When you think about it, literally almost all of mathematics can be given in terms of ultrafilters. Large Cardinals? Ultrafilters. Continuity? Ultrafilters. Compactness? Ultrafilters. Forcing? Ultrafilters. Social choice theory? Ultrafilters. And the list goes on...
Hey! Call me Lucy. I might make an introduction blog later, but I first wanted to make a blog-post about ultrapowers.
Ultrafilters are a concept from set theory, I'll try my best to explain what they are and why they're defined as they are.
First, a quick overview of what we will do: we will extend the real number line by adding new numbers through the use of an ultrapower, these new numbers are called "hyperreals". Roughly, this means that we will have infinite sequences [a₀,a₁,a₂,...] of real numbers representing hyperreal numbers, where similar sequences are regarded as equal. We will also show a surprising theorem: although there are seemingly more hyperreals than reals, hyperreals look the same as real numbers "from within".
If we have a sequence of reals like this: [0,1,1/2,1/3,1/4,...] (I'll call this sequence "ε"), the hyperreal that it represents can be viewed as the "limit" of the sequence. Since a large number of entries of this sequence is smaller than any positive real number r > 0, ε will be smaller than any positive real number r, but since also a large number of entries is larger than 0, ε will be larger than 0. ε is thus an infinitesimal hyperreal number. This is mostly just intuition though, so don't worry if you don't entirely get it.
Two hyperreal numbers x = [x₀,x₁,x₂,...] and y = [y₀,y₁,y₂,...] are equal if x_i = y_i for a large number of indices i. But what does "large" mean in this context?
Well, that's where the ultrafilter comes in. Ultrafilters split a family of sets into sets that are "large" and sets that are "small". In this case, we split sets of natural numbers (numbers 0, 1, 2, 3, etc) into large sets and small sets, so we have an ultrafilter on ℕ, the set of natural numbers. Ultrafilters are identified by the family of large sets: if some set A is in an ultrafilter U, then it is large, and if it's not, then it is small.
We do want our notion of "large sets" and "small sets" to make sense: for example, a hyperreal should always be equal to itself, so we want the whole set of natural numbers, {0, 1, 2, 3, 4, ...} (which is the set of indices for which a sequence is equal to itself), to be large.
Obviously, it would make sense that if a set A is large and B is larger than A, then B is also large. Thus, if A ∈ U is a member of an ultrafilter U ("∈" is the membership symbol), and if B ⊃ A contains everything A contains too ("⊃" is the superset symbol), then B ∈ U is a member of the ultrafilter as well.
We also want hyperreal equality to be transitive, thus if [x₀,x₁,x₂,...] = [y₀,y₁,y₂,...] and [y₀,y₁,y₂,...] = [z₀,z₁,z₂,...], then we want [x₀,x₁,x₂,...] = [z₀,z₁,z₂,...]. If A = {i ∈ ℕ | x_i = y_i} is the set of points at which x and y are equal and B = {i ∈ ℕ | y_i = z_i} is the set of points at which y and z are equal, then C = {i ∈ ℕ | x_i = z_i}, the set of points at which x and z are equal, includes the set A ∩ B = {i ∈ ℕ | x_i = y_i ∧ y_i = z_i}, the set of points at which x is equal to y and y is equal to z. It thus makes sense to have our ultrafilter be closed under intersections: if two sets A and B are large, then the set of points that are both in A and in B, called the "intersection" of A and B (denoted A ∩ B), is a large set as well (and thus also in the ultrafilter).
It would also make sense that, if two hyperreal numbers are nowhere equal, then they aren't equal. So the empty set, {} = ∅, is small.
The five axioms above describe a filter:
A filter F on κ is a family of subsets of κ.
A filter F on κ must contain the whole set κ.
A filter F on κ must be upwards closed, thus for every large set A ∈ F, and every larger set B ⊃ A, B ∈ F is large as well.
A filter F on κ must be downwards directed, thus for every large set A ∈ F and every large set B ∈ F, the intersection of A and B, A ∩ B ∈ F, is large as well.
A filter F on κ may not contain the empty set.
However, these are the axioms of a filter, and not of an ultrafilter. Ultrafilters have one additional axiom.
Suppose we have the hyperreal [0,1,0,1,0,1,...]: an alternating sequence of 0's and 1's. Is this equal to 0 = [0,0,0,0,...], or to 1 = [1,1,1,1,...], or is it its own thing? (Note: the 0 in 0 = [0,0,0,0,...] is a hyperreal and the 0's in 0 = [0,0,0,0,...] are real numbers, so they're different (kind of) numbers both called "0"). If it is its own thing, then is it smaller than 1? If it is smaller than 1, then it must be smaller on a large set of indices, meaning it's equal to 0 on a large set of indices, meaning it's equal to 0. If it's not smaller than 1, well, it can't be larger, so it'd only make sense if it's equal to 1, but no axiom about filters says it should! That's why we have this last axiom for ultrafilters, which makes them "decisive": for every set A, it is either large (thus, A ∈ U), or small, meaning that its complement, Ac = {i | i ∉ A}, the set of all points that aren't in A, is large.
And so we have our six axioms of an ultrafilter:
An ultrafilter U on κ is a family of subsets of κ, these subsets are called "large sets".
κ is large.
U is upwards closed.
U is downwards directed.
∅ is not large.
For every set A ⊂ κ, either A ∈ U or Ac ∈ U.
But we're still missing one thing. We can take our ultrafilter U to be the set of all sets of natural numbers that contain 6. ℕ is large, as it contains 6. It is upwards closed: if A contains 6 and B contains everything that A contains and more, then B also contains 6. U is downwards directed: if both A and B contain 6, then the set of all points that are in both A and B still contains 6. The empty set does not contain 6, and every set either does contain 6 or does not contain 6. With this ultrafilter, two hyperreals x and y are equal simply when x₆ and y₆ are equal, so we don't get cool infinitesimals and infinities, and that makes me sad :(
These kinds of boring ultrafilters are called principal ultrafilters. Formally, a principal filter on κ is a filter F on κ for which there is some set X ⊂ κ so that any set A ⊂ κ is large only if it contains everything in X. This filter is often denoted as ↑X. If you want a principal filter U to be an ultrafilter, X needs to be a singleton set, meaning it only contains a single point x. Proving this is left as an exercise for the reader.
Let U be a non-principal ultrafilter on ℕ. This post is getting a bit long, so I won't show why such an ultrafilter exists. Now, we can take the ultrapower of ℝ, the set of real numbers, by U. This ultrapower is often denoted as ℝ^ℕ/U. Members of this ultrapower are (equivalence classes of) functions from ℕ to ℝ, meaning that they send natural numbers/indices to real numbers (the sequence [x₀,x₁,x₂,...] maps the natural number i to the real number x_i). These functions/sequences/equivalence classes are called hyperreal numbers. Two hyperreal numbers, x and y, are equal if {i ∈ ℕ | x(i) = y(i)}, the set of points at which they are equal, is large (i.e. a member of U). We can also define hyperreal comparison and arithmetic operations: x < y if {i | x(i) < y(i)} is large, (x + y)(i) = x(i) + y(i) and (x · y)(i) = x(i) · y(i). Every real number r also has a corresponding hyperreal j(r), which is simply [r,r,r,r,...] (i.e. j(r)(i) = r for all i).
In general, if M is some structure, κ is some set and U is some ultrafilter on κ, then we can take the ultrapower M^κ/U, which is the set of equivalence classes of functions from κ to M, where any relation R in M (for example, "<" in ℝ) is interpreted in M^κ/U as "R(x₁,...,xₙ) if and only if {i ∈ κ | R(x₁(i),...,xₙ(i))} ∈ U is large" and any function f in M (for example, addition in ℝ) is interpreted in M^κ/U as "f(x₁,...,xₙ)(i) = f(x₁(i),...,xₙ(i)) for all i ∈ κ".
A quick note on equivalence classes: in M^κ/U, points aren't actually functions from κ to M, but rather sets of functions from κ to M that are all equal on a large set of values. Given a function f: κ → M, the equivalence classes that f is in is denoted [f]. In this way, if f and g are equal on a large set of values, then [f] and [g] are actually just equal.
The hyperreal [0,1,2,3,4,...], which sends every natural number i to the real number i, is often called ω.
This part of the blog will get a bit more technical, so be warned!
In the beginning of this blog-post, I mentioned that hyperreals look the same as real numbers. I'll make this statement more formal:
For any formula φ that can be built up in the following way:
φ ≡ "x = y" for expressions x and y (expressions are variables and "a + b" and "a · b" for other expressions a and b)
φ ≡ "x < y" for expressions x and y
φ ≡ "ψ ∧ ξ" (ψ and ξ are both true) for formulas ψ and ξ
φ ≡ "ψ ∨ ξ" (ψ or ξ is true (or both)) for formulas ψ and ξ
φ ≡ "¬ψ" (ψ is not true) for an formula ψ
φ ≡ "∃x ψ(x)" (there exists a value for x for which ψ is true) for a variable x and an formula ψ
φ ≡ "∀x ψ(x)" (for all values of x, ψ is true) for a variable x and an formula ψ
We have that ℝ ⊧ φ (φ is true when evaluating equality, comparison and expressions from within ℝ, where variables can have real number values) if and only if ℝ^ℕ/U ⊧ φ (φ is true when evaluating equality, comparison and expressions from within ℝ^ℕ/U, where variables can have hyperreal number values).
In other words: ℝ and ℝ^ℕ/U are elementary equivalent.
So, how will we prove this? Well, we will use induction: "if something being true for all m < n implies it being true for n itself, then it must be true for all n (where m and n are natural numbers)". Specifically, we will use induction on the length of formulas: we will show that, if the above statement holds for all formulas ψ shorter than φ, then it must also hold for φ.
However, we won't use the exact statement above. Instead, we will use the following:
Given a formula φ(...) and hyperreal numbers x₁,...,xₖ, ℝ^ℕ/U ⊧ φ(x₁,...,xₖ) if and only if {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} is large.
Now, why does this imply the original statement? Well, when k = 0, {i | ℝ ⊧ φ} can only be ∅ or ℕ. It being ∅ is equivalent to φ being false in ℝ and, if the statement is true, also equivalent to φ being false in ℝ^ℕ/U. And it being ℕ is equivalent to φ being true in ℝ and, again, if the statement is true, it is also equivalent to φ being true in ℝ^ℕ/U. We thus have that φ being true in ℝ is equivalent to φ being true in ℝ^ℕ/U.
Note: M ⊧ φ simply means that the formula φ is true when interpreted in M.
Now, why do we need this stronger statement? Well, it makes induction a lot easier: given that this statement holds for all ψ shorter than φ, it's easier to prove it also holds for φ.
Now, we can actually do the induction.
First, if φ ≡ "x = y", then we need to show that (1) ℝ^ℕ/U ⊧ φ(x,y) iff (2) {i | ℝ ⊧ φ(x(i),y(i))} is large. This follows immediately from the definition of equality in ℝ^ℕ/U, the same holds for "<".
Now, if φ(x₁,...,xₖ) ≡ "ψ(x₁,...,xₖ) ∧ ξ(x₁,...,xₖ)", we have that {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i)) ∧ ℝ ⊧ ξ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i))} ∩ {i | ℝ ⊧ ξ(x₁(i),...,xₖ(i))}. Since {i | ℝ ⊧ ψ(x₁(i),...,xₖ(i))} is large iff ψ(x₁,...,xₖ) is true in ℝ^ℕ/U, and {i | ℝ ⊧ ξ(x₁(i),...,xₖ(i))} iff ξ(x₁,...,xₖ) is true in ℝ^ℕ/U, and U is closed under intersections, we have that {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} is large iff φ holds in ℝ^ℕ/U. A similar argument works for ∨.
If φ(x₁,...,xₖ) ≡ "¬ψ(x₁,...,xₖ)", then we can just use the ultraness of the ultrafilter.
If φ ≡ "∃y ψ(y,x₁,...,xₖ)", then {i | ℝ ⊧ φ(x₁(i),...,xₖ(i))} = {i | ℝ ⊧ ∃y ψ(y,x₁(i),...,xₖ(i))} = {i | ∃y ∈ ℝ. ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} = ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))}. We have that the set {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} for y ∈ ℝ is large iff ℝ^ℕ/U ⊧ ψ(j(y),x₁,...,xₖ). If this set is large for some y ∈ ℝ, and thus if ℝ^ℕ/U ⊧ φ(x₁,...,xₖ), then ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} is larger than that set, so it is large as well. For the converse direction, if ∪_{y ∈ ℝ} {i | ℝ ⊧ ψ(y,x₁(i),...,xₖ(i))} is large, then we can create a hyperreal z where ψ ⊧ ψ(z(i),x₁(i),...,xₖ(i)) for all i for which ℝ ⊧ ∃y ψ(y,x₁(i),...,xₖ(i)), and we have ℝ^ℕ/U ⊧ ψ(z,x₁(i),...,xₖ(i)), and thus ℝ^ℕ/U ⊧ φ(x₁(i),...,xₖ(i)). Again, a similar argument works for ∀.
(Sorry if you couldn't follow along, I'm not good at explaining these things in an intuitive way.)
This result can be extended to show that M^κ/U is elementary equivalent to M for every structure M, every set κ and every ultrafilter U on κ.
Now, this result might be surprising, as we have a new number ω in ℝ^ℕ/U. Surely, there is a formula that states the existence of this number, right?
Well, it turns out, such a formula does not exist! You can try something like "there is no natural number n so that 1+...+1 w/ n 1's is greater than ω", but ω+1 is a natural number in the hyperreals, so such a natural number does exist. Similarly, any formula you can come up with, as long as it is created using the rules above (using conjunction, disjunction, negation, qauntification, etc), cannot state the existence of an infinite number ω.
But if ℝ^ℕ/U and ℝ are seemingly indistinguishable, might there already be an undetectable infinite real number in ℝ? Well, maybe~ :3 But it's undetectable anyways, so you don't have to worry about it.
Before I end this blog-post, I want to give some more intuition on what filters & ultrafilters actually are. To me, ultrafilters, and filters in general, are like "limits of sets". The principal filter ↑X has X as limit, while non-principal filters and ultrafilters have limits that aren't really sets, but look like ones. For example, you might have the set of prime numbers in your filter, and then the limit of that filter will be a "set" in which all numbers are prime numbers. And if your ultrafilter is non-principal (so for every n, there is a set A ∈ U in the filter that does not contain n), then the limit of that ultrafilter will be a "set" in which all numbers don't actually exist. In the case of filters, this "set" can be any "set" (though it still isn't really a set), but in the case of ultrafilters, this limit looks like a singleton set (i.e. it only has one "element": ω).
I don't know if my intuition of filters and ultrafilters will help anyone, tho, but I think it's cool!
A space is compact iff every ultrafilter on that space converges
I'm recording this exercise as a reference; recall that an ultrafilter on a space is said to converge to a point in that space if the ultrafilter contains the filter of opens containing that point.
\((\Leftarrow)\). Suppose first that there is an ultrafilter \(\mathcal{U}\) on \(X\) which has no limit point. Suppose there is a finite subcover of \((U_x)_{x \in X}.\) Then there are \(U_1, \dots, U_k\) such that \(\bigvee_{i \leq k} U_i = X\), so that \(\bigvee_{i \leq k} U_i \in \mathcal{U}.\) Since the complement of \(\mathcal{U}\) is an ideal, some \(U_j\) among the \(U_1, \dots, U_k\) was in \(\mathcal{U}\), a contradiction.
Therefore, the cover \((U_x)_{x \in X}\) has no finite subcover.
\((\Rightarrow)\). Suppose that every ultrafilter converges but \(X\) is not compact.
Fix an open covering \((U_x)_{x \in X}\) of \(X\) with no finite subcover.
For each finite subset \(F \subseteq X\), let \(V_F \overset{\operatorname{df}}{=} \bigvee_{x \in F} U_x.\) The collection \((V_F)_{F \underset{\operatorname{fin}}{\subseteq} X}\) is closed under \(V\). Expanding this collection to
\[\left(W_{F,Y} \overset{\operatorname{df}}{=} V_F \cap Y \right)_{F \underset{\operatorname{fin}}{\subseteq} X, Y \subseteq X}\]
yields an ideal.
Therefore, \(\left(X \backslash W_{F,Y}\right)_{F,Y}\) is a filter which is contained in some ultrafilter \(\mathcal{V}\). By assumption, this converges to some point \(x \in X\). Therefore, \(U_x \in \mathcal{V}.\) By construction, \(\mathcal{V}\) also contains \(X \backslash U_x\). This means \(\mathcal{V}\) was not an ultrafilter, a contradiction.
Let me just break down the process for forcing a failure of the singular cardinals hypothesis via large cardinal ultrafilters. First, we need to ensure that κ remains a cardinal, a.k.a. is preserved. This is checked by means of setting κ as a large cardinal (a measurable, to be more precise), and then seeing if there is a forcing notion which makes κ singular without collapsing it. This is accomplished using iterated forcing (first stage of construction) by constructing a model M[G] in which 2^κ = κ^++. The forcing in particular is Add(δ,δ^++); it creates an extension M[G] satisfying 2^δ = δ^++ for all inaccessibles δ leq κ in the ground model M. The argument that κ is measurable in M[G] is done via a large cardinal assumption on κ in M, and is proven (follows from) the existence of the Mitchell order of ultrafilters on κ, especially when o(M) = κ^++.
The second stage involves singularizing the said measurable cardinal. In this, we attempt to find a forcing notion P so that in the resulting generic extension M[G], Cof(κ) = ω. The forcing notion used is called Prikry forcing; it beats Cohen forcing at this in that Cohen forcing has no property which implies preservation of (especially singular) cardinals < κ; it instead collapses cardinals < κ to, say, ω.
Did you know we use 2 filters for each and every treatment provided to patients at KidneySolutions? We are not aware of any other centers in Lagos that do this. Also- we use brand new filters and bloodlines each time patients are treated. Some centers manually wash filters and bloodlines and that is so dangerous for patients. Always ask how many filters are used during dialysis and if filters are reused if you don't get dialysis at KidneySolutions. #kidneysolutions #filterreuse #ultrafilters #dialysis
