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23:02
Just curiosity, I was wondering wether there are spaces such that every open cover (not using the whole space as a set of the cover) is countable, or above some cardinal and I stumbled upon those other spaces when thinking about the former
(I quickly realized that there are no spaces like that, if a space admits a covering with open sets that doesn't use the whole space as a set it clearly has a finite cover of open sets too)
Hello.
hi
Is $\{a\}=a$ ? Just wondering.
no
No, it's a set containing $a$
23:06
i think it is one of the axioms or something
I read that $f(\{a\})=f(a)$
$\{\{3,4,5\}\}\neq\{3,4,5\}$, assuming I didn't mess up the brackets on my phone the one on the left has a single element and the one on the right has 3
That's correct, how is f(A) defined when A is a subset of the domain of f?
@DHMO it follows from the axiom of regularity (in ZFC)
@aless i see
I have to go now, but if you recall that $f(A)=\{f(a):a\in A\}$ you should see why $f(\{a\})=f(a)$ @Mahmoud
Thanks @Alessandro $:)$
23:18
@DHMO but isnt factoring out "hidden" devision? since $a+b=x(\frac{a+b}{x})$
@Mahmoud $\{$apple$\}$ is like a bag containing an apple. you wouldnt call a bag an apple or?
No, it's not the same thing,
exactly ;)
Thanks :D
@saturatedexpo no it isn't
you just need to use distribution, which is one of the axioms
ah ok ;)
the "gathering" haha
23:28
This function is strange $f : \mathbb R \rightarrow \mathbb R | x^2+y^2-2\sqrt{xy}=0$
edit the / to a \
strange is subjective
It shows the symbol of infinity :D
can you prove that it is a function? @Mahmoud
@DHMO exactly what i thought
23:31
It takes elements from $\mathbb R$ And maps them in a way expressible with the equation above. Right ?
no
I didn't know if it maps into or onto
the normal notation is: $a\mapsto b$ in plaintext: a\mapsto b
Isn't it a function then ?
(for the elements)
23:32
But we use $\rightarrow$ for the domain and range. Right ?
@Mahmoud the first statement is false so the second statement does not follow
the notation is wrong then
What statement it's just the definition of a a function.
@DHMO well, some profs are vague on that, $f:R\to R$ doesnt mean necassary for them that its bijective
23:33
presumably mahmoud is considering the set of all ordered pairs for which the relation holds. the first thing to do would be to investigate if every x value has one and only one corresponding y value.
Yes, @arctictern I didn't know how to express $y$ as an expression of $x$.
based on W|A's plot, y is not a function of x
in any case you can solve for y in terms of x with algebra
move the radical to one side, square, solve the polynomial equation
@arctictern by symmetry its also no function for y right?
so no function
x^4+2x^2y^2-xy+y^4 = 0 then
well, "y is not a function of x" does not imply "x is not a function of y," but the graph in this case indicates neither x,y is a function of the other
23:37
Why ?
@arctictern by symmetry
since x and y are interchangeable
y is not a function of x iff x is not a function of y
What do we call this now ?
relation
since its a subset of RxR
@DHMO this statement is false for relations. for instance the graph of x=constant has y not a function of x, but x is a function of y.
Thanks.
23:38
@arctictern x and y are interchangeable here. he said "by symmetry"
now, if you or saturated meant symmetry in the original equation...
yes
Heya tern!
its all cool guys
Ted is here
23:39
Hello @TedShifrin Long time no see :D
Not cool :)
lets throw a party, i order 1 beer
Hi @Mahmoud.
If the beer is for me, I'll take a martini :)
how to show that $\Bbb R$ only has 2 ideals?
it's a field
23:40
It's a field
its the preciousest of rings!
how to show that a field only has two ideals?
suppose I is an ideal that contants a nonzero element. then think about this.
You show that in a ring $R$ if and ideal $I$ contains a unit then $I=R$
I'll chime in: It's a field. ... Precious, but boring as a ring!
23:41
And you get the fact about fields as a consequences
Good evening @Ted!
Hi @Alessandro
I have a long way to go in order to understand this.
@Alessandro how would i show that?
@Mahmoud what?
Definition.
@DHMO review the definition of ideal and unit
23:43
ideal is absorber right
Can't study math without truly knowing the definitions!
Start with the easy case where you have the identity in the ideal instead of a generic unit
it absorbs ambient multiplication, yes
unit is exists multiplicative inverse
That should show you how to prove the general case too
23:43
@TedShifrin or at the very very least having them somewhere ready?
@Alessandro oh! thanks
No, KNOW.
that is a bold statement, i follow
@Alessandro, ages ago we had an unresolved question, but I don't recollect.
is it true that every nonzero element in a field is a generator?
23:46
what do you mean by generator?
Hi @Ted
What do you think?
i suspect yes
is an element in R^2 of the form (a,b)?
if you mean Fx={fx: f in field}=F for all nonzero x in F, then yes
23:46
Good night, Balarka!
nice
I'm full of unresolved questions @Ted, do you remember the topic? I remember discussing minimal generators of $\sigma$-algebras with you but maybe it wasn't that?
@Ted: Back home, finally?
Oh yeah ... what did you decide, @Alessandro?
@Balarka: No, driving all day tomorrow.
23:48
1 in L -> forall x in R x•1 in L -> R=L
That I can't really understand the paper answering my question right now, but it's on my "list of things I should read"
u in L -> u/u in L -> 1 in L -> R=L
Ah, and our next query? :)
Better to write $u^{-1}$ for $u\ne 0$.
what would a field look like if the additve inverse doesn't exist?
@TedShifrin i like to abuse notations
a division semiring
23:50
i see
Your teacher doesn't like it.
I want to prove that $|\bar{F}:F|=\infty$, where $\bar{F}$ is the algebraic closure of $F$, implies that the degree of the irreducible polynomial in $F[x]$ is unbounded (Mike says it's true) but I'm not sure where to begin @Ted
i am my teacher
I can't think of any examples that are not subsemirings of [0,infinity) though
@arctictern then use ordinals?
23:52
dunno how you intend to use them
never mind, just use the complex
C\0 right
in C you get rotation around 0, so you'll end up with additive inverses just using multiplication
C-{0} is not a semiring, they need 0
what? no way
However I should really go to sleep now since I have lectures early in the morning, goodnight @Ted and everybody
in any case, C\{0} is not even closed under addition, which is also a requirement
23:53
you can't get 0 from non-zero
1+(-1)
@DHMO 2 + (-2)
@Alessandro, I don't know what you mean by "the irred poly".
in that case
Night!
23:54
what would be an example?
Hi @Danu.
of a division semiring that is not a subsemiring of [0,infinity)? I can't think of any. But if you want less interesting examples, just intersect [0,infinity) with any subfield of R.
Q>0?
why do we need 0?
definition
23:55
??
oh, identity
@TedShifrin Is there any shortcut to computing Todd classes? Even doing the first one was already a pain.
you have the ability to look up "semiring" on wikipedia.
might as well look up the def of things if you're going to talk about them
maybe
Sadface.
23:56
Formal power series computation.
Yeah, that's what I did.
what would a field with 3 operations look like?
But look at Hirzebruch for all such matters.
i know exponent is not commutative
@TedShifrin Do you think it might be a good idea to read that book at some point? It's referred to all the time...
23:57
so basically (R,*,^) is a ring right
In particular, nobody seems to be willing to prove HRR :D
I meant irreducible polynomials by the way, "the" was not the right article there
@DHMO that question is ill-defined. without saying anything about the third operation, it's just a field with a completely random and unrelated third operation on the field's underlying set.
Look at it ... don't read it thoroughly.
@arctictern vague questions make for creative ideas
but anyway
3 abelian groups?
23:58
Alright.
like (R,+) is an abelian group
(R,*) is an abelian group
Oh @Alessandro; I'll ponder that on my 8+ hour drive tomorrow.
uselessly vague does not make for creative ideas, they make for frustration in getting someone to add more definition to the question
In other news, I realized that I don't really know how to induce connections on $\bigwedge^k E$. Can you give me a hint? I was thinking perhaps I can just antisymmetrize the definition for the tensor product somehow?
(R,^) is an abelian group
23:59
(R,^) is not a group
and then we can explore relations between the three operators
look, ^ is another operator
R is not the reals

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