$; then induct using $A_1\times \mathrm{\cdots }\times A_n\leftrightarrow (A_1\times \mathrm{\cdots }\times A_{n-1})\times A_n$.

• Caution. Finite is essential — see the next theorem.

An uncountable set

• $\mathrm{★}$ Thm. Let $X=\{0,1\}$. Then $X^{\omega }$ is uncountable.

  Proof. (Cantor diagonal.) Let $g:\mathbb{N}\to X^{\omega }$ be any function and write $g(n)=(x_{n1},x_{n2},x_{n3},\mathrm{\dots })$. Define $y=(y_1,y_2,\mathrm{\dots })$ by

  Then $y\in X^{\omega }$ and $y\ne g(n)$ for every $n$, since the two differ in coordinate $n$. So $g$ is not surjective, and by the criterion $X^{\omega }$ is not countable.

  Note. Draw the array and circle the diagonal. Emphasise that no cleverness in choosing $g$ can help — the argument defeats every $g$ at once. That uniformity is the point, and it is the same move that will reappear for the uncountability of $ℝ$.

• Rmk. So countable products of countable sets need not be countable, in sharp contrast to finite products. [Mk §19] — the same contrast reappears for product topologies, where the box and product topologies agree on finite products and diverge on infinite ones.

• Rmk. $ℝ$ is uncountable, by essentially the same diagonal argument applied to decimal expansions.