Baire space N is the unique, up to homeomor-phism, nonempty Polish zero-dimensional space, for which all compact subsets have emptyinteri
N^N is Polish (separable, completely metrizable), zero-dimensional (admits a basis of clopen sets). All compact sets have empty interior (contain no open sets).
is said to have the
either countable or contains a nonempty closed, no-isolated-points set
(x) ..= (d(x,xn
f(x) is the distance between x and x_n for n \in N. f is injective: i think pretty clear. f, f^{-1} is continuous
Every separable metrizable space embeds into the Hilbert cube IN.
any dense set containing metrizable space can be a subset of the Hilbert cube [0,1]^N. So all Polish spaces are precisely the G_delta subspaces of ^N. The compact metrizable spaces are exactly the closed subspaces of I^N
G
no :)
Tychonoff
this bitch needs to get out of my life
f : D →X by {f(
so here we can define a map from D to a polish state X such that {f(x)} is the intersection
now on, for s ∈N<N, we will denote by Ns the basic open sets of N, i.e
basic open sets of the Baire Space are elements x in the Baire space such that x contains s. So e.g. s = {1,2,3} then N_s = {x \in N^N that contains {1,2,3}}
x ∈N
all the elements \N^\N such that the intersection of A_{x|n} is not empty
the As are only nonempty when s ∈ 2<N, then we just write (As)s∈2<Nand call it aCantor schem
gotcha!
imn→∞diam(Ax|n) = 0, for x ∈N
the diameter of A_{x|n} is 0 for any element x.
As, for s ∈N<N,i ∈
this is just the tree property. A_{1,2} is a subset of A_{2}
saj = ∅
the i and j extensions of a A_s is empty for any distinct i and j, where s \in N. E.g. A_{1,2} intersection A_{1,3} is empty. So in some way we can index A_s
X is a family (As)s∈N<Nof subsets of X
Luzin scheme -> fundamentally a special set of subsets
he Cantor space C is the unique, up to homeomorphism, perfectnonempty, compact metrizable, zero-dimensional space
what is this theorem saying? this is THE space that is perfect (closed, no isolated points), nonempty (obvs), compact (finite subcover), metrizable (admits a metric), zero-dimensional (has a basis of clopen sets)
Let X be a second-countable topological space. Then every basis B of Xhas a countable subbasis
Makes sense. (U_n) n \in N is a countable basis of X. Now B is also a basis B of X. Each U_n must belong to some element B_n of B. Then this subcover must be countable and a basis
n ∈N : ∃V ∈V such that V ⊇U
all the elements n where there exists some set in the open cover V such that V contains U_n. So I are all the indices of U_n contained in some V
second-countable
can write X as a countable union of open subsets
It is clear (why?) that Hausdorff zero-dimensional spaces are totally disconnected, i.e.the only connected subsets are the singletons. However the converse fails in general even formetric spaces.
Hausdorff--any two points can be completely separated + zero-dimensional--admits a basis of clopen sets. so we can partition any set into two unless it contains precisely one point. so only connected subsets are singletons.
zero-dimensional
disconnected, has a basis of sets that are closed and open
determined
this game has a winning strategy
Observation 1.3. If X is a topological space with a compatible metric d, then the followingmetric is also compatible: for x,y ∈X, D(x,y) = min(d(x,y),1
lowkey still want to prove this
or all n ∈N, n = {0,1,...,n −1} is Polish with discrete topolo
every subset is open---then we take the discrete metric (0 if two points are the same, 1 if two points are different); the dense countable subset is simply N or the entire set
e work with Polish topological spaces as opposed to Polish metric spaces because wedon’t want to fix a particular complete metric, we may change it to serve different purposes;all we care about is that such a complete compatible metric exists
never have to choose a particular metric
For any Polish space X, either X is σ-compact, or else, Xcontains a closed subset homeomorphic to
Recall: a Polish space is separable and completely metrizable (contains a dense + countable subset, and every Cauchy sequence converges).
Homemorphic: exists some continuous bijective function \(f\) from X to N, where f inverse is also continuous.
Baire space
The Baire space \(N = \N^\N\)with the discrete topology on N.
σ-compact
In mathematics, a topological space is said to be σ-compact if it is the union of countably many compact subspaces.[1]
nching, then [T] is compact.
compact -> every open cover of [T] has a finite subcover of open sets. what are our open sets here though?
Any finitely branching infinite tree T on A has an infinite branch, i.e.[T] 6= ∅.
that intuitively makes sense I guess
ely bra
<-> does not branch down to an infinite number of nodes ?
T(s) ..= {a ∈A : saa ∈
T(s) are precisely all the elements of a such that s^a is also part of the tree T
2.B. Infinite branches and closed subsets of
How do we define clsoed subsets in this space?
Note that if A is countable, then AN is Polish. Examples of such spaces are the Cantorspace and the Baire space, which, due to their combinatorial nature, are two of the most usefulPolish spaces in descriptive set theory
these are both trees
yn ∈Y such that x|n ⊆yn
there is some tree y_n such that x|n is a subset of y|n?
x ∈ [TY ].
fix some x in T_Y
Y ⊆ [TY ]
follows automatically
subset Y ⊆AN is closed if and only if Y = [TY ]
a subset Y is closed if and only if the set of infinite branches through Y is precisely equal To Y
f x /∈[T], then there is n ∈Nsuch that s = x|n /∈T.
suppose some x is not in [T], then there is some tree s of length n that is not in T. But then that implies that N_s intersect [T] is empty---enough to show openness in a product topology?
Ns ..= {x ∈AN : s ⊆x
all possible trees that are subsets of X
Thus we obtain a subset of AN from a tree. Conversely, given a subset Y ⊆ AN, we canobtain a tree TY on A by:TY = {x|n : x ∈Y,n ∈N}.
don't understand this definition I think I need to know what x|n is
[T]
set of infinite branches through T
AN.
all possible tuples (including infinite) of elements of A
product topology
don't know what this is
To answer this question, we give A the discrete topology
simply every open sets
[T] the set of infinite branches through T
all possible branches that go through the tree T
For s,t ∈ A<N, we also write sat to denote thetuple obtained by appending t at the end of s
append t to the end of s
For s ∈ A<N and a ∈ A, we write saa to denote the extension of s to a tuple of length|s|+ 1 that takes the value a at index |s|.
simply add "a" to the end of s
Definition 3.1 A σ-algebra Aon a set X is a family of subsets of X with thefollowing properties:X ∈A,(Σ1)A ∈A=⇒ Ac ∈A,(Σ2)(An)n∈N ⊂A=⇒ ∪n∈NAn ∈A.(Σ3)A set A ∈Ais said to be measurable or A-measurable
A family A of subsets of X such that: the whole set (X) is in A, if B is in A then B' is in A; if
The old saying: a doughnut (torus) is homeomorphic to a coffee cup. (Look thisup if you haven’t heard of it.)
why?
ny spaceMis homeomorphic to itself through the identity map.
this is the most trivial case, it should be true
discrete space
I don't know what a discrete space is here, which might be a problem
A functionf:M→Nis ahome-omorphismif it is a bijection, and bothf:M→Nand its inversef−1:N→Marecontinuous.
a continuous function between two spaces that is a bijection.... interesting. why do we have to have the inverse also be continuous?
When do we consider two groups to be the same? Answer: if there’s a structure-preserving map between them which is also a bijection. For metric spaces, we do exactlythe same thing, but replace “structure-preserving” with “continuous”
it's interesting to think that a good way of considering learning mathematical concepts might be from a more personal, "memorisation"-based way.... something that I usually think of as antithetical to mathematic reasoning. but there is a certain element of that maybe... homeomorphism is to metric spaces as isomorphism is to groups, etc.
ifx1,x2, . . . is a sequence inMconverging top, thenthe sequencef(x1),f(x2), . . . inNconverges tof(p)
okay, this is kind of what I was unconsciously thinking, but I couldn't quite word it right. basically if x_1, x_2 converges to p, then f(x_1), f(x_2) in N converges to f(p). wait - that's pretty cool!
Nonetheless, this definition iskind of cumbersome to work with, because it makes extensive use of the real numbers(epsilons and deltas).
very true!
LetM= (M,dM) andN= (N,dN) be metric spaces. A functionf:M→Niscontinuousat a pointp∈Mif for everyε >0 there exists aδ >0 suchthatdM(x,p)< δ=⇒dN(f(x),f(p))< ε
Okay, this is a bit tough to understand. Let's see, we have two metric spaces equipped with distance functions. We say that a function f from M to N is continuous at p if for whatever distance we want epsilon (on the real number line) we can find a delta ... so long as the distance between x and p is less than that delta, then the distance between f(x) and f(p) will be less than epsilon. makes sense! basically here, instead of absolute values we just sub in our distance functions, for both M and N (i.e. for both x and f(x))
Question 2.3.1.Can you guess what the corresponding definition for metric spaces is?
It probably is very close to the above....
A functionf:R→Ris continuous at a pointp∈Rif for everyε >0 thereexists aδ >0 such that|x−p|< δ=⇒ |f(x)−f(p)|< ε.
ah yes, the classic definition
However, even though eachxiis inQ, this sequence does NOT converge whenwe view it as a sequence inQ!
Right... I guess because \(\sqrt2\) is obviously irrational that 1, 1.4, 1.41, 1.414 just doesn't pan out... recall each finite decimal can be expressed as a rational
Definition 2.2.1.Let (xn)n≥1be a sequence of points in a metric spaceM. We saythatxnconvergestoxif the following condition holds: for allε >0, there is an integerN(depending onε) such thatd(xn,x)< εfor eachn≥N.
This part kind of makes the whole abstract distance function make sense. Basically, you want to expand just being able to describe limits for sequences and series on the real number line... you could describe for example the limit of a certain curve going to a point in space, etc. since "distance" is such an abstract concept
Another limitation of our current approach is that all of theembeddings used are fully “black box,” where the dimensionshave no inherent meaning.
This is definitely a substantial limitation!
The projection of a nor-malized word vector onto a cultural dimension is calculated with cosine similarity, as is the angle between cultural dimensions.
So essentially this is just a projection transformation onto a certain direction vector (given by these word pair differences) - I wonder what other linear transformations would have meaningful interpretations with word embeddings.
Somewhat surprisingly, these questions can be answered by performing simple algebraic operationswith the vector representation of words. To find a word that is similar tosmallin the same sense asbiggestis similar tobig, we can simply compute vectorX=vector(”biggest”)−vector(”big”)+vector(”small”).
I thought this was impressive! It really shows the power of linear algebra, which I suppose is the motivation behind word embeddings - as soon as you can represent words as vectors, all the tools of linear algebra such as the dot product, vector addition, etc. become available. I'm surprised that even these subtler syntactic relationships can be manipulated and represented with these embeddings.
This could result in what Crawford(2017) and Blodgett et al. (2020) call anallocational harm, when a system allo-allocationalharmcates resources (jobs or credit) unfairly to different group
Glad to see some critical consideration of biases implicit in embeddings trained in natural text.