12:00 AM
Yeah, but that's not very elaborate.
Do keep in mind that when you're going from the subsets of $\{n_1,n_2\}$ to the subsets of $\{n_1,n_2,n_3\}$, the thing you need to add (or not add) is $n_3$, not $n_2$. In case that wasn't clear.

you could say that to obtain the subsets of {n1,n2,n3} that don't contain n3, that's like finding the possible combinations of {n1,n2} - you're removing n3 from the list of possible elements for subsets. so you're going to the subsets of {n1,n2}

@heather Yep, indeed

and then, you add n3 to each subset to get the subsets that contain n3, I'm guessing, and then add the two, and you get the subsets of {n1,n2,n3}.

Yep, exactly
Again, how does this affect the number of subsets?

@heather don't worry
I just spent 30 minutes on a thermo problem
I used the wrong ideal gas constant

12:04 AM
it doubles them

@heather yep.
Is it clear why adding each additional element (e.g. $n_4$ would be next) doubles the number of subsets?

yes
so $n$, the number of subsets of $\{x_1,x_2,x_3,...,x_n\}$ equals the number of subsets of $\{x_1,x_2,x_3,...,x_{n-1}\}$ times two?

@DavidZ I don't think that's really a proof

@heather yep
@0celo7 Sure it is. Proof by induction.

@DavidZ For that you actually need to prove by induction...
You seem to have done it in a special case, then said it's "clear" for the others

12:06 AM
@0celo7 Yes, that is what we're doing.

and the number of subsets of {} is 1, so we do that times two $n$ times to get to n=2, for example.

@0celo7 Oh, I see what you mean. I meant that this technique forms the basis of a valid proof. But I'm not going to make heather come up with the full, formal proof that way.
@heather Yeah, exactly

or really, $2^n$?

@heather Yep

huh, I wonder how hard it would be to come up with a formal proof from here.

12:09 AM
Here is the way one proves it normally. Let $X_n=\{1,\dotsc,n\}$. Then there is a bijection $\mathcal P(X_n)\to \{0,1\}^{X_n}$ given by $A\mapsto \chi_A$ (characteristic functions). For brevity, write $2=\{0,1\}$. Then $2^{X_n}=\prod_{i=1}^n \{0,1\}$. But as each one of these has cardinality two, we have $|2^{X_n}|=2^n$. As $\mathcal P(X_n)\cong 2^{X_n}$, $\mathcal P(X_n)$ has the same cardinality, namely $2^n$.

um.

Here $\Pi$ is a Cartesian product.

you lost me at bijection.

@heather What @0celo7 posted is the non-inductive version of what I just walked you through.

@DavidZ You need induction to compute the cardinality of $\prod_{i=1}^n\{0,1\}$.
But for that set, it's clear that adding one more factor just doubles what you have.

12:13 AM
@0celo7 Eh, whatever, induction might be "hidden" in there somewhere, but I still classify the proof as non-inductive

@heather Then you need to read more of Halmos before measure theory

i've heard of bijection - i just don't know what it means. =( I'll keep reading Halmos.

Don't you have Munkres?

i do.
i've been trying to finish my other math books.

which ones?

12:23 AM
halmos, a copy of an abstract algebra book by durbin i got ahold of, calculus vol II by apostol.
and one or two chapters in calculus vol I by apostol i need to work on too.
i'm very slow.

if you're reading that stuff at your age, I think you're fine

hmm, I just came across a fact about complementation that I don't quite understand:
$A\subset B$ iff $B'\subset A'$
where it is assumed that all sets mentioned are subsets of one set $E$
and that all complements are formed relative to that $E$

Yeah, prove it

so if A is a subset of B, and the set B' which is such that B+B'=E, is the subset of A' which is such that A+A'=E - I just don't see that making sense.
@0celo7 okay.

The important thing is tha $A$ and $B$ are switched

12:30 AM
yeah, that's what I don't get.

If $A$ is contained in $B$, then the points not in $B$ are certainly not in $A$
The converse is similar
@heather draw a venn diagram

i think i get it
the venn diagram helped
okay, so A+A'=E by the definition of complement, and same with B+B'=E. and then we say that the elements of B' aren't in B, by def of complement, and if they are in A' because B' is a subset of A', then they must not be in A either, because the elements of A' aren't in A by the definition of complement. and then because A' and B' share elements, and A and A' add up to make E, and B and B' add up to make E, then A and B must also share elements.
no, I don't get it, because I can't get my mind to understand that $A\subset B$

No, that is not how we prove things
Proposition. $A\subset B$ iff $B'\subset A'$.

would you do a two column proof? like in geometry?
you know, I'm just going to look at your proof.
(from earlier)

Proof. $(\Rightarrow)$. Let $x\in B'$. Then $x$ is not in $B$. Consequently, $x$ is not in $A$, for if it were, then it would be in $B$, as $A\subset B$. We conclude $x\in A'$.

12:37 AM
so "let" is like the given
no, it isn't though.
You use it to set up some situation.
that x is in B'.

$(\Leftarrow)$. Let $x\in A$. Then $x$ is not in $B'$, for if it were, then $x\in A'$, as $A'\subset B'$. Since $x\notin B'$, $x\in B$. $\Box$
That is how you write the proof.

what do the arrows at the beginning mean?

You need to prove two directions, and I am indicating which direction I am proving.

prove...in directions?
is that like proving that it is true, and that the converse isn't?

@DavidZ halp

12:40 AM
i thought geometry was supposed to teach you how to do proofs ::glares at highschool math curricula makers everywhere::

no, those are not actual proofs

oh.
well then.
how about I try to do one with (A')' = A?
Proposition. $(A')'=A$

@ACuriousMind Bit late, but thanks a ton for that paper. It's really been helpful.
I'm surprised papers exist merely to explain things intuitively (at least, that's what your paper seems to be)

@ACuriousMind you're good at logic, can you please explain iff to heather

12:48 AM
Proof. ($\Rightarrow$). Let $A\subset E$ and $A'\subset E$ and complements of each other. $A'$ does not equal $A$ by definition of a complement. If we ask for the complement of $A'$, or $(A')'$, we are asking for the set of all elements $x$ such that $x\notin A'$ and $x\in E$. $A$ by definition is the set of all elements $x$ such that $x\notin A'$ and $x\in E$. This is the same as the definition for $(A')'$, therefore $(A')'=A$.
@0celo7, like that?

@heather Oh, if by that you mean what you think I mean that's not at all how "real" proofs work; it's a total failure of math education

you could argue that math education is a total failure, but okay @ACuriousMind =)

↑ boy, are mathematicians messed up

@heather For a proof like that you'd write $(\subset)$ and $(\supset)$

12:49 AM
@EmilioPisanty topologist's sine?

@ACuriousMind that's called a Warsaw circle.

@dmckee Oh, remember when you explained protostellar contraction and stuff?

a topologist's sine, closed with an arc
which of course would give you a delightfully pathological circle

@heather In order to see what 0celo7 means by directions, we should examine what the statement "$A\subset B$ iff $B^c\subset A^c$" really means.

via this question:
7

For a continuous scalar field on a circle, there is a diameter of the circle such that the endpoints of the diameter have the same value. If you think of the scalar field as "temperature", then what this says is that there are points on opposite sides of the circle that are the same temperature: ...

12:51 AM
@ACuriousMind okay.

@heather How would you explain to me what "iff" means?

@SirCumference Vaguely.

@ACuriousMind if and only if...so like, only if something is true. i never got why you needed the extra "and only if" - if we're super sure this is true, why were we using the if when we weren't sure.

@dmckee Er, well, this is an unrelated question, but you're the only person I've seen here who described something in stellar physics
Do you know how the mass-loss rate $\dot{M}$ of the Sun changes with its age?

I like how these steam tables go up to 2000 degrees
how is that even possible
@SirCumference presumably it depends on the mass of the sun
and its temperature, etc
so that's kind of clear

12:55 AM
@heather they're logically distinct things.

@heather Ah, 'tis as I suspected. The difference between 'if' and 'if and only if' is crucial and not neglegible. Consider the following statements: "The ground is wet if it rains" and "The ground is wet if and only if it rains".

the first: if it rains, but it can be wet if it doesn't rain?

@0celo7 Wood et al. just says $\dot{M} \propto t^{-2.00 \pm 0.52}$

and the second: if it rains and ONLY IF it rains it is wet - never wet unless it rains?

"A happens if B happens" is different to "A happens if and only if B happens". The first one leaves open the possibility that A happens even if B doesn't.

12:56 AM
@SirCumference Main sequence star grow ever so slightly brighter across their hydrogen burning life. That's a pretty minor change due to mass conversion. But when they go red giant they start shedding mass. Not much for small stars, a non-trivial bit for stellar mass stars and a lot for stars big enough to be headed for a supernova.

@heather Exactly - the first statement just says that upon the observation of rain we may conclude the ground is wet. The second claims that additionally whenever the ground is wet, we may conclude it rains.

@0celo7 looks like ACM has it covered

yes, @ACuriousMind thanks

In fact, we don't need take 'iff' as its own primitive operation - instead of "A iff B" you can just say "(If A, then B) and (if B, then A)".

@dmckee Main sequence stars do lose mass, though. I'm interested in how quickly. I've checked out Reimers' law for post-main sequence stars, but I'd like to know for main sequence stars.

12:58 AM
Well, it sound like you're more up on the subject than I am.

so iff is like saying that if and the reverse if is true? by which i mean, if you say a iff b, it's really saying a if b, and b if a?

yes

Howdy @obe

hi

And in that formulation, you can see the "directions": Saying "A iff B" is equivalent to two separate 'if'-statements, so in order to prove "A iff B" you can first prove that "If A, then B" and then prove that "If B, then A". Altogether, you have then shown that "A iff B".