Mathematics - 2017-01-31 (page 1 of 5)

Secret

00:00

@SteamyRoot O sorry, I skipped some writing there. Yes, it should be $A_1 \supset B_1 \supset \dots$ where $\dots$ is the nth case shown above

@TedShifrin I thought proving f is injective and then deuduce A and B has the same cardinality is what is needed in the proof? (Meanwhile I do got an onion diagram drawn, I am currently fetching it because it is on paper...)

Ted Shifrin

No, you are given $f$ is injective and you're going to define $h$ and show $h$ is the desired bijection.

Arrow

Hello
Let $\pi:\mathrm TX\to X$ be a tangent bundle with a chosen connection. Let $\gamma$ be a path in $X$. Say a section of $\gamma ^\ast \pi$ is constant w.r.t the connection if it's a parallel transport along $\gamma$. Is $\gamma $ being geodesic equivalent to $\dot \gamma$ being constant in this sense?

Ted Shifrin

Isn't that the definition of a geodesic?

Secret

Arrow

I saw the definition of covariant derivative being zero, but I don't know how to show they're equivalent

Ted Shifrin

00:11

(You aren't talking about specifically a Levi-Civita connection for a Riemannian manifold, where geodesics have a different definition.)

@Secret: The picture isn't right.

The ring between $A_1$ and $B_1$ maps precisely to the ring between $A_2$ and $B_2$.

Parallel transport of the section means that covariant derivative is 0, @Arrow.

Semiclassical

hi chat

Arrow

That makes sense, I just don't know how to show it

Those statements are equivalent, right? It's not just an implication

Ted Shifrin

Yes.

I don't know where you're starting if it isn't basically definition.

Rehi @Semiclassic

Arrow

I think it's probably very easy, just not easy enough for me. I'll write out what I'm struggling with.

Ted Shifrin

You starting with a covariant derivative or a notion of parallel transport? You deduce each from the other.

Mike Miller

00:15

Hi.

Ted Shifrin

Re g'night @MikeM

Secret

Actually, I am only given the recursion relation of $f$ on the sets $A_n$, $A_1$, $B_1$ and $B_n$ and that $f$ is injective. How to show that $f(A_n-B_n)=A_{n+1}-B_{n+1}$ since $f$ is not necessary linear wrt its arguments, thus I cannot do it via $f(A_n-B_n)=f(A_n)-f(B_n)$?

Ted Shifrin

OK, that's a good place to start. Just do $n=1$ for starters. $f(A_1)=A_2$ and $f(B_1)=B_2$ by definition.

Why does something in the difference map to something in the difference? What hypothesis on $f$ must you use?

Arrow

Let $\pi:\mathrm TX\to X$ be a tangent bundle with a connection. Let $\gamma$ be a curve in $X$. The covariant derivative of a section $h$ of $\gamma ^\ast \pi$ is $$\nabla_\gamma h(t_0)=\lim_{t\to t_0}\frac{\underset{\gamma:t_0 \to t}{\mathrm{tra}}^{-1}(h(t))-h(t_0)}{t}.$$

I want to show $\nabla_\gamma \dot\gamma \equiv 0$ is equivalent to $\dot\gamma $ being a path lift of $\gamma$ chosen by the connection.

Ted Shifrin

tra is parallel transport along $\gamma$?

Arrow

00:18

Yes, sorry

Mike Miller

@Arrow Pick an immersed path in the manifold. By pulling back the bundle, section, and connection to the path, you reduce your entire question to understanding why those two notions - covariant derivative is zero iff it's a parallel transport - to working on the bundle over the path itself. Then this is an initial value problem.

Ted Shifrin

So if $h$ is parallel to start with, the numerator is dead 0.

Mike Miller

Clearly if $h$ is parallel transport along the path, the covariant derivative is zero. But since the solution to an ODE on the path is uniquely determined by its value and derivative at 0...

Semiclassical

hi @BalarkaSen

Ted Shifrin

yells at Balarka for not being asleep

Balarka Sen

00:19

hi @Ted, @SemiC

my internet's being dumb

Arrow

@MikeMiller I think I understand the second explanation. For the first, why does the path need to be an immersion?

@TedShifrin thank you

Mike Miller

@Arrow For a good notion of parallel transport. You could make it a piecewise smooth piecewise immersion if you wanted.

If it's not piecewise an immersion you no longer have a well-defined parallel transport.

Arrow

Sorry, could you explain why?

Semiclassical

@BalarkaSen One of the things you raised when I was rambling about my stuff is on my mind atm. Namely, whether one can always find an algebraic perturbation which resolves some double points but not others.

Ted Shifrin

@Semiclassic: I thought I gave an argument for that (but you have to allow higher degree than perhaps you want).

Semiclassical

00:23

(I'm not thinking hard about it, right now, since I"m in post-dinner mode)

Ted Shifrin

I'm in pre-cocktail mode.

Balarka Sen

@MikeMiller I fiddled around for a while and I think 3, 4, 234432, 12344321 generates a Z^4 inside B_5.

Semiclassical

All I remember arguing is that I had examples where I could do that.

Balarka Sen

I think in general you could play the same palindrome game to get Z^n in B_n+1

Fargle

Jesus. I need to get more smarterer.

Balarka Sen

00:24

err, 3 and 4 doesn't commute

Ted Shifrin

@Fargle: You stopped working on my books, so you must be smarterer already :)

Fargle

@TedShifrin Cessation should not be conflated with completion. >_>

Balarka Sen

well, whoops.

Fargle

(Would that it were.)

Ted Shifrin

I just meant that ignoring me should make you smarterer. :)

Balarka Sen

00:25

@Semiclassical yeah?

Fargle

Ahhhhh.

Arrow

@TedShifrin you said geodesics are defined differently for the Levi-Civita connection. Why? The definition using parallel transport seems very intuitive and works for all cases

Semiclassical

No conclusion yet. Maybe once I'm past the post-dinner yawns.

Ted Shifrin

Geodesics are defined (for many people) as paths of (locally) least length.

Mike Miller

@Arrow What does $\nabla_{\gamma'} \gamma'$ mean when $\gamma'$ can be zero?

Arrow

00:27

@MikeMiller I think I got it. Thanks again

Mike Miller

Sure.

Are you working towards something or just learning what seems fun?

Arrow

Just trying to understand the basic ideas in differential geometry

Ted Shifrin

Learning is fun?

Mike Miller

Hah. I think most differential geometers wouldn't call the categorical approach I think you're taking the basic ideas

Ted Shifrin

Remove all geometry from geometry — why not? :P

Semiclassical

00:31

geometry - geometry = categories?

Balarka Sen

yup

Arrow

I just take the categorical approach because it helps me stay organized

Semiclassical

sounds legit.

Balarka Sen

algebra - algebra is also categories

categories - categories is a 2-category

Fargle

Crap. If I see "universal property" one more time...

Ted Shifrin

00:32

LOL

Mike Miller

@Arrow unlike the rest of the room, I won't pick on you

Ted Shifrin

I was answering questions, damn it.

Fargle

Categories are neat. They just fly over my head right now.

Arrow

Haha I don't mind this kind of picking actually. I think I can make a solid case for categories even though I know very little math.

Ted Shifrin

Fargle: we've had this chat before. Your head is too high for that. But I won't dwell on it.

Arrow

00:33

Question. Suppose we have a bundle (just a continuous map) with a Hurewicz connection/parallel transport defined on it. We can define geodesics as the lifts of paths downstairs specified by the connection. Is this a useless notion? It seems that specifying geodesics is simply specifying parallel transport

Fargle

@TedShifrin We have? Whoops.

Mike Miller

@Arrow I think everyone should learn math the way they find most comfortable or helpful. In particular, if people find categorical language helpful, great! If they don't, also great! I find it helpful exactly as far as I need to use it, but the general language not so much.

Ted Shifrin

I have never heard of a Hurewicz connection. What's dat?

Mike Miller

@TedShifrin Parallel transport.

Ted Shifrin

Oh.

SteamyRoot

00:34

Woo, poster is finished

Ted Shifrin

Geodesics are downstairs, though, Arrow.

Arrow

Ahh, right. All clear now.

Ted Shifrin

Was that sarcastic or genuine?

Secret

Hmm... given any set $C \supset D$, $g: C \to D$ only preserves unions and inclusions (as $g(C) \cap g(D)$ can be $=\emptyset$ even when $C \cap D \neq \emptyset$) but their preimages preserves unions, intersections (hence differences), and inclusions. So in order to avoid the case $g(C) \cap g(D)=\emptyset$ it means $g$ is injective.

So going back to the current problem, since $f$ is injective, intersections and hence differences $A-B=A \cap B^c$ is preserved, hence $f$ preserve set differences...

Mike Miller

Remember that a geodesic is one such that the horizontal lift is parallel. But it still lives downstairs.

Arrow

00:36

Genuine! I wouldn't be sarcastic towards people who help me

Ted Shifrin

I didn't read your first paragraph, @Secret, but the second is right.

Just checking, @Arrow :)

SteamyRoot

@Secret $C \cap D \neq \emptyset$ but $g(C) \cap g(D) = \emptyset$ ?

Arrow

@MikeMiller regarding categories, I also feel that they're useful for "elementary" things too. For instance the inductive proof of homotopy invariance of singular homology basically uses naturality to reduce to simple spaces only. And from there the acyclic model theorem hints itself, which is also pretty useful

Balarka Sen

@Arrow I think of both the examples you give as homological algebra/diagram chasing, not category theory. The thing from category theory I find most useful in day to day life is Yoneda lemma

Arrow

Do feel algebraic topology flows freely with little abstract nonsense?

Secret

00:38

Steamyroot: I'll fetch a diagram to illustrate that when I get back to my main computer

SteamyRoot

Uhhh...

Let $x \in C \cap D$. Then $g(x) \in g(C)$ and $g(x) \in g(D)$.

Ted Shifrin

@Secret: No you won't :)

SteamyRoot

If $g$ is a function but not a map, sure.

But I think Munkres always means map when he says function, so I doubt that's the case.

Mike Miller

@Arrow I don't get a lot out of that. You could do that, sure, or think about things with model categories or whatever. But defining the chain homotopy is all about triangulating a product simplex, which maybe that's what you're saying?

Ted Shifrin

Whoa.

Arrow

00:40

@BalarkaSen I see. I think Yoneda is pretty deep even though its proof is simple.

Mike Miller

But I don't think about that categorically.

Ted Shifrin

@Steamy: What do you think "function" means?

Mike Miller

(Alternatively you could do cubical homology and the homotopy invariance is suuuuper trivial.)

Ted Shifrin

I never remember what Yoneda lemma is. I saw it only in an algebraic geometry class 40+ years ago.

Arrow

Triangulating the product simplex is what Hatcher does in his book, but e.g Rotman and tom Dieck take an inductive approach

Balarka Sen

00:41

@MikeMiller OTOH it should be harder to prove excision/Mayer-Vietoris in cubical

Arrow

Haven't look at cubical stuff yet

Balarka Sen

lots to subdivide

Or rather, harder to keep track of the combinatorics

Mike Miller

I don't remember what any of these books say. Just how I know how to prove things.

SteamyRoot

@TedShifrin A function $f : X \to Y$ is a relation $f \subseteq X \times Y$ such that every $x \in X$ appears in at most one tuple $(x,y) \in f$.

Arrow

Maybe I'll get to that point someday.. haha

Ted Shifrin

00:42

No, that's not a function.

Mike Miller

Mayer-Vietoris is not significantly worse, really. The main technical disadvantage is you have to mod out by degenerate cubes.

SteamyRoot

Hmmm?

Ted Shifrin

If the domain is $X$, there must be exactly one ordered pair $(x,y)$ for every $x$.

SteamyRoot

We call that a map o.O

Ted Shifrin

Well, in the US they're identical.

Mike Miller

00:43

That's just not true.

They're identical everywhere. At best what was given is a partially defined function.

SteamyRoot

Not in Europe (or at least Belgium)

function is not necessarily defined everywhere, map is.

Ted Shifrin

Only in algebraic geometry do you have not-everywhere-defined maps.

SteamyRoot

Well, we kind of use it so we can be lazy with properly defining our domains... hehe

Ted Shifrin

I think that's horrible, actually.

Arrow

@MikeMiller, @BalarkaSen, just out of curiosity, do things like topoi interest you for their own sake?

SteamyRoot

00:45

it's perfectly fine to say $f: \mathbb{R} \to \mathbb{R}: x \mapsto \frac{1}{x}$

Ted Shifrin

@Steamy: Not fine to me.

Arrow

e.g the phrase "a topos is a mathematical universe" etc

Balarka Sen

@TedShifrin If F is a functor from an arbitrary (small) category to Sets, then natural transformations hom(A,-) --> F are in 1-1 correspondence with F(A), given by sending each nat. transfromation T to the image of the identity hom A -->A in F(A).

Ted Shifrin

You've told me before, @Balarka. I never remember it.

SteamyRoot

Well, it's mostly just another different naming convention I suppose :P

Ted Shifrin

00:47

I suspect there are a few things I've told you that you never remember, though, so I don't feel too bad :P

Balarka Sen

yep, you shouldn't

Ted Shifrin

@Steamy: Map ordinarily is used in the US to connote a morphism in whatever category one's in. Function needn't have that specificity.

Balarka Sen

@Arrow I don't even know what they are, and the standard references don't really tell me why I should care

SteamyRoot

Hmmm... I think we use "map" in that context too. Been a few years since I last touched any category theory, though.

Arrow

Gotcha

SteamyRoot

00:49

Either way, we pretty much always stick to maps because functions are just awful if you try to do well-defined things.

Balarka Sen

if I ever get a reason to be interested, then sure! why not

Arrow

Ah, something I was wondering about - what's the relationship between Gaussian curvature and the definition of curvature of a connection?

SteamyRoot

@Arrow in Riemannian geometry?

Arrow

@SteamyRoot Yep

Ted Shifrin

@Arrow: What's your definition of curvature of a connection?

SteamyRoot

00:53

The Riemannian curvature is $R(X,Y)Z = ([\nabla_X,\nabla_Y] - \nabla_{[X,Y]})Z$

Arrow

This thing ${\displaystyle F^{\nabla }(X,Y)(s)=\nabla _{X}\nabla _{Y}s-\nabla _{Y}\nabla _{X}s-\nabla _{[X,Y]}s}$

Ted Shifrin

OK, so we have to unwind a bunch to answer your question.

Arrow

@SteamyRoot I like your presentation much better

SteamyRoot

Then you can define the sectional curvature $K(X,Y) = \langle R(X,Y)Y,X\rangle$ for orthogonal unit vectors $X,Y$

And if the dimension of $T_pM$ is $2$, the sectional curvature coincides with the Gaussian curvature

Balarka Sen

sectional curvature should be something like $\langle R(X, Y)Y, X\rangle$ in an orthogonal frame

Ted Shifrin

00:55

Gaussian curvature is a scalar function that's built out of the Riemannian curvature for either a hypersurface of a Riemannian manifold or an even-dimensional manifold in general.

orthonormal, @Balarka

Balarka Sen

Ah, right

SteamyRoot

@Arrow another presentation is $R(X,Y) = \nabla^2_{X,Y} - \nabla^2_{Y,X}$.

Ted Shifrin

In general, Gaussian curvature is built out of the curvature 2-form matrix, @Arrow. Do you want me to go on?

Balarka Sen

I don't get the Riemann curvature tensor

Arrow

I would actually like some intuition for the curvature tensor

Balarka Sen

00:56

I know the "parallel transport along an infinitsimal square" interpretation but I still don't get it

Arrow

The definition of Gaussian curvature is intuitive I think

Ted Shifrin

Well, so you have to have intuition for curvature of surfaces.

Arrow

Measures stretching. In Gaussian case relies on embedding into Euclidean spaces and sees how much normals "spread out"

is that reasonable?

Ted Shifrin

Sectional curvature of a 2-plane (doing Riemannian connection still) at $p$ is the Gaussian curvature of the surface you get by following geodesics emanating at $p$ with tangent vectors in that 2-plane.

SteamyRoot

Well, hey, every Riemannian manifold can be isometrically embedded in some Euclidean space.

Ted Shifrin

00:57

For a hypersurface in $\Bbb R^{n+1}$, the Gaussian curvature is in fact the jacobian determinant of the Gauss map.

@Steamy: For the intuitive one I just gave, you need a hypersurface. High codimension means you'll have to do some average process.

SteamyRoot

Hmm, right.

Meh, that's the sad thing about Nash' embedding theorem. The codimension gets ridiculously high quickly...

Balarka Sen

@TedShifrin Yeah, but why should anyone come to think of the Riemann tensor - symbolically - at all? And why should that agree with the Gaussian curvature of the 2-plane after exponentiating it onto the manifold? That sounds like a technical result I have no reason to believe

Ted Shifrin

Very.

It's a good moving frames exercise, @Balarka. I put that on my final exam in the grad course :P

Balarka Sen

Hm

Arrow

@TedShifrin where might I find an explanation of this good exercise?

Ted Shifrin

01:01

Here's another good exercise for you, @Balarka. If $M$ is a surface in $\Bbb R^n$, prove that the Gaussian curvature $K=\sum K_\mu$, where $e_\mu$ form an orthonormal basis for the normal space and $K_\mu$ is the projection of $M$ into the $\Bbb R^3$ spanned by $T_pM$ and $e_\mu$. :)

Balarka Sen

Interesting; bookmarked

Ted Shifrin

Good question, @Arrow. I don't know all the modern differential geometry texts. I worked it out for myself years ago. And I no longer have my huge library to check.

Sadly, there's not enough geometry in so many of the books ...

I think my first exercise was proved by Riemann, actually, using normal coordinates.

Arrow

wow

Ah, I have a somewhat annoying question about the covariant derivative in terms of parallel transport again

$$\nabla_\gamma \gamma ^\prime(t_0)=\lim_{t\to t_0}\frac{\underset{\gamma:t_0 \to t}{\mathrm{tra}}^{-1}(\gamma ^\prime(t))-\gamma ^\prime(t_0)}{t}.$$

Mike Miller

@Arrow I don't care about them at all. But I don't object to people who do like them.

Balarka Sen

At some point I should stop being a sucker at differential geometry

Arrow

01:06

In the above definition we're identifying the fibers with their tangent spaces, right?

Ted Shifrin

@Arrow: As Mike already pointed out in part, you should be writing that as $\big(\nabla_{\gamma'(t_0)} \gamma'\big)(t_0)$.

Arrow

@MikeMiller gotcha.

Mike Miller

My friend Kevin likes them, I think. He took a course on them or something.

Ted Shifrin

Yes, no doubt, @Arrow.

Arrow

@TedShifrin I don't understand. The covariant derivative by definition is with respect to a curve... even if its value at a point depends only on the derivative

Ted Shifrin

01:08

More generally, covariant derivative is a directional derivative in a direction.

The right functoriality to understand is how it behaves with respect to the direction (linearity, for example).

Arrow

Isn't that less general?

Ted Shifrin

No.

Arrow

A tangent vector is by definition obtainable from a germ of a curve

Ted Shifrin

But I want a linear map, just as I do in calculus.

Arrow

Linear in the section of $\gamma ^\ast \pi $?

Ted Shifrin

01:10

No, linear in all the directions tangent to $M$.

Arrow

Ah, so you're saying this doesn't make sense with curves?

Ted Shifrin

I'm saying you have a lot to prove from where you are to get what I'm saying.

You're talking about directional derivatives one direction at a time, without having a notion of derivative as a linear map.

Interestingly, there was a question like that about directional derivatives on vector spaces, asking to prove linearity in the direction if one knows a bit more. I'd never seen that before.

Arrow

Then I don't follow. Without any linearity or anything, a covariant derivative comes from a parallel transport (suppose also without any linearity conditions). It's defined along curves. If we want to look at the covariant derivative in a certain direction then we can look at the derivative of the curve at the point in question.

If we want linearity then we must work with covariant differentiation with respect to a tangent vector.

Is this correct or am I missing the point?

Ted Shifrin

Well, your definition gives one tangent vector at a time. How in the world do you see linearity as a function of them all at once?

That's a hugely important part of the structure of covariant derivatives.

Arrow

Ok, I don't understand. First of all I don't understand what you mean by "gives one tangent vector at a time"

Ted Shifrin

01:15

You pick a curve $\gamma$ with tangent vector $v$ and do what you're doing to compute $\nabla_v s$.

Arrow

Yes

Ted Shifrin

It is trivial to see that $\nabla_{cv}s = c\nabla_v s$.

But how the hell do I prove that $\nabla_{u+v} s = \nabla_u s+\nabla_v s$?

Arrow

I thought this comes from assumptions on the connection

Ted Shifrin

You're defining your connection only in terms of parallel transport on curves.

So there's a real theorem needing to be proved.

Arrow

What underlies it?

Ted Shifrin

01:18

LOL ... the "better" way of looking at connections, I suppose. I'm saying the formula I just wrote up there (linearity) is a theorem you need to prove somehow.

Arrow

What's the "better" way / how would you go about this?

Ted Shifrin

Off the top of my head, I don't know how to get there from where you are.

Arrow

Ok. I think I'll be able to find this in a book somewhere.

Ted Shifrin

The better way is to define a covariant derivative as depending on a vector field on $M$ and a section of $E$.

But I actually have dinner company coming soon, so I need to skedaddle and cook.

Arrow

I thought that's worse because those are global objects, so less general

Okie

Ted Shifrin

01:20

You can localize them as much as you want. :)

But that's how people actually compute things, not your way.

Arrow

Ok. I'll try to fill in some gaps. Thank you very much!

Ted Shifrin

Talk to you soon :)

Lelo

01:48

So, if a family has two children, what is the probability that both children are girls? Wouldn't be $\binom{2}{2}*0.5*0.5$?

*Assuming both genders are independent and are equally likely.

Fargle

@Lelo Correct. Another way to see it: in order of birth, you could have either BB, BG, GB, or GG with equal chance, so P(GG) = 0.25.

Lelo

I see. Thanks :)

Secret

02:03

@SteamyRoot Sorry, I must have accidentally swapped $=$ with $\neq$ in my previous message. I now get the same conclusion you have wrote there

Lelo

Hmm now I'm trying to find the probability that both children are girls if you know that there is at least one daughter. I thought it would be $0.5$ but it was incorrect. So confused.

Secret

sorry mistake, inclusions are always preserved, thus this particular type of intersection will always be preserved regardless of whether g is injective

Lelo

02:23

Nevermind I got it. It was $1/3$ because the order of the set DID matter.

Rajesh Dachiraju

3

Q: A question on a limit arising in Fourier series

Let $$a_k = \frac{1}{k\pi}\sum_{j=0}^m\alpha_j\sin(kT_j)$$ $\alpha_j,T_j \in \mathbb{R}, 0<T_j<\pi,m \in \mathbb{N}$. I'd like to know the limit if it exists, $$\lim_{n\to\infty}\frac{1}{\log(n)}\sum_{k=1}^n\left|a_k\right|$$

Chandler Bing

0

Q: When do the needles of clock meet (Where did I go wrong)?

I tried finding it out mathematically by taking their velocities in degrees per second, and finding out the L.C.M of them which should give me the time period in which all the needles intersect. Speed of seconds needle : 6°/sec Speed of minutes needle : 0.1°/sec Speed of hours needle : 0.00833...

least-common-multiple

There is something wrong, but I don't know what help mee !

user400188

02:44

What is the difference between ⊆ and ∈ ? Apart from the name.

arctic tern

{1,2} is a subset of {1,2,3}, and 1 is an element of {1,2,3}

the words "subset" and "element" are different

(if you don't have latex in chat going for you then see "LaTeX in chat" in room description)

user400188

I'm switching browser right now to see it

arctic tern

guess there wasn't really a reason for latex for this question

user400188

Hmm ok; if so ⊆ means element what does the underscored bellow it signify? I have seen some without it

arctic tern

⊆ does not mean element, and the partial "equals" sign underneath serves the same function it does in the "less than or equal to" symbol: it means the the two things could be equal

{1,2,3}⊆{1,2,3} for instance

user400188

02:50

I'm confused about the difference again. ∈ allows for the following case you see {1,2,3}∈{1,2,3}. It also allows for {1}∈{1,2,3}

so how does ⊆ differ?

Hello?

Secret

I don't quite understand in the Munkres solution why the map $l$ will be a surjection (thus concluding it is a bijection). It is true that in the previous line we concluded that $|f(A)|=|C|$ and it is given that $f$ is injective and $f(A) \subset C$ but I see no reason why $f$ will also be surjective

For example. consider $A=\mathbb{N}$ and $f(A)=2\mathbb{N}$. It is easy to see that $f$ is indeed injective as required, but $2\mathbb{N} \subset \mathbb{N}$ despite $|2\mathbb{N}|=|\mathbb{N}|$, thus $f$ is not surjective?

arctic tern

@user400188 {1,2,3}∈{1,2,3} is wrong

x∈A means x is one of the things in A, while A⊆B means every element of A is also in B

user400188

Why? I was under the impression that ∈ allowed for the case in which the two things were equal to another. It just meant the same set or smaller.

arctic tern

@user400188 no. you're making stuff up.

A⊆B allows for A=B

user400188

what do you mean by one of the things? Do you mean only 1 element is allowed?

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$Mathematics$ Mathematics