topology homework for celene

Asking me to teach you math is a bad idea unless you're sure I won't be evil about it

Recently I was, as one often does, arguing with my girlfriend about the correspondence between the notion of a utility function and the various preferences, values, hopes, wants, wishes, desires, inclinations, or similar abstract folk-psychological or axiological objects (hereafter “values”) that a person might have¹¹ My view, for the most part, is that while utility functions and expected utility maximization and so on are very worthwhile frameworks to study and suggest many principles that ought to end up being approximately true with some caveats, they also aren’t the be-all-end-all.

I don’t know what my utility function is, “a mapping which takes a point in the quantum state space for the entire quantum multiverse and maps it to some real number, such that states with higher utility are more preferred” is certainly not a sort of object that I have any real conception of how to even begin to productively think about—and the “quantum” and “multiverse” parts aren’t even the main problem—literally any behavior a person could take is exactly the behavior prescribed by an incredibly large set of possible utility functions, I frequently need quite a lot of reflection to figure out which of two options I prefer, and the reflection is often more about introspection than about attempting to calculate simple empirical facts.

But like, yeah, obviously no piece of mathematics yet found can really fully deal with the complexity of Reality, and nonetheless they can be insightful and useful to discuss.. I tried to explain that the idea of my “true utility function” expressing indifference between two randomly chosen world states seemed a little bizarre to me. There are many things I simply do not have the bandwidth to form opinions on, but which I would form some sort of convergent preference about if I had a good reason to consider it.

Probably I did not explain this spectacularly clearly, because Celene’s takeaway was that I believed utility functions ought to be injective—specifically, she said something about how a set of possible worlds of cardinality ℵ₂ couldn’t possibly inject into R²² I haven’t managed to get her to yet internalize that the Continuum Hypothesis is, while arguably a somewhat subjective question, certainly at the very least in heavy contention.. Now, she’s not wrong that injectivity is ridiculous. But it seems to me that, if you have very much topological intuition, you shouldn’t really be misunderstanding people as asking for injective utility functions when the domain is anything like all possible fully-detailed states of Reality, unless maybe if you’re explicitly modeling it as discrete. It follows from the pretty trivial³³ Well, relative to other stuff in formal topology, anyways. one-dimensional case of Borsuk-Ulam that, if there is any single closed curve on which the utility function is continuous, then it can’t possibly be injective. This is a weaker condition than being nowhere continuous, but it’s not much weaker. Injectivity is an obviously ridiculous condition, but mostly because “the world state might have some real-valued parameters” and not because of anything having to do with sets too large to inject into R!

So I ranted about this to Celene a little, because she likes it when I do that, and eventually I ended up telling her that I would write up some topology homework for her to try to work through. Because I am evil, I am going to just write up a big list of problems that I expect will be very difficult to get through if you have very little formal mathematics experience. Celene once learned some Lean, which I hope at least helps a little for thinking about simple boolean logic statements and such.

In Zermelo-Fraenkel (ZF) set theory, the axiom of power set states

where z⊆x (“z is a subset of x”) is defined by

Verbally, these are read as “for all sets x, there exists a set y such that for all sets z, z is an element of y if and only if z is a subset of x” and “z is a subset of x if and only if, for all sets w, w being an element of z implies that w is an element of x.” For any given x, we call the corresponding y the power set of X, denoted 𝓟(X).

Exercise 1: Describe the definition of a subset and the axiom of power set in more colloquial phrasing, while still being reasonably precise. For example, you might use the phrasing “A contains the Bs” to mean that the set C is an element of A if and only if C is a B—where the “if” is explicitly stated, and the “only if” is a Gricean implicature via the maxim of quantity.

Exercise 2: Determine which of the ZF axioms can be (most directly) used to establish that any set X has a unique power set 𝓟(X).

The union of A and B is the set A∪B such that, for all x, x is an element of A∪B if and only if x is an element of A or an element of B. If we make an exception and permit an unbounded quantification in set-builder notation, we can write this definition as A∪B :=⁴⁴ The expression “X := Y” means “X is defined to be equal to Y”. {x : x∈A or x∈B}.

Proving that such a set exists requires using the axiom of pairing to create the set {A, B} and the axiom of union to create the set ⋃{A, B} = A∪B, so arguably one ought to first define the union of a family⁵⁵ This literally just means set. We use “family” because “set of sets” is confusing. of sets.

Exercise 3: Invent a plausible formal definition for ⋃X, the union of the sets which are elements of X.

One of the most essential concepts to develop when building up from raw set theory is the notion of a function. However, functions are difficult to define without first defining the Cartesian product. The Cartesian product furthermore relies upon a concept of an ordered pair. But a definition of ordered pairs in terms of functions is far more convenient than a definition in terms of raw set theory. Consequently, we will give a temporary definition of ordered pair, and then immediately scrap it for a better one after we have defined functions.

The Kuratowski pair (a, b)_K is the set {{a}, {a, b}}.

This is a suitable notion of an ordered pair because it satisfies the condition that {{a}, {a, b}} = {{c}, {c, d}} if and only if a = c and b = d.

Exercise 4: Marvel at how awful it is to work with these in raw set theory.

The Kuratowski Cartesian product of A and B, written A×_KB, is the set of all Kuratowski pairs of an element of A and an element of B,

A function f with domain X, codomain Y, and graph ↦∈𝓟(X×_KY) (where we write x↦y to denote (x, y)_K∈↦) must satisfy both of the properties

1. (Totality, cosurjectivity): for all x∈X, we have x↦y for some y∈Y.

2. (Single-valuedness, coinjectivity): if x↦y₁ and x↦y₂, then y₁=y₂.

or, equivalently, the single property

3. (Cobijectivity): for all x∈X, there is precisely one y∈Y such that x↦y.

Exercise 5: Explain why these are equivalent conditions. If you can, write a formal proof of this fact. Equivalence of the propositions P and Q, “P ⟺ Q” or “P if and only if Q”, can be proven by proving both P ⟹ Q and its converse Q ⟹ P. The implication P ⟹ Q can be proven by assuming P to be true, and deriving that Q must then be true.

If X, Y, and ↦ satisfy the required properties, then we define

For x, y such that x↦y, we define f(x) := y. We write f: X → Y to denote that f is a function with domain X and codomain Y.

A function that fails coinjectivity, such as the function which maps a complex number to both of its square roots, is known as a multi-valued function. A function that fails cosurjectivity, such as the reciprocal function on R, is known as a partial function. It is sometimes preferred to use a domain of R^×, the set of nonzero reals, rather than working with a partial function.

A function f: X → Y is known as injective if for all y∈Y, if x₁ ↦ y and x₂↦ y then x₁ = x₂; as surjective if for all y∈Y, we have x↦y for some x∈X; and as bijective if it is both injective and surjective.

Exercise 6: A function f: X → Y is invertible if there exists a function f^-1: Y → X such that (for all x∈X and y∈Y) f^-1(f(x)) = x and f(f^-1(y)) = y. If f is invertible, then f^-1 is known as the inverse of f. Prove that a function is invertible if and only if it is bijective.

Exercise 7: A partial function must instead be defined as f := (X, (Y, ↦)_K)_K instead of as f := (Y, ↦)_K. Why is this necessary for partial functions in particular? (If you’d rather not use Kuratowski pairs, you could wait until you have a definition of 3-tuples so you can define it as f := (X, Y, ↦).)

The set of all functions f: X → Y is written {f: X → Y} or Y^X. There is a canonical isomorphism between 2^X = {0, 1}^X and 𝓟(X), which maps the subset A⊆X to the function f: X → 2 where f(x) = 1 if x∈A and f(x) = 0 otherwise.

Let f: X → Y be a function, x∈X, y∈Y, A⊆X, and B⊆Y:

• The image f(x) of x is the point f(x).

• The image f(A) of A is the set {f(a) : a∈A}.

• The image of f is the set f(X) = {y∈Y : y = f(x) for some x}.

• The preïmage f^-1(B) of B is the set {x∈X : f(x)∈B}.

• The fiber f^-1(y) of y is the set f^-1({y}), but probably no one will get too mad if you call it a preïmage.

Fibers and preïmages are well-defined even if f is not invertible. The notation f(A) for the image of A conflates f: X → Y with a function f: 𝓟(X) → 𝓟(Y); the notation for a preïmage is even worse, since f^-1: 𝓟(Y) → 𝓟(X) is generally not an inverse of f: 𝓟(X) → 𝓟(Y)—I think maybe they’re inverses only when f: X → Y is bijective. And the fiber introduces a function f^-1: Y → 𝓟(X)! It’s fine though. You can write f[A] or f^-1[B] or f^-1[y] if you really need to disambiguate, but honestly it'll almost always not really matter.

For a natural number n and a set X, an n-tuple on X is a function T: n → X, where n is interpreted as the von Neumann ordinal⁶⁶ To be clear, 0 = {0, 1, 2, …, -1} = { }, 1 = {0, 1, 2, …, 0} = {0}, 2 = {0, 1, 2, …, 1} = {0, 1}, etc. The “no numbers past n-1” condition takes precedence over “it starts with 0, 1, and 2” condition.

We write the tuple T using the notation (T(0), T(1), T(2), …, T(n-1)). There is a unique 0-tuple on X, the empty tuple ( ), which is given by the unique function ∅⁷⁷ The symbol ∅ denotes the empty set, { }. Note that ∅ = 0 when 0 is interpreted as a von Neumann ordinal. → X, the empty function (X, ∅)_K. A 1-tuple is often called a singleton, and is not in any regard appreciably different from the set containing the tuple’s only entry. A 2-tuple, 3-tuple, etc is often called an ordered pair, triple, quadruple, quintuple, sextuple, septuple, octuple, nonuple, decuple, etc. The terms “ordered pair” and “unordered pair” distinguish between 2-tuples and sets with two elements.

Exercise 8: For natural numbers n and m, how many elements of {f: n → m} are there?

Alternatively, an n-tuple is a surjective function with domain n. The codomain of an n-tuple is then the set containing each entry in the tuple. Using the image of the function as the codomain feels like a weirder definition when saying something like “the Cartesian plane is the set of 2-tuples on R”, but is inconvenient if you’re drawing the entries of the tuple from disparate sets—as is common when you wish to define some mathematical object by using a few specific other objects as parameters, such as when a function is defined as a (Kuratowski) pair of the codomain and the graph.

An indexed family of sets {A_i}_i∈I is an I-tuple where i↦A_i. We refer to I as the indexing set, and we call its elements indices. Any set can be used as an indexing set.

The union ⋃A_i is given by exercise three, where it is written ⋃{A_i}. Be careful not to confuse yourself too much about the distinction between the sets A_i and the family {A_i} containing the sets A_i. This distinction hardly makes any difference, just don’t accidentally apply a formalism that gives you the unary union of {A_i} when you really wanted the I-ary union of all the sets A_i. The empty union is the empty set.

The intersection ∩A_i is given by

The empty intersection is only a legitimate concept when your sets A_i are constrained to be subsets of some particular set X, in which case it is X. For example, if you are concerned with the intersections of various intervals of real numbers, then it makes sense to view the empty intersection as the set R.

Two sets are disjoint if their intersection is the empty set. A family of sets is pairwise disjoint if every (unordered) pair of two sets from the family is disjoint.

The Cartesian product ∏A_i is given by

The Cartesian product A×B is a special case where the indexing set is 2 = {0, 1}.

We can interpret ∏A_i as the set of I-tuples where the i-th entry is an element of A_i. A Cartesian product has a corresponding family of canonical projections, which are functions π_i: ∏A_i → A_i given by π_i(a) = a(i), returning the i-th entry of a.

The Cartesian product of n copies of a set X is often written Xⁿ, because the Cartesian product of n copies of a set X is indeed the set of functions from n to X (which is furthermore the set of n-tuples on X).

Exercise 9: What is the empty Cartesian product?

The coproduct or disjoint union ∐A_i is given by

The coproduct is distinct from a union in how it handles overlapping sets. The union of {1, 2} and {2, 3} is {1, 2, 3}, but their disjoint union is {(1,0), (2,0), (2, 1), (3, 1)}. It is appropriate to use a disjoint union when we care about keeping track of which set an element originated from.

The coproduct ∐A_i has a corresponding family of canonical injections, which are functions ι_i: A_i → ∐A_i given by ι_i(x) = (x, i), returning the element of ∐A_i corresponding to the instance of x in the set A_i.

Exercise 10: What is the empty coproduct?

A topological space is a pair (X, τ), where X is the set of points in X and τ∈𝓟(𝓟(X)) is the set of open subsets of X and satisfies the following properties:

• (Closure under union) Any union ⋃A_i of open sets is open.

• (Closure under finite intersection) For any intersection ∩A_i of open sets which is indexed by a finite set I, ∩A_i is open.

Let A⊆X. A is open if A∈τ. A is closed if A^C = X \ A = {x∈X : x∉A} is open. A is clopen if A is both closed and open. The interior int(A) of A is the union of all open subsets of A. The closure cl(A) of A is the intersection of all closed supersets of A, and is also written with an overline which can’t be done outside a LaTeX environment. The exterior ext(A) of A is the union of all open sets disjoint from A. The boundary ∂A of A is cl(A) \ int(A).

Exercise 11: Invent as many possible definitions of these concepts in terms of each other as you can. One example to get you started: A is open if int(A) = A.

Exercise 12: What criterion must the set of closed subsets of X satisfy, in order for X to be a topological space?

When it is clear from context, we may refer to subsets of X simply as “sets”, and leave it implicit that we care about sets of points of X in particular.

Frequently, mathematicians simply refer to the set X itself as a “topological space”, leaving the underlying pair (X, τ) implicit. If one is considering multiple different possible topologies on the same set X, then it is often useful to explicitly write out (X, τ₁), (X, τ₂), and so on.

Let x∈X. A neighborhood of x is a set N whose interior contains x. Equivalently, a neighborhood of x is any superset of an open set containing x.

Let X and Y be topological spaces. A function f: X → Y is continuous if any of the following equivalent conditions hold:

• Preïmages of open sets are open. If V⊆Y is open (in Y’s topology), then U=f^-1(V) is open (in X’s topology).

• Preïmages of closed sets are closed.

• For all A⊆X, f(cl(A)) ⊆ cl(f(A)).

• For all B⊆Y, f^-1(int(B)) ⊆ int(f^-1(B)).

• For any x∈X and any neighborhood V of f(x), there is a neighborhood U of x such that f(U)⊆V⁸⁸ This should remind one of the ε-δ definition of continuity, if they are familiar with it..

Exercise 13: Try to prove these to be equivalent to each other.

It turns out that sequences are too good for topological spaces unless the topological spaces are “sequential”. We will instead define the concept of a net, which is sort of like a sequence except it’s allowed to sort of spread out and rejoin.

A binary relation on A is a function ~: A² → 2. If ~(a, b) = 1, then we write a ~ b. Otherwise, ~(a, b) = 0 and we write a≁b.

A preörder on A is a binary relation that is reflexive and transitive.

An upward-directed set (A, ≤) is a preördered set such that for any a, b ∈ A, there exists an element c ∈ A such that a ≤ c and b ≤ c. Such an element c is called an upper bound of {a, b}.

A net in S for a set S⊆X is a function from some upward-directed set (A, ≤) to S. A net is often written (x_a)_a∈A. We say that (x_a) is eventually in S if there is some b∈A such that, for any upper bound c of b, x_c∈S⁹⁹ If this is confusing, you can analogize this to saying that, say, a sequence {a_i} of real numbers is “eventually zero” if there’s some N such that for all n > N, a_n = 0.. We say that x is a limit point of (x_a) or that (x_a) converges to x if, for every neighborhood N of x, (x_a) is eventually in N.

Let S⊆X. S is open iff no net in S^C converges to any point in S. S is closed if every limit point of any net in S is an element of S.

Exercise 14. Characterize interior, closure, etc in terms of nets.

Exercise 15. Show that a function is continuous if and only if, for every x ∈ X and every net in X, x being a limit point of (x_a) implies that f(x) is a limit point of the net (f(x_a)).

The topological space R is very useful, but unfortunately we cannot access it yet, as Celene has no clue how to define R. We can’t actually define R as the metric completion of Q without defining metric spaces whose metric is Q-valued or something—which maybe isn’t an issue—but Dedekind completeness is a reasonable approach. For today, let’s just get to the topology on Q.

For a set M, an (n-ary) operation on M is a function Mⁿ → M. A nullary operation is called a constant, because M⁰ is the set containing the empty tuple {()} and so a function from M⁰ selects a single element f(()) of M. An example of a unary operation is -: Z → Z, which maps x to -x. An example of a binary operation is +: N² → N, addition.

An algebraic structure on a set M is a tuple (M, f₁, f₂, f₃, …) consisting of X and zero or more operations on M. A variety of algebras is consists of a signature and a set of equational laws. A signature is a tuple of natural numbers (n₁, n₂, n₃, …) prescribing that operations (f₁, f₂, f₃, …) on M be n₁-ary, n₂-ary, n₃-ary, etc. An equational law is an equation consisting of the (perhaps nested) application of operations (f₁, f₂, f₃, …) to variables (x, y, z, α, …) which represent elements of M. For M to be an instance of a variety of algebras, every equational law must be true under any possible assignment of variables to elements of M.

A magma (M, •) is a set equipped with any binary operation. The signature of the variety of magma is (2), and the set of equational laws is the empty set.

A semigroup (S, •) is a magma that satisfies the associative law, “x•(y•z) = (x•y)•z”.

A monoid (M, •, e) is a semigroup (M, •) with an identity element. The signature of the variety of monoids is (2, 0), because the identity element is a constant. In addition to the associative law on the binary operation, “x•e=x=e•x” must be added as an equational law.

A group (G, •, –^-1, e) is a monoid (G, •, e) with an additional unary operation that returns the inverse of an element. We add the equational law “g•g^-1=e=g^-1•g”.

An abelian group (G, +, -, 0) is a commutative group, satisfying “g+h = h+g”.

A rg (R, +, 0, •) is a commutative monoid (R, +, 0) under addition and a semigroup (R, •) under multiplication which satisfies distributive laws “a(b+c) = ab + ac” and “(a+b)c = ac+bc”, where juxtaposition denotes •.

A rig (R, +, 0, •, 1) is a rg (R, +, 0, •) which is a monoid (R, •, 1) under multiplication.

A rng (R, +, -, 0, •) is a rg (R, +, 0, •) which is an abelian group (R, +, -, 0) under addition.

A ring (R, +, -, 0, •, 1) is a rig (R, +, 0, •, 1) and a rng (R, +, -, 0, •).

A commutative ring is a ring where multiplication (R, •, 1) forms a commutative monoid.

Skew fields and fields do not form a variety, because they cannot be described with equational laws.

A skew field or division ring is a ring (F, +, -, 0, •, 1) such that 0 ≠ 1 and there exists a unary operation –^-1 on F \ {0} such that (F \ {0}, •, –^-1, 1) is a group.

A field is a skew field that is a commutative ring.

The axiom of infinity says that there is a set containing all of the natural numbers—or more specifically, there is some set which contains ∅ and is closed under the mapping that takes a set n to the set {n∪{n}}, which is the successor of n if n is a finite von Neumann ordinal. It might have a bunch of extra elements that aren’t natural numbers.

Let’s throw out all the other elements and only look at the natural numbers, which is a thing you can in set theory. You now have the set N.

Exercise 16. Give an inductive definition of + and •. Prove that N is a commutative rig.

Exercise 17. Construct a ring Z and exhibit an injective function φ: N → Z such that we have φ(0_N) = 0_Z, φ(1_N) = 1_Z, φ(a +_N b) = φ(a) +_Z φ(b), and φ(a •_N b) = φ(a) •_Z φ(b). This is known as an injective rig homomorphism. If you accidentally added some numbers other than the normal integers to your ring, you can simply consider the subring containing the elements φ(N) and -φ(N) to recover the normal integers.

Exercise 18. Construct a field Q and an injective ring homomorphism φ: Z → Q.

A partial order on a set X is a preörder that is antisymmetric: if a ≤ b and b ≤ a, then a = b.

A total order on a set X is a partial order that is total: either a ≤ b or b ≤ a. If we define a < b to mean “a ≤ b and a ≠ b”, then a total order satisfies trichotomy: for elements a and b of a totally ordered set, exactly one of a < b, a = b, or b < a is the case.

An ordered field is a field F with a total ordering such that

• (Compatibility with addition) If a ≤ b, then a+c ≤ b+c.

• (Compatibility with multiplication) If a ≤ b and 0 ≤ c, then ac ≤ bc.

An element r of an ordered field is positive if r > 0, and negative if r < 0.

Exercise 19. Define a total order on Q such that Q is an ordered field.

Let X be a (totally ordered) set, and let 𝓑 be a family of subsets of X. The topology τ generated by 𝓑 is defined as

Exercise 20. Prove that τ is in fact a topology on X.

This is the smallest or coarsest topology on X where every set in 𝓑 is open, where “smallest” is defined by the subset relation on 𝓟(X). If 𝓑 generates τ, we say that 𝓑 is a subbase for τ. If any open set U in τ is a union of open sets in 𝓑, we furthermore call 𝓑 a base or basis for τ.

An open ray on a totally ordered set X is a set of the form {x : x < a} or {x : x > b}. In Q, these can be written using the interval notation (-∞, a) and (b, ∞). The distinction between an ordered pair and an open interval can generally by made from context, but ]a, b[ is a less ambiguous but far uglier notation which can be used if absolutely needed.

The open rays generate a topology τ on X known as the order topology. The open rays, when combined with the open intervals (a, b) := {x : a < x < b}, form a base for the order topology.

For a family of topological spaces, the product topology is the smallest / coarsest topology on ∏X_i such that each canonical projection π_i is continuous¹⁰¹⁰ This condition is very natural from a category-theoretic perspective.. If not otherwise specified, one should assume that the relevant topology on a product of topological spaces is the product topology.

A topological field is a field F that is also a topological space (F, τ) such that addition and multiplication are continuous as functions F² → F.

Exercise 21. Show that, in a topological field, all polynomial functions (such as p(x) = ax² + bx + c for coefficients a, b, c in F) are continuous. It may be helpful to prove the lemma that the composition of two continuous functions is continuous.

Exercise 22. Prove that Q is a topological field under the order topology.

Since Q is a totally ordered set and two is finite, it may help to know that the open boxes (a, b) × (c, d) form a base for the product topology on Q². Or maybe not, I have no idea, it would maybe make a lot of sense if just working with π₁ and π₂ is the simplest approach by far. The open boxes do not, however, form a base for the product topology on Q^ω, where ω is the first infinite ordinal. The topology generated by the open boxes is called the box topology, and is equal to the product topology for finite products but finer (has more open sets) for infinite products¹¹¹¹ I think this is in some ways analogous to how the direct sum and direct product of abelian groups or of vector spaces are equivalent for finite sums and products, but not diverge for infinite sums and products.. The issue is that taking products of infinitely many intervals on different Qs lets you get some messed up sets; a base for the product topology on Q^ω consists of sets similar to the open boxes, but where only finitely many of the terms in the product can be a precise interval rather than just the whole space Q.

Exercise 23. Prove that for any two distinct points x and y in Q, there exist a neighborhood U of x and a neighborhood V of y such that U and V are disjoint. This criterion defines a Hausdorff space.

Exercise 24. For a set E in Q, a neighborhood of E is any set U such that E is a subset of int(U), or equivalently such that U is a neighborhood of every point in E. Prove that for any two disjoint closed sets E, F in Q, there are disjoint neighborhoods U and V of E and F. This criterion defines a normal space.

I assume getting through all this when you have done no real university-level mathematics must take, like, months. Good skill Celene, let me know if you need help or more problems.