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10:00
wow this is beautiful
see this is why I like discrete stuff :P
because 1 = 6 - 12 + 7
@MikeMiller Hm, I have actually never thought about the distinction carefully.
Let me have a thought about this while smoking a cigarette
@BalarkaSen I have talked about this before I think but it's somewhat subtle, the statement holds up to dimension 4 and no further
It's straightforward in dimension 2 and not too bad in dimension 3
the figure-eight is an immersion of the circle
I'm trying to visualize why their cones are "the same"
can they be deformed into each other under "reasonable restrictions"?
@MikeMiller Ah I see, 2 dimension is straightforward because links are good.
10:03
their cones immersed in R^3, that is
The cone on a nontrivial knot is not locally flat
I imagine there's no way to upgrade the simplicial structure on S^5 thought as double suspension of a simplicial homology 3-sphere to a PL structure
So it is not an immersion
Yeah that's right B
oh wow there's a distinction? oh no
does the chaos happen near the vertices or something
10:05
@LeakyNun Define a PL manifold for me
Don't look it up just tell me what you think it is
hmm...
This isn't trying to dunk on you or something this is a guided exercise
so a manifold is where the transition maps are smooth
Since Mike isn't trying to dunk on you I will
so for PL you replace "smooth" with "piecewise linear"
but you also need to replace "open sets" with "open polyhedrons"?
10:07
why?
piecewise linear makes sense for open subsets of R^n with no triangulation
oh cool
you can tell me a definition if you like
so we need a second-countable Hausdorff space
with PL transition maps between charts
where a chart is a homeomorphism between open subsets of the space and R^n
is that a correct definition
10:09
That's a good definition, and I like that you're using R^n instead of open subsets thereof --- that will make the guiding easier
Now let X be a simplicial complex which is also a manifold. The goal is to find conditions so that X is canonically a PL manifold
so our first goal --- find canonical charts
is the hollow cube a 2-manifold? I suppose it is
the corners look sharp to me
that's just a sphere
but topologically it doesn't matter
project radially
ok let's go on
10:11
(It's also a PL 2-manifold!)
So let's pick a vertex in the simplicial complex X
@BalarkaSen (as is any 2-simplicial complex etc)
(Why?)
I want you to suggest me a definition --- write Star(v) for all of the simplices which are 'close to v'
(Mike said it above lol)
10:11
Suggest a definition for 'close to v'
contains v?
yeah, that's exactly it
or if you use open cells, take the closure
Simplicial complexes are combinatorial, and the cells are closed cells
10:13
it turns out Star(v) is actually well controlled by something else
This is our "canonical neighborhood", remember
are we only looking at n-cells
All cells
The thing to study now is its boundary --- this is called Link(v), and I'd like you to give me a combinatorial definition of it
so remove v = vi from [v0 ... vn] where [v0 ... vn] in Star(v)?
no
yes
is this correct?
10:16
There is a less cumbersome phrasing by adding v instead of removing it
It is the collection of simplices which do not contain v, but so that $\sigma \cup \{v\}$ is still a simplex
do "simplex" and "cell" mean the same thing?
Who said cell
when someone says cell i think cw complex
oh ok
10:17
simplicial complex is a completely combinatorial object (which is then realized into something topological)
it's literally a finite set equipped with a collection of subsets
the finite set is the vertices, the subsets are the collections of vertices which span a simplex
I see
Here is your first real exercise in this business
Show that there is a PL homeomorphism Star(v) ~ Cone(Link(v))
oh no
Hint: check this for simplices first
We'll continue when you have shown this
what's the PL structure on Cone([v0 ... vn])?
10:22
I suppose I'm asking you to tell me that too
Draw a picture for n = 1, n = 2
[v0 ... vn] is canonically embedded in R^(n+1) right
so I should embed Cone([v0 ... vn]) in R^(n+2) or something
[v0, v1] is the line connecting (1,0) and (0,1)
so its cone should be the triangle with vertices (1,0,0) and (0,1,0) and (0,0,1)
AKA, the standard 2-simplex
and similar for n=2
so we just want to show that two (n+1)-simplices are PL-isomorphic
for your hint
Or rather your definition answers the hint
I guess
10:25
The triangulation of the cone of a simplex is just the simplex one dimension up
and now it is piecewise linear; each piece is a simplex in Star(v), on which it is linear
Yup
though you have the names backwards, Link(v) is the boundary
(There is the combinatorial condition to check --- that there is exactly one (k-1)-dim simplex in Link(v) for every k-dim simplex in Star(v) containing v --- but this follows from the definition of a simplicial complex, that a simplex is determined by its vertices)
So you know that C(Link(v)) = Star(v)
What condition should I put on the simplicial complex Link(v) so that this automatically gives us charts?
(PL charts, that is to say)
Link(v) should be PL-isomorphic to S^(n-1)?
that doesn't make sense
by S^(n-1) I mean the boundary of Δ^n
does it make sense now?
Correct statement
10:34
yay
If the given simplicial complex is a manifold, to get a PL structure, it suffices to know Link(v) is a topological manifold, in fact.
connected?
if disconnected then not manifold
(unless dim 1)
@BalarkaSen Cheeky
is this an exercise
10:36
Maybe
If you are willing to invoke some heavy machinery
Which you like
what machinery
hey it's @loch again
telling you the machinery tells you the proof of the exercise
the exercise is basically knowing which three theorems to cite, two of homotopy theory and one of manifold topology
Mike is exposing my exercise as a bad exercise
so I am to prove that if the given simplicial complex is a manifold and Link(v) is a topological manifold then Link(v) is PL-isomorphic to the boundary of Δ^n?
Right.
And then you are to sit back and amaze at what modern topology has been able to achieve in the last few decades
10:39
Actually there is a first step or two which is independent of those theorems
So maybe it's actually too difficult instead of too easy
But it's one of thoss
anyway what was the next exercise @MikeMiller
What was our goal again? I forgot
Like what did you want to know that spawned this
@LeakyNun Definition: a combinatorial manifold is a simplicial complex where every link is PL homeomorphic to a sphere. A combinatorial manifold has a canonical PL structure, and a PL manifold has a combinatorial triangulation. If a simplicial complex is NOT combinatorial, it is not PL homeomorphic to any PL manifold --- even if it was a topological manifold
(This is because links are PL homeomorphism invariants)
33 mins ago, by Mike Miller
Now let X be a simplicial complex which is also a manifold. The goal is to find conditions so that X is canonically a PL manifold
this is our goal
Oh well you did that.
That was what I wrote out above
I thought there was something about surfaces
lemme look at my knot theory paper...
10:45
Oh I was trying to get you to figure out why PL manifold = triangulated manifold in dimension 2.
Give me a second to work out the correct hints
hmm I cited a theorem that says a homeomorphism R^3 -> R^3 is isotopic to the identity iff it is orientation-preserving
I suppose it's irrelevant
I forget if that's straightforward or difficult but it's irrelevant to this discussion
Isn't that the Alexander trick
Oh no disk
There's nothing to start with
Yeah I want to comb to infinity
Oh that's not an iaaue
Schoenflies
Good call
Nice.
10:48
Ok yeah so you just comb to 0 and to infinity thinking of it as a homeo of S^3
It's tricks no hard work
Yes, straightforward
Ok I need to think about hints 1m
I also cited Theorem 1.10 in Burde and Zieschang which says that two PL knots are C^infty-equivalent iff they are PL-equivalent
I don't know very many machinery lol
It's all fine in dimension 3 and down, you can assume stuff. PL = DIFF
Ok your next job is a calculation
Let X be a simplicial complex which is topologically a manifold
Show that Lk(v) has the homology of a sphere one dimension down
I can give a hint if necessary but I think you know everything you need for this
10:53
@BalarkaSen overall in your geometrical journey is it painful or fun?
More fun and less painful than algebra
Algebra sucks balls
touche
Algebra is easy but unenlightening for me
That is to say, fbe algebra I do
Which is the easy algebra
The algebraist can do the hard stuff
@MikeMiller Normal varieties are geometrically understood as the ones which have connected links in codimension 2
This is bliss, you say to yourself
Star(v) is a compact neighbourhood of v
10:55
Turns out the theorem is Zariski's Main Theorem
People tend to have difficulty with algebra. Like, all of my classmates in the last two algebra classes I've taken have said the class was very difficult
which is 50 pages of commutative algebra
and Star(v) is homeomorphic to C(Lk(v))
and any neighbourhood of the apex in C(Lk(v)) contains a copy of C(Lk(v))
since the given simplicial complex is a manifold, C(Lk(v)) is homeomorphic to a compact nbhd of 0 in R^n
and Lk(v) is the boundary of C(Lk(v))
eh... what
boundary in the sense of manifold
no it isn't a manifold
Not a priori at least
ah one can make it in the sense of topological boundary if I include a larger copy of C(Lk(v))
the point is, C(Lk(v)) is a very nice thing
so what does the boundary of a connected compact nbhd of 0 in R^n look like homologically
that's the wrong question, it can look like anything
aha, C(Lk(v)) is contractible
contractible compact nbhd of 0 in R^n...
11:04
I think you're close so I won't hint but if you get frustrated let me know
if I have a contr. cpt. nbhd. of 0 in R^n, is its boundary necessarily homologically a sphere?
hmm
How do you plan to relate the boundaries back bro
It's a good plan but how do you do it
they're homeomorphic
I guess I shouldn't care about the topological boundary
just treat (C(Lk(v)), Lk(v)) as a topological pair
it's a good pair (in the sense of Hatcher)
morning frens
so by LES it suffices to show that Σ(Lk(v)) is homologically... a sphere...
I feel like I know nothing about R^n
11:10
"What is Manifolds?"
homologically
the local homology groups look like spheres
excision etc
So work with relative homology
Or even local homology
hmm
C(Lk(v)) - v def. ret. to Lk(v)
(sorry I was away)
i.e. Star(v) - v def. ret. to Lk(v)
so H(Star(v), Star(v) - v) = H(Star(v), Lk(v))
the LHS is H(S^n) since the complex is a manifold
so H(Lk(v)) = H(S^(n-1))
@MikeMiller yay
Yup, and you used excision there
I see
11:25
Here is a more direct argument, which also says something about pi_1
There is a copy of R^n contained in C(Lk(v)), and a copy of C(Lk(v)) contained in that (scale by some very small real epsilon) and then another copy of R^n contained inside that (scale by the same constant epsilon), all containing v
So we have a sequence of inclusions R^n \ 0 -> Star \ v -> R^n \ 0 -> Star \ v
Where each of the composites of two are just scaling by epsilon
So any one kf thw inclusions is a homotopy equivalence
Whence Lk(v) is a homotopy sphere
wow
that's genius
what do we do now
Well I'm going to go to bed but what I've realized is that this was all rather unnecessary if your goal is to show that Lk(v) is always a circle when X is a triangulated surface
you can tell what happens when you cone a non-manifold point of Lk(v)
It's like coning on a chicken feet
Because I mean, come on, Lk(v) is a graph and if any of those vertices have valence other than 2 you get a local model which is not a manifold --- eg show that Y x I is not a surface where those spaces are both just the letters Y and I
It seems a little irritating to prove that's not a manifold? But it follows immediately from for instance invariance of domain
11:40
“where those spaces are both just the letters Y and I”
@BalarkaSen whats your fav counterexample
why does this work for dim 3
why does what work, ctrexample in what
oh this stuff, Mike was sketching some proof elsewhere that I didn't get
simplicial complex is PL manifold
you mentioned some weird S^5
how can it not work for S^5
S^5 is double suspension on a homology 3-sphere
so it admits a simplicial structure where the links are not manifolds
what is a homology 3-sphere
A 3-manifold with the homology of a 3-sphere
11:45
In algebraic topology, a homology sphere is an n-manifold X having the homology groups of an n-sphere, for some integer n ≥ 1 {\displaystyle n\geq 1} . That is, H 0 ( X , Z ) = H n ( X , Z ) = Z {\displaystyle H_{0}(X,\mathbb...
why is its double suspension S^5
thats a theorem
its not easy
double suspension on any homology sphere is a sphere
what is this icosahedron nonsense
you are slowly descending into lmgtfy
that was going to be my next response
@LeakyNun Let $M$ be a 3-dimensional simplicial complex which is a topological manifold. Then the link of every vertex is a manifold (and hence a sphere, as it has the homotopy type of). Proof: let $v$ be a vertex in the link. I will write $\text{Lk}_2(v)$ to denote taking the link of $v$ inside of the complex $\text{Lk}(v)$. Then we have that $C(\text{Lk}_2(v)) \times I$ is a manifold, because $M$ is a manifold, and so $\text{Lk}_2(v) \times I^2$ is a manifold.
Now $\text{Lk}_2(v)$ is a 1-complex, so a graph. If it had a vertex of valence $1$, this would give boundary points, so that's a no-go. Write C(n) for the tree with one central vertex of valence n. If $\text{Lk}_2(v)$ had a vertex of valence > 2, using the non-surjective embedding $I \to C(n)$, we would get an embedding of $I^3$ into our manifold which is not locally surjective, hence not a homeomorphism, contradicting the invariance of domain theorem.
@BalarkaSen
11:53
the shape of the universe is a Poincaré sphere???
Did you want to play a game of blitz chess Leaky?
not now
Ok
@MikeMiller Nice, this makes sense
Therefore $\text{Lk}_2(v)$ is a disjoint union of circles. But $C(\sqcup_n S^1) \times I$ is not a manifold, because the homotopy type of the complement of the origin is $\Sigma(\sqcup_n S^1) \simeq \vee_n S^2$
@BalarkaSen Now a very slight modification of this argument shows that a triangulated 4-manifold has links which are manifolds up to codimension 3; so the singularities look like suspensions of surfaces, but now you need to argue that $C(S) \times I$ is not a manifold unless $S$ is the sphere.
But the homotopy type of the complement of the origin is again the suspension of $\Sigma S$, which is only homotopically a sphere when $S$ is a sphere.
So we now know that in a triangulated 4-manifold, the link of a vertex is a simply connected homotopy sphere which is a manifold. But by Perelman this is PL homeomorphic to the standard sphere.
11:59
Gotcha
@LeakyNun lol.
12:17
@LeakyNun by the way boundary of any compact contractible oriented manifold is a homology sphere, because of Poincare duality. $H^k(M, \partial M) = H_{n-k}(M) = 0$ unless $k = n$, and $H^k(M, \partial M) \cong H^{k-1}(\partial M)$.
the nontrivial point is of course that there exists such manifolds which are not topologically balls, and there does: Mazur's manifold
12:35
Fun, Mazur's construction of the manifold is very simple. Take $S^1 \times B^3$ and $B^2 \times B^2$ and tube them togather along the boundary by taking a tubular nbhd of $S^1 \times \{0\}$ in $S^1 \times S^2$, a tubular neighborhood of $S^1 \subset S^3$, and then identifying them off
@BalarkaSen Really? are you sure it's not a nontrivial knot?
Unless your last circle in the 3-sphwre was secretly knotted
Otherwise that looks like B^4 to me
He explicitly says it's an unknot apparently. Take the curve $S^1 \times 0$ in $S^1 \times B^2 \cup B^2 \times S^1 = \partial(B^2 \times B^2)$
That's the curve he uses
Oh I guess the point is you choose some curve on the boundary of $S^1 \times B^3$ which represents the generator of $H_1$
In differential topology, a branch of mathematics, a Mazur manifold is a contractible, compact, smooth four-dimensional manifold (with boundary) which is not diffeomorphic to the standard 4-ball. The boundary of a Mazur manifold is necessarily a homology 3-sphere. Frequently the term Mazur manifold is restricted to a special class of the above definition: 4-manifolds that have a handle decomposition containing exactly three handles: a single 0-handle, a single 1-handle and single 2-handle. This is equivalent to saying the manifold must be of the form S...
I think $S^1 \times \{0\}$ will just give you a ball
Right, that was what I was missing
He uses a funky curve later on to actually get a non-simply connected homology sphere as boundary
dsm
dsm
12:54
Is there a simple way to show why if a linear map is an isometry it's also invertible? I want to say the following, but I don't think it's valid. Suppose $T$ is an isometry: $(Tv|Tw) = (v|w)$ and suppose $Tv = Tw$, then $(Tv|Tw) = (Tw|Tv) \Rightarrow (Tv-Tw|Tw-Tv) = (T(v-w)|T(w-v)) = (v-w|w-v) = 0$, and then conclude $v=w$ so it's one-to-one and hence invertible
but I don't think the last part necessarily implies $v=w$
@dsm the step $(Tv|Tw) = (Tw|Tv) \implies (Tv-Tw|Tw-Tv) = 0$ is wrong
In the case of $S^1 \times \{0\}$ the boundary is clearly $S^3$ tubed with $S^1 \times S^2$ along the trivial curves, which is just... $S^3$
My bad
dsm
dsm
@LeakyNun ahh, because $(\cdot|\cdot)$ isn't bilinear, it's Hermitean?
it is bilinear
OK, this is a fun way to think about it in general I guess. You tube the unknot of $S^3$ with a knot in $S^1 \times S^2$ which generates $H_1$.
12:58
and Hermitian also implies bilinear
wait
that step might not be wrong lol
eh
how do you justify that step
So the fundamental group of the boundary would just be the fundamental group of the knot complement in $S^1 \times S^2$
I guess
@BalarkaSen what do you mean when you say "tube them together"?
I have an embedded $S^1$'s in two oriented $3$-manifolds. Take tubular neighborhoods in each, which are both diffeomorphic to $S^1 \times B^2$. Identify them togather
dsm
dsm
the $(\cdot|\cdot)$ is conjugate linear in the first term, i.e. $(cv|w) = \bar{c}(v|w)$. the non-degenerate Hermitian form doesn't have to be bilinear. I suppose I'm justifying $(Tv|Tw)=(Tw|Tv) \Rightarrow (T(v-w)|T(w-v)) = 0$ by the linearity of $T$
it looks like you're using the identity $(a|b)-(c|d)=(a-c|b-d)$ which is wrong
dsm
dsm
13:06
ahh, I see. hmm. back to the drawing board on showing $T$ is invertible then. although I didn't like my above route anyways XD
ok, slightly different. suppose $Tv=Tw$, then $(v|w) = (Tv|Tw) = (Tw|Tw) = (w|w)$. so $(v|w)-(w|w) = (v-w|w) = 0$. but I still don't think this implies $v=w$, because hermiticity just states $\forall v\neq 0 \exists w\in V \text{s.t.} (v|w)\neq 0$. hmm, maybe I need to take a different approach.
feels close though
13:22
1
Q: Locally euclidean and first countability

topologicalmagicianSuppose $X$ is a topological space that is locally euclidean of dimension $n$. Show that $X$ is first countable. My attempt:Let $p\in X$ and $U$ a neighborhood of $p$. By assumption, there exists a neighborhood $U'$ of $p$ such that $U'$ is homeomorphic to a first countable space.Hence $U'$ is f...

any feedback?
13:46
@LeakyNun I seem to make fewer mistakes playing bullet games than blitz games lol lichess.org/xbXPkh83/white#1
I didn't realise how fun chess can be
that bishop hang by your opponent tho
A+
Hahaha yeah, that was a good start
His rating is the highest I've versed before
dsm
dsm
@LeakyNun ok, I'm gonna go a round-about approach. if $T$ is a linear map, it can also be represented as partial application of a $(1,1)$ tensor; i.e. $T\equiv u\otimes f$ with $u\in V$ and $f\in V^*$ such that $T(v) = (u\otimes f)(v) = f(v)u$. if $T(v) = T(w)$ then $f(v) = f(w)$. the dual $f$ was arbitrary, so let it be the metric dual. then $f(v)=f(w)\Rightarrow f(v)-f(w)=(v-w|\cdot) = 0$. by hermiticity, and since this is valid for any argument, we must have $v=w$. I feel decent with that.
I lost 5 times in a row or something to a 1750 player on my first or second day
bullet is 2 hard
13:52
I used to play chess with my father when I was very young, and he'd always tell me I was taking too long. Then he introduced a rule that he claimed was a standard rule, that if I take longer than 5 seconds to move, he'd get to move again lol
I've been conditioned for bullet
I can try a 5+3 blitz with you if u want
Other than now, I'd probably only played a game or two per year for years
I'm keen
Although I'm worse at that
im probably worse than u
dsm
dsm
@LeakyNun let me know if you see cracks in that. for now, I'm going to bed feeling assured that every element $T\in\text{Isom}(V)$ is invertible. cheers
14:09
@dsm you need a sum of "pure tensors"
@dsm also you don't know that $f(v)u = f(w)u$ implies $f(v)=f(w)$
also the $f$ isn't arbitrary; it depends on $T$
14:52
Thanks for playing @BalarkaSen
I'll probably play again in like 5 hours
The turn 12 thing I was thinking of was if you castled I'd have pawn to c4 discovered check, and you'd be able to block with bishop, and I'd trade
yah I was afraid of ec4
I played horribly lol
But after that I'd have c takes bishop on d3
You were fine haha
We'll be pro's in no time
Wait what does ec4 mean?
like pawn to c4
How can a pawn get from e to c?
Oh right
e is notation for pawn?
I think so!
could be wrong
14:56
Well gg for now :D
Oh wow, we played for over an hour
lol
actually they dont use anything for pawns
earlier they used to do "p"
strange
Hello, I have a question about the wave equation for pde.
u_tt-4u_xx=0 t>0
u(x,0)=1-x^2 when |x|<=1 and u(x,0)=0 when |x|>1
u_t(x,0)=4 when 1<=x<=2 and u_t(x,0)=0 else.
I'm looking for all points when u is not C^2 (twice differentiable).
I tried with D'Alembert but because I have multiple cases I get like 5-6 cases and I'm not sure this is the right way.
@MikeMiller @LeakyNun OK, so the picture is super-clear now. Given a knot $\Gamma$ in $S^1 \times S^2$ which represents the generator of $H_1(S^1 \times S^2) = \Bbb Z$, glue $S^1 \times B^3$ to $D^4$ by tubing an nbhd of $\Gamma$ to an nbhd of the unknot in $B^3$. Call this $W$. The boundary $\partial W$ is $S^1 \times D^2$ glued to $S^1 \times S^2 \setminus N(\Gamma)$ by a pasting meridians to longitude.
So $\pi_1 \partial W$ is essentially the fundamental group of the knot complement $S^1 \times S^2 \setminus \Gamma$ but with a parallel curve of $\Gamma$ collapsed. $H_1$ vanishes because
unknot in *$S^3$
The funny part is $W \times I \cong B^5$. This is because in the product you're tubing $S^1 \times B^3 \times I = S^1 \times B^4$ to $B^4 \times I = B^5$ along a tubular neighborhood of the corresponding curves (we're crossing with $I$ so tubular neighborhoods stay tubular neighborhoods), but embedded circles in $\partial(S^1 \times B^4) = S^1 \times S^3$ and $\partial(B^5) = S^4$ are unknots, so it's just a trivial tubing
So basically it's the curve $S^1 \times 0 \subset S^1 \times S^3$ in $S^1 \times B^4$ tubed to a round circle $S^1 \subset S^4$ in $B^5$, which is $B^5$.
This says $W$ is contractible
15:28
Consider the set $X=\{(a,b)\in\Bbb C^\times\mid \frac{\overline{a}}{a} = \frac{\overline{b}}{b}\}$ and endow this with an equivalence relation $(a,b)\sim (c,d)$ if there exists a positive real number $r$ such that $bd^{-1}=a^{-1}cr$. Can someone help me show that this (hopefully) has only two equivalence classes?
This will tell me a certain second cohomology lol (which is known to be $\Bbb Z/2\Bbb Z$)
I think the property defining the set tells us that $a,b$ must either have the same angle, or their angle differs by $180^\circ$
And the equivalence relation tells us that $(a,b)\sim (c,d)$ if and only if $arg(a)+arg(b)=arg(c)+arg(d)$
So if each pair fall into the 'same angle' case, then all 4 of them have the same angle, but if one of them falls into the same angle, and the other doesn't, then you get some strange relation
It seems like the sort of thing Balarka could answer, where he'll say something like "This is just $\Bbb CP^1$ quotient hte action of blah blah"
So a random simple question:
I am trying to show that the roots of $\cos (ax)$ is dense as $a \to \infty$
15:44
Can you answer my question @Secret, it seems right up your alley
hmm...
$\bar{a}b = a \bar{b}$
$ab=cdr$
so $ab$ are those equal to their conjugate
that means $ab$ is real
Since $r$ is real, it follows $cd$ is also real
there are like, continuumly many equivalent classes, one for each r?
continuumly
hahahaha
amazing
In the last days, some of you have been claiming that a probability distribution is a synonym for a probability measure. However, a p.m.f. for discrete r.v.s can also be considered the probability distribution for that same r.v. In fact, the p.m.f. of a discrete r.v. $X$ is defined as $p_X(x) = \mathbb{P}(X=x)$, which implies that the p.m.f. can also be considered a probability distribution and it is also a measure. The same can be said about p.d.f.s for continuous r.v.s
(This wasn't spiteful laughing, I genuinely think continuumly is a cool word)
if you do not assume any CH or GCH, continuumly is a good word to denote the $\mathfrak{c}$ of the reals
15:58
I just thought the word was cool hehe
@Secret This doesn't follow, consider $a=e^{\pi/6 i}$ and $b=2e^{\pi/6 i}$
@nbro The pmf is a measure, the pdf isn't
@BalarkaSen But the pdf is the derivative of the cdf of the continuous r.v., so, in a way, it can also be considered a distribution.

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