Mathematics - 2017-02-14 (page 2 of 6)

DHMO

09:00

@AliCaglayan in a measure of subsets of R, must a point have measure 0 or infinity?

I suppose not; I can construct a measure that is 1 if the set contains 0

Ali Caglayan

Don't think there is a restriction

DHMO

@AliCaglayan do you have any results pertaining to measures on R?

Ali Caglayan

Well measures distribute over unions as sums

ignore that

DHMO

that's true to measures in general

Ali Caglayan

only when countable

yeah if I can remember there is not much else they do

DHMO

09:05

yes, only when countable

Ali Caglayan

non negative and empty set has measure 0

So there are definitely measures that have singletons with whatever measure you want

DHMO

that's also true to measures in general

Ali Caglayan

A countable union of points will have measure 0 in R

For example the rationals

DHMO

only in Lebesgue measure

Ali Caglayan

then that answers your question

DHMO

09:08

Can I define a measure called "the number of rational numbers"?

Ali Caglayan

yes

just the counting measure

DHMO

I don't think that's equivalent to the counting measure

Ali Caglayan

why not?

DHMO

I'm talking about $\mu(S) = |\{x \in S:x \in \Bbb Q\}|$

Ali Caglayan

well if S is open then this is pretty much infinity most of the time

DHMO

09:11

is this a valid measure?

Ali Caglayan

yes it fits the axioms

DHMO

I see

@AliCaglayan do you have any exercise about measure in general or Lebesgue outer measure for me?

Ali Caglayan

I did measures a long time ago

I do not have any analysis books with me

So thats a no sorry

DHMO

eh, I'm a beginner so any exercise would do

Ali Caglayan

But if its anything other than analysis I might have something

DHMO

09:14

alright

what does "Lebesgue-measurable" mean?

Ali Caglayan

Simply means sets that can be assigned a Lebesgue measure

DHMO

how do we determine that?

Ali Caglayan

You look for sets that aren't in the Lebesgue sigma algebra

DHMO

how do you find sets in the algebra?

Ali Caglayan

Are you not following a book?

DHMO

09:18

I am not

Ali Caglayan

Its probably best to

I might unintentionally be imprecise causing confusion for you

Here is a question talking about immeasurable sets math.stackexchange.com/questions/925371/…

DHMO

thanks

Alessandro Codenotti

@DHMO fix $x_0$, let $f:2^{\Bbb R}\to\bar\Bbb R$ be a function such that $f(A)=1$ if $x_0\in A$ and $0$ otherwise. Is $f$ an outer measure?

DHMO

$f:2^{\Bbb R}\to\bar{\Bbb R}$

@AlessandroCodenotti what is the difference between a measure and an outer measure?

SteamyRoot

09:36

Outer measures are defined on all subsets

DHMO

oh, lol

SteamyRoot

but (in general) they're only subadditive

You can distill a measure from an outer measure, though

DHMO

@AlessandroCodenotti
1. non-negativity: by definition the image is {0,1}.
2. null-empty set: x_0 cannot be in {}, so f({}) = 0.
3. countable subadditivity: if A_1,A_2,...,A_n does not contain x_0, then both sides are 0; if they contain x_0, then the left hand side must be 1 while the right hand side is at least 1

@SteamyRoot Isn't monotonicity derived from countable subadditivity?

SteamyRoot

If you think it is, prove it? ;)

DHMO

let A subset B. Then let C = B-A. by subaddivity, f(A+C) <= f(A) + f(C), so f(B) <= f(A) + f(C) and I can't continue. interesting

but, monotonicity is a result of countable additivity

is subadditivity a result of countable additivity also?

SteamyRoot

09:45

countable subadditivity is a result of countable additivity, of course.

DHMO

I see

SteamyRoot

If "=" holds, then "$\leq$" also holds...

DHMO

@SteamyRoot in additivity, the sets are required to be disjoint

Alessandro Codenotti

the point is that for outer measure $\leq$ holds for non measurable sets, while they're additive on their $\sigma$-algebra of measurable sets

SteamyRoot

Sure, but you can easily enough create a bunch of disjoint sets with the same union and such

DHMO

09:48

@AlessandroCodenotti am I correct?

SteamyRoot

As for the monotonic following from subadditivity: think of functions on $\{0,1\}$ for example.

Alessandro Codenotti

@DHMO where?

DHMO

12 mins ago, by DHMO

@AlessandroCodenotti
1. non-negativity: by definition the image is {0,1}.
2. null-empty set: x_0 cannot be in {}, so f({}) = 0.
3. countable subadditivity: if A_1,A_2,...,A_n does not contain x_0, then both sides are 0; if they contain x_0, then the left hand side must be 1 while the right hand side is at least 1

@AlessandroCodenotti here

Alessandro Codenotti

yes, you need monotonicity too but that's easy to show

DHMO

thanks

SteamyRoot

09:54

Actually, isn't that the Dirac measure?

Alessandro Codenotti

now the interesting question is what are the measurable sets with respect to this measure?

@SteamyRoot Could be, I've never seen this measure called with a specific name

SteamyRoot

Really?

That's really surprising to me. It's one of the standard examples in measure theory, and it's how you can justify integrating over a delta "function".

Alessandro Codenotti

in the same way pick a set $X$ and define an outer measure $\varphi:2^X\to[0,+\infty]$ such that $\varphi(A)=0$ if $A=\varnothing$ and $1$ for all other subsets of $X$ (it's easy to check that this is a measure), what are the measurable sets though?

@SteamyRoot According to wiki that's called the Dirac measure, weird, I wonder why our prof didn't tell us that is a name

DHMO

if E and A contain x0, then f(E)=1, f(ExA)=1, f(Ex-A)=0.
If E contains x0 but not A, then f(E)=1, f(ExA)=0, f(Ex-A)=1.
So E is measurable.

If both E and A do not contain x0, then f(E)=0, f(ExA)=0, f(Ex-A)=0.
If E does not contain x0 but A contains x0, then f(E)=0, f(ExA)=0, f(E-A)=0.
So E is measurable.

Every set is measurable.

Alessandro Codenotti

what's ExA?

DHMO

10:00

$E \cap A$

Alessandro Codenotti

So what's Ex-A?

DHMO

$E \cap A^c$

Alessandro Codenotti

yup looks good

DHMO

thanks

and how is it interesting?

SteamyRoot

Every set being measurable is kinda nice :P

DHMO

10:04

Is there an outer measure where no set is measurable except the empty set and $\Bbb R$?

SteamyRoot

Also, this means physicists are actually doing nothing wrong when they go $\int f(x)\delta_{x_0}(x)dx = f(x_0)$, even though they don't know why.

There's at least one set which is always measurable :P

Alessandro Codenotti

at least $2$ I'd say?

DHMO

@AlessandroCodenotti which two?

Alessandro Codenotti

you can find out by yourself

SteamyRoot

Oh, yeah, of course

But if you have one, you automatically have the other...

DHMO

10:08

@AlessandroCodenotti $\phi(A) + \phi(A^c) = 1 + 1 \ne 1 = \phi(\Bbb R)$, unless $A = \varnothing$ or $A = \Bbb R$

but why must $\Bbb R$ be measurable in every outer measure? @AlessandroCodenotti

Alessandro Codenotti

@DHMO after proving that the measurable sets of an outer measure form a $\sigma$-algebra the proof that the whole space is always measurable will be easy

SteamyRoot

In every outer measure, or in every outer measure defined on $\mathbb{R}$ ?

Alessandro Codenotti

actually you don't even need all of that, it's easy to show that $A$ measurable implies $A^C$ measurable

and since $\varnothing$ is always measurable...

SteamyRoot

yup

DHMO

@AlessandroCodenotti wait, $\Bbb R$ is not measruable: Take $A=\{1\}$, $\phi(\Bbb R \cap A) = \phi(\{1\}) = 1$ and $\phi(\Bbb R \cap A^c) = 1$

Alessandro Codenotti

10:13

that means that $A$ is not measurable

DHMO

oh, lol

Alessandro Codenotti

$\phi(A)=\phi(A\cap E)+\phi(A\cap E^C)$ for all $A\in2^X$ means that $E$ is measurable

DHMO

in that case I understand why $\varnothing$ and $\Bbb R$ must be measurable

@AlessandroCodenotti Earlier I created an outer measure $f(A) := |\{x \in A: x \in \Bbb Q\}|$

what are the measurable sets?

I feel that all sets are measurable lol

Alessandro Codenotti

Hmm, if your measure were defined on $\Bbb Q$ than all subsets would be measurable, I'm not sure on $\Bbb R$ but it shouldn't be much harder

DHMO

$f(A \cap E) + f(A \cap E^c)$
$= |\{x \in A \cap E: x \in \Bbb Q\}| + |\{x \in A \cap E^c: x \in \Bbb Q\}|$
$= |\{x \in A \cap E: x \in \Bbb Q\} \cup \{x \in A \cap E^c: x \in \Bbb Q\}|$
$= |\{x \in A \cap E \lor x \in A \cap E^c: x \in \Bbb Q\}|$
$= |\{x \in A: x \in \Bbb Q\}|$
$= f(A)$

The risky step is $|X|+|Y|=|X \cup Y|$

Alessandro Codenotti

10:22

yeah I think everything is measurable

DHMO

which is proved here

Alessandro Codenotti

I'm not sure what paradox are you referring to

DHMO

@AlessandroCodenotti but my measure isn't translation-invariant

I was referring to Banach-Tarski paradox, which also requires translation-invariance

Alessandro Codenotti

@DHMO right

DHMO

@AlessandroCodenotti Can you find the translation-invariant sets under my measure?

all open sets in $\Bbb R$ are translation-invariant for example

Alessandro Codenotti

10:27

probably all the sets containing an open interval

DHMO

Consider any basis in $\Bbb R / \Bbb Q$. It does not contain any open interval

It is translation-invariant having measure always $1$.

Am I correct?

Alessandro Codenotti

it had measure $0$ before translating though

DHMO

how?

Alessandro Codenotti

it doesn't contain any rational

DHMO

How can it generate $\Bbb Q$ if it does not contain any rational?

Alessandro Codenotti

10:30

ah, wait I was thinking about $\Bbb R\setminus\Bbb Q$

DHMO

$\Bbb R \setminus \Bbb Q$ has measure $\infty$ if translated by a number within, and $0$ otherwise. Am I correct?

Alessandro Codenotti

yep

I'm not sure what you're saying about a basis of $\mathbb R/\Bbb Q$ or why is it relevant here. How did you decide that it has measure $1$?

DHMO

@AlessandroCodenotti well, it has to contain exactly one element of $\Bbb Q$ in order to generate $\Bbb Q$

and I used intuition for the translation...

Fun fact: "pardon" stems from "per-" and "donare", meaning "for" and "give" respectively, which is created exactly from a Germanic form of "for- give"

so our language contains "pardon" and "forgive"...

@AlessandroCodenotti well, my intuition tells me that after translation, it is still a basis of $\Bbb R / \Bbb Q$. Is my intuition correct?

@MartinSleziak your bounty is expiring lol

Martin Sleziak

@DHMO Every bounty expires after one week. What's unusual about that?

At least I'll have something to post in the bounty room.

Jan 23 at 7:22, by Martin Sleziak

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to promote it here. So here is link to the relevant meta post and here is the room.

DHMO

@MartinSleziak could you answer my question?

Martin Sleziak

10:39

I probably should not visit chat. People are always laughing out loud at me.

@DHMO I am sure I have answered several of them.

DHMO

@MartinSleziak I asked if a basis of $\Bbb R / \Bbb Q$ is still a basis after translating

Martin Sleziak

Do you mean Hamel basis of $\mathbb R$ as a vector space over $\mathbb Q$?

Alessandro Codenotti

that should be true

Martin Sleziak

Or what is $\mathbb R/\mathbb Q$?

Alessandro Codenotti

I don't see why can't it have measure $0$ after translating

@MartinSleziak yes

Martin Sleziak

10:42

If $B$ is a Hamel basis of $\mathbb R$ and $1\in B$, then after translation $B-1$ you no longer have a basis.

Since all elements of $B-1$ are irrational.

Are there some restrictions on $a$, if you are asking whether $B+a$ is a Hamel basis?

DHMO

@MartinSleziak oh...

in that case, must it contain exactly one rational?

Martin Sleziak

Oh, that's non-sense what I wrote, right?

DHMO

I don't think it's nonsense

since there must be a rational that is in $B$

you can translate it by $-B$ and end up with a set containing $0$

Martin Sleziak

So must Hamel basis contain a rational number?

DHMO

so you must remove the zero and the remaining elements would be irrational, so they cannot generate $\Bbb Q$, right?

@MartinSleziak wait...

They don't have to contain a rational number?

Martin Sleziak

10:46

Rational combination of irrational numbers can be rational. $\sqrt 2+(1-\sqrt 2)$.

I was too hasty.

DHMO

yes, I just figured that out

Alessandro Codenotti

must there be one? What if my basis contains $1+\sqrt{2}$ and $\sqrt{2}$?

lol we all realized at the same time

Martin Sleziak

Still, if you are asking whether $B+a$ is Hamel basis, you definitely want $-a \notin B$, as DHMO said.

DHMO

@MartinSleziak I guess I should give you the original question: is there a set such that it contains exactly one rational number under any translation?

Martin Sleziak

That reminds Vitali's construction.

DHMO

10:49

you have too much knowledge lol

Martin Sleziak

That's non-sense.

When "constructing" Vitali set we take exactly one element from each coset $a+\mathbb Q$, right?

DHMO

It looks like the Vitali set is the set I am finding

@MartinSleziak what is $a$?

Martin Sleziak

$a\in\mathbb R$

DHMO

isn't that the same as $\Bbb R / \Bbb Q$?

Alessandro Codenotti

@MartinSleziak right. But there's some more work needed

Martin Sleziak

10:53

The coset $a+\mathbb Q$ an element of the quotient group $\mathbb R/\mathbb Q$.

DHMO

I think I confused Hamel basis with quotient group.

Can I just pick a basis of the quotient group $\Bbb R / \Bbb Q$?

Alessandro Codenotti

actually maybe now, I'm misremembering

Martin Sleziak

Since we have $V\cap (a+\mathbb Q)$ is singleton for each $a\in\mathbb R$, this also means that $(V-a)\cap \mathbb Q$ is singleton.

@DHMO I do not know what is basis of a group.

DHMO

@MartinSleziak I guess I need to study everything all over again

@user400188 welcome

user400188

Hello again

DHMO

10:55

@MartinSleziak I mean, the set of the quotient group contains equivalence classes right

pick one from each class and that forms the basis

@user400188 so, should we continue?

Martin Sleziak

Yes, the underlying set of the quotient group $\mathbb R/\mathbb Q$ is $\{a+\mathbb Q; a\in\mathbb R\}$.

user400188

Sounds like a good Idea; but first, I have a question on The Axiom of Extensionality

Secret

Ok I am back, I think what DHMO found in the MSE is very similar to what I had in mind (and very similar to that incoherent mess of sentence when I first asked about topology before I read munkres). So to begin (reposting the diagram)....

3 hours ago, by Secret

DHMO

@user400188 go ahead

Secret

once again label the figures as
1 2 3
4 5 6
7 8 9
(also please ignore the typo "emptyset not shown", as DHMO and I discussed earlier)

DHMO

10:56

@MartinSleziak oh, so I was correct except for using the term "Hamel basis" lol

thanks then

@AlessandroCodenotti so there you are, a translation-invariant set containing no open interval under my measure

@MartinSleziak just to confirm, does it contain any open interval?

no, because any open interval must contain at least 2 rational numbers, lol

user400188

So the Axiom of Extensionality: $\forall A\forall B(\forall X[X\in A\leftrightarrow X\in B]\rightarrow A=B)$

DHMO

yes

user400188

what happens when I write the if as $\lnot A or B$ instead of $A\rightarrow B$ ?

DHMO

as I said... the $\implies$ there means $\iff$... because it is a definition

user400188

You said that then later said that was a mistake

DHMO

10:59

if you replace it with a real $\implies$, that means there are examples of $A=B$ for which $A$ and $B$ do not share elements...

Martin Sleziak

In case there are any takers - this was asked in another room:

in Linear & Abstract algebra, 21 hours ago, by BAYMAX

I was reading the proof of a theorem which states that "if $|G| > 1 $ and $G$ is simple,Abelian then $G$ is prime cyclic ." Can someone help me in this !

DHMO

@user400188 the mistake is with the axiom of pairing, not here

user400188

ah ok. Its strange that they use it that way.

Secret

1 and 9 are the easiest. 9 is the discrete topology. Here we can see a basis for 9 are the singletons. By using the discussion about hausodoff spaces in munkres as a starting point, since we can find neighbouhoods such that a,b,c are disjoint form each other, a,b,c are in the intuitive sense "isolated form each other

Whereas for 1, the only basis is the whole set, so it the neighbourhood. This contains all points. Therefore the singletons a,b,c are not disjoint from each other. If I understood correctly, that means a,b,c are in some sense "near each other"

DHMO

@MartinSleziak just to make sure: is the $\implies$ in the axiom of extensionality the same with $\iff$?

Martin Sleziak

11:01

I'm not sure what you mean.

DHMO

@MartinSleziak the axiom of extensionality: $\forall A \forall B: (\forall x: x \in A \equiv x \in B) \implies A = B$

Martin Sleziak

Are you asking about $\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \Rightarrow A = B)$ vs. $\forall A \, \forall B \, ( \forall X \, (X \in A \iff X \in B) \Leftrightarrow A = B)$?

DHMO

@Secret we say that a,b,c are connected

@MartinSleziak yes

Martin Sleziak

I think that you do not need any axioms to get $A=B \Rightarrow (x\in A \Leftrightarrow x\in B)$.

DHMO

@MartinSleziak heh? how?

Alessandro Codenotti

11:03

@DHMO yup, Vitali is weird. Keep that set in mind when you get to the Lebesgue measure

DHMO

@AlessandroCodenotti now, why isn't it measurable under Lebesgue?

Kasmir Khaan

Hi guys , i need help with this question , am gonna calculate the line integral around the circle x^2+y^2 =1 and y>=0 , y>=x , F is not difined at (0,0)

user400188

What are you suggesting Martin? That A=B implys by itself the other structure?

Alessandro Codenotti

that's not a short proof @DHMO

DHMO

@KasmirKhaan and what is the function?

@AlessandroCodenotti never mind then

Secret

11:04

Now for 5, we have a similar situation to 9, except a is isolated from b,c while b,c are connected (near each other(?)). A similar thing can be said about a,b in 8 as their neighbourhoods are disjoint from each other, thus a,b, are disconnected from each other

DHMO

@Secret X and Y are said to be connected if every open set containing X also contains Y, and vice versa

Kasmir Khaan

DHMO am going to make other path , elliptical using this function r(t) = (cost, sint/ sqrt3 )

Martin Sleziak

@DHMO Or to be more precise, this has nothing to do with axioms of ZFC, this is already part of logic. Have a look, for example, at axioms of Hilbert system. Wikipedia lists this as I9.

Kasmir Khaan

the problem is that they set the bounds on that ellips from pi/3 to pi

Martin Sleziak

@user400188 See above ^

Kasmir Khaan

11:06

but i find it from pi /4 to pi

I dont see how they did it

DHMO

@user400188 so I was wrong. The definition of $=$ itself includes the fact that every property has the same truth value

user400188

I don't belive you ever said it wasnt. At least I didnt notice it if you did.

DHMO

@Secret "their neighbourhoods are disjoint from each other" is imprecise, if not incorrect

@user400188 so are we clear on this matter?

user400188

Not exactly. In I9, I see a := symbol to define '=', howver to my understanding =, := and $\leftrightarrow$ all have the same meaning

DHMO

@user400188 I don't think $:=$ there means $\iff$

and $=$ certainly has a different meaning with $\iff$

Secret

11:11

However, when I tried to apply a similar logic to 2 (and by extension to the other figures), I ran into a problem

Using the logic I just mentioned, a is disconnected from b,c. But there's also a neighbouhood around a that makes a only be disconnected from c and of course the neightbourhood that is the whole set where all 3 points are connected to each other.

The only way out is I think when we discuss whether points are "near each other" intuitively speaking, we seemed to be talking about different scales. So if we take unions, we "zoomed out" in the scale thus some points that seemed dis…

user400188

@DHMO I doubt it does. Its just that the few times I have come accross it it has meant the same thing

DHMO

@user400188 where?

user400188

Also = has always meant the same thing to me in logic

It was in the chat actualy. I asked what it meant when soemone brought it up and they said := just meant $\leftrightarrow$; although it was pronounced differently.

It was on the topic of propositional logic mind you

DHMO

I am minded

user400188

I'm fairly sure in that context they are the same

DHMO

11:15

@Secret I guess I am wrong, as I always am

I seem to have invented the word "connected"

The term "isolated" is real though

@Secret I don't know what "near" means though

Secret

Is "two different points lying within the same neighborhood" the same as the intuitive notion of "two points are next to each other"?

DHMO

@Secret two points must lie within a same neighbourhood, namely the underlying set

Martin Sleziak

What I meant that one of logical axioms says that if you have a formula $\phi(x)$ and if $A=B$, then $\phi(A)$ and $\phi(B)$ are equivalent.

Secret

The answer you refer to me also reference to the idea of "two different scales"

Martin Sleziak

I might be mistaken, but I think that this is what I9 in that Wikipedia article says.

DHMO

11:19

@MartinSleziak I agree

Secret

user400188

You say that; but you used "if" for the first $\rightarrow$ and equivilant for the second.

@MartinSleziak ^

DHMO

@user400188 yes, because the other direction is embedded in the definition of $=$ and is a separate axiom

so the axiom of extensionality deals with the $\implies$ case

and I9 deals with the $\impliedby$ case

Secret

But then, the other two neighbourhoods of $a$ shown above will mean $a$ is somehow not next to $b,c$ and next to $b,c$ at the same time?

DHMO

@Secret As I said, you need to define "next to"

user400188

11:21

ahhhh. I see what you mean now. Thanks @DHMO and @MartinSleziak

DHMO

@user400188 so are we clear?

user400188

Ill need to write that hilbert thing down next to my definition of Extensionality

DHMO

axiom of extensionality

user400188

We are clear.

DHMO

then should we go to the next axiom?

user400188

11:23

One last question before we do: would it be logicaly correct to "AND" I9 with the Axiom of Extensionality?

DHMO

sure

user400188

I'm thinking if I did so then the one directional $\implies$ would go away

DHMO

you would end up with $\forall A,B: (\forall x: x \in A \equiv x \in B) \iff A = B$

user400188

I'll go through thw working in my own time. I jjust wanted to make sure it was allowed.

Secret

If the space has some given metric d, then we can say "a is next to b" if d(a,b) < $\delta$ where $\delta > 0$ is some number. But in general topology, we don't necessary have a metric, thus I don't know how the daily life intuition of "next to" as formalised above is reflected by open sets of different sizes and types in topology

DHMO

11:26

Can we prove that $S \subset \Bbb N \land |S| < |\Bbb N| \implies S\text{ has a maximum}$?

@Secret then every number is next to each other...

@user400188 so should we go to the next axiom?

user400188

Sure

Jasper Loy

The jokes in this chat are so not funny, lol.

user400188

Sorry I took so long I was getting lead for my pencil

Jasper Loy

I used normal pencil then mechanical pencil and now I am going back to normal pencil again, because mechanical pencil you always need to put in the lead and sometimes it doesn't work very well. =)

DHMO

@user400188 Axiom of Union: $\forall X \exists Y \forall u [u \in Y \equiv \exists z[u \in z \land z \in X]]$

user400188

11:35

I've never had a problem with mine. Plus I've found the lead useualy lasts longer than the lead does beofre you need to sharpen it on a regular one.

Jasper Loy

The problem with pen is that you keep running out of ink, lol.

user400188

Just to confirm $\equiv$ means $\leftrightarrow$ ?

DHMO

yes

user400188

Sorry I had to move rooms just then. So I was gone for a while

at the moment im trying to translate the axiom into english

DHMO

sure

user400188

11:45

For all x and u, there exists a Y suh that u is an element of Y; if and only if there exists a z such that u is in z and z is in X

DHMO

you cant swap the quantifiers!

user400188

WHat do you mean? I only changed the order of $\exists Y$

DHMO

you cant do that

user400188

Is that only the case when there is more than one?

I'm pretty sure $P(x)\land [\exists y \in Y Q(y)]\equiv\exists y \in Y [P(x))\land Q(y)]$

DHMO

hmm

that isnt swapping quantifiers

alright

user400188

11:56

Indeed. So its only quantifiers that cant be swapped?

DHMO

i dont know

user400188

I sort of get why it might not work. $\forall A\exists B$ seems to cover less cases than $\exists B\forall A$ becuase one takes all cases of A and choses a b in them; while the other takes 1 or more cases of B and finds all the A in them.

or at least that is how they seem to read

DHMO

yes

user400188

I've just never truthed that definition beucase I made it up myself. And that it sounds to much like using subsets which Quantifiers are not.

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$Mathematics$ Mathematics