Mathematics - 2016-09-30 (page 2 of 4)

mercio

10:00 AM

well $15+17=32$ I think

so I got one

so $2^{8n+5}$

Lozansky

correct :)

mercio

I'm not sure how that's funny though

Balarka Sen

How do you prove there's nothing else however?

mercio

$2^4 = -1$ and $2^8 = 1$ so the remainders are $8$ periodic

Balarka Sen

Ah, good point.

Lozansky

10:03 AM

i have one that is maybe a bit more interesting

In ancient Byzantium, a game similar to basketball was played that gave $a$ points for a field goal and $b$ points for a free throw. Surviving manuscripts do not record the values of $a$ and $b$. However, it is known that in this game exactly $14$ natural numbers were impossible point totals and that one of the impossible scores was $22$. Determine the number of points given for field goals and free throws in Byzantine basketball. Assume that $a>b$.

mercio

canyouwriteitinmaths

Balarka Sen

I think I am done with number theory for now though

mercio

I wish it was easier to visualize $\Bbb P^n(\Bbb C)$

because it seems beautiful

but we can't see it

;w;

Balarka Sen

I assume you know the cell decomposition?

mercio

a point, then a $2$cell then a $4$cell etc ?

Balarka Sen

10:08 AM

Right.

Which brings the question - how do you want to visualize it? I am quite satisfied with the pictures I have.

mercio

I want to see how a degree $n$ equation defines a curve that loops $n$ times

for example

Balarka Sen

Make it intersect with a generic line? You'd get n points in the intersection (by Bezout's theorem, if you wish), which would tell you it wraps n times, yeah?

Also a degree n equation gives you a hypersurface, not a curve.

But yeah, I get you.

mercio

I was thinking of complex projective plane first

let's not be too hasty

Balarka Sen

Fair enough.

mercio

like, what's the fundamental group of projective plane with a degree $n$ curve removed

Balarka Sen

10:12 AM

Oh, that'd be much more complicated.

fundamental group of quasiprojective varieties are not well understood. but I think it's doable in dimension 2, hmm.

mercio

what's the fundamental group of the space of smooth degree $n$ curves

Lozansky

in wolfram alpha

Balarka Sen

I do believe the fundamental group depends on your equation of the curve. Just the homology class is not sufficient, I do not think

Lozansky

how do you plot in the complex plane?

mercio

what do you mean by that @Lozansky

also did you know that when you rotate a line of slope $i$ by a real angle it stays of slope $i$

Lozansky

10:16 AM

say i have a function $z(t) = a+t(bi+d)$ and i want to plot it for some interval

if i just plug it in, wolframalpha plots the real part and the imaginary part separetely

mercio

isn't that just a segment

Lozansky

yes

mercio

anyway I wouldn't know I never use wolframalpha for plotting

and I don't have mathematica

Balarka Sen

@mercio I am forgetting what the space of degree $d$ hypersurfaces in $\Bbb P^n$ is. A degree $d$ hypersurface is given by $\sum a_{i_0 \cdots i_n} X_0^{i_0} \cdots X_n^{i_n} = 0$ with $i_0 + \cdots + i_n = d$, yeah? So it's $\Bbb P^{\binom{n+d}{d}}$, I think.

Now you want smooth curves in there. That's harder.

mercio

yeah you have to remove the hypersurface corresponding to vanishing of the discriminant or something

Balarka Sen

10:20 AM

Yes, remove the hypersurface given by det of Jacobian = 0.

These are interesting questions I don't know the answers to.

mercio

I'm not expecting you to know them o.o

Lozansky

if i want to integrate $\int_{2}^{i} \frac{dz}{1-z^2}$ for the path where $z$ goes from $2$ to $i$, can't i just evaluate $\frac{1}{2}(log(z+1)-log(1-z) |_{2}^{i}$?

mercio

be careful with complex logarithms

Balarka Sen

@mercio Recently I heard the following question. Suppose you have a complex abelian surface, aka $E_a \times E_b$ where $E_a, E_b$ are elliptic curves. Remove a finite family of elliptic curves from $E_a \times E_b$ which are of the form $mE_a + nE_b$. What's the fundamental group?

I think this is open or something.

People don't know much about fundamental group of quasiprojective varieties, as opposed to projective varieties.

Lozansky

well i would use $ log(z) = Log|z| + iArg(z)$

mercio

10:25 AM

would you be able to evaluate your thing to compute the integral for any path from $2$ to $i$ ?

@BalarkaSen that looks like a stupid hard question

but beautiful ;w;

Lozansky

its independent of the path?

mercio

WRONG

Lozansky

damn it

its not continous at z = 1

mercio

so you will have to explain why whatever expression for $log$ you took will work for the actual path

Lozansky

so we avoid that point

or well $z = \pm 1$

mercio

10:28 AM

if I write a giant scribble from $2$ to $i$ while somehow avoiding $1$ and $-1$, I don't always get the same result for the integral

Balarka Sen

@mercio yup

Lozansky

hmm yeah we cant allow $z+1$ or $1-z$ to be real negative either

mercio

well

how do you even start at $2$ then ??

Lozansky

good question

:D

mercio

you just need to find some expressions for those logs that work on the path

Lozansky

10:32 AM

hmm do you mean finding a suitable branch?

mercio

yes

Balarka Sen

@BalarkaSen shoulda been $\binom{n+d}{d} - 1$ now that I look back, but w/e

Lozansky

can you do that just from noting that $-1<z<1$?

then the argument is between $-\pi$ and $0$

ugh this is confusing

mercio

did you just write $-1 < z < 1$

Lozansky

yes

mercio

10:37 AM

what does it mean ?

Lozansky

well that would be a line segment on the x-axis

mercio

I thought your integral was from $2$ to $i$

Lozansky

yes

mercio

which is quite different form that segment on the real axis

you can give a branch of $log$ by telling where we should pick the arguments yes

so you only need to show a way to pick arguments for $z+1$ and $1-z$

when $z$ moves along the segment from $2$ to $i$

in fact you will only use the difference in the arguments

Lozansky

so the value of $z+1$ goes from $3$ to $ i+1$ and the value of $1-z$ goes from $-1$ to $1-i$?

mercio

10:43 AM

yes

so you could solve both logs by saying that you will pick the arguments in ... $(-5\pi/4 ; +3\pi/4)$

for example

or you could draw those segments and measure the angle they make with $0$

I mean the angle at $0$ of the triangles

Lozansky

how did you find that interval?

mercio

I drew a line from $0$ that didn't cross the two segments

do you prefer a spiral instead ?

Also you can choose different branches for each log

like the principal log for the first segment, and $i\pi + \log(-z)$ for the second segment

you don't even have to be crafty

in fact for whatever path you have, at worst you can just split it up in different pieces so that on each piece there is one of those two that work

Lozansky

i dont understand what you mean by drawing a line from $\mathbb{0}$ that doesn't cross the two line segments

mercio

you can never define a logarithm function on a set that loops around $0$

so the so-called principal branch of the logarithm is defined on all of C except the negative reals

that works because if you do a loop around $0$ you will cross the negative reals at one point

Lozansky

yeah

mercio

10:51 AM

and usually we want to get definitions that work for the largest possible subsets of C

so here, we need one that works for the two segment

and if I choose to remove a halfline that doesn't cross the segments, I will be able to define a log on all of C except that halfline and which will be helpful for your integral

so i chose to remove the diagonal that goes in the second quadrant

and then i chose a way to compute arguments continuously

Lozansky

the diagonal?

mercio

in order to define that log

the points whose arguments are in $3\pi/4$ mod $2pi$

that's the line $(0 ; -1+i)$

then for every point in $C$ that is NOT on this line, they have an argument that fall in the interval $(-5\pi/4 ; 3\pi/4)$

Lozansky

well how did you get -1+i?

mercio

I chose it so that the line $(0;-1+i)$ didn't intersect the segments

the halfline*

I could have picked $-7+12i$

Lozansky

yeah

mercio

10:56 AM

or $15-3i$

Lozansky

it just had to be in the second quadrant?

mercio

no

ith as to not intersect the two segments

or $2+19i$

Lozansky

yeah okay but any (half)line in the second quadrant satisfies that, right?

mercio

yes except the horizontal line

Lozansky

mhm

does the interval have to be of length $2\pi$?

mercio

11:00 AM

there is no reason why it shouldn't

as I said you usually want to define arg and log on biggest regions possible

but noone is stopping you for leaving them undefined on a larger chunk of the plane

Lozansky

okay so now when I calcuate the logs, i choose the argument that lies in that interval?

mercio

now you define log with that argument

then its derivative is $1/z$ where it's defined, and the segments are included in the domain of log, so you can use it to find an antiderivative of your function on the segment between $2$ and $i$

Lozansky

11:18 AM

i got $-\frac{1}{2}(Log3+i \pi)$

does it make sense that it's negative both real and imaginary

Balarka Sen

Hi @Anubhav

Anubhav

Hii

Sylent Nyte

Hey guys, I need some help

Anubhav

11:34 AM

@BalarkaSen I read your one comment, where you asked whether any sphere bundle can be realized as a unit bundle of some v.b?

So did you get any ans for that?

@SylentNyte say it, we'll try to help you

Sylent Nyte

I have a question, a pattern continues to infinity and the pattern is in rows; 1, 2 3, 4 5 6, so triangle numbers, I need to find where the number 999 is, how do I do this?

I used the equation n(n+1)/2 = 999 so therefore n^2 + n - 1989 = 0

But I'm not sure how I find what row the number is in using this

Anubhav

find the value of n first

Sylent Nyte

How do I do that without a calcukator

Anubhav

integer value [n] will be the answer

Do you know how to solve a quadratic equation

replace n by $x$ and find the solution for it

Sylent Nyte

I don't know how I'll find a solution

Anubhav

11:39 AM

do you know how to solve a quadratic equation?

Sylent Nyte

If I replace n with x I get x^2 + x - 1989 = 0

Balarka Sen

@Anubhav Yes, it's not true.

Anubhav

Why?

@BalarkaSen

Balarka Sen

Essentially equivalent to asking that Diff(S^n) is homotopy equivalent to O(n+1).

Anubhav

How??

I didn't get

Balarka Sen

11:40 AM

That's more or less obvious. Look at the transition maps.

Transition map of a linear sphere bundle comes from O(n+1).

Anubhav

yes that is obvious...

wait let me think for a sec

How come diff(S^n) is comming into the picture?

Balarka Sen

That's the range of the transition maps of a sphere bundle, innit?

Lozansky

In the integral $\int_{2}^{i} \frac{dz}{1-z^2}$ where the path is the semicircle $|z-1|=1, Im z \leq 0$ and the imaginary axis

Can I choose the branch $(-5\pi/4, 3\pi /4)$?

Anubhav

Sphere bundle is what? essentially fibers are sphere ?

Balarka Sen

Yes. The transition maps are (U_i \cap U_j) x S^n ---> (U_i \cap U_j) x S^n, homeomorphisms on the second coordinate.

Or diffeomorphisms, w/e

Anubhav

11:45 AM

Yes.

On the other hand for unit vector bundle we have $O(n+1)$

Balarka Sen

Right.

Hmm, now that I think about it there's a slight work involved to show that this is equivalent to O(n+1) and Diff(S^n) being homotopy equivalent.

I mean, transition functions are just associations of the overlaps of the charts to elements of O(n+1) and Diff(S^n) respectively. That isn't sufficient to prove anything.

Anubhav

I agree

Balarka Sen

I think what we need here is that the respectively bundles are classified by maps to BO(n+1) and BDiff(S^n), the classifying spaces.

Anubhav

ooppsss... I have no idea about this classifying bundles

Isn't there any easy way?

Balarka Sen

I don't think so. But what I said about classifying spaces is the formalization of the previous idea, I believe.

Anubhav

11:50 AM

What is classified bundle?

I saw this in Hatcher, but didn't read

tell me some reference

Balarka Sen

Classifying spaces is in Hatcher. But I am not sure if they speak of their relation with fiber bundles.

I am not actually entirely sure of sphere bundles being classified by BDiff(S^n) part either, to be honest.

The BO(n+1) part is true and not hard.

Anubhav

Ok. I'll think and discuss this problem with you in the night

I've to go now

Balarka Sen

If one can show this, we'd end up with a 1-1 correspondence [X, BO(n+1)] <---> [X, BDiff(S^n)] for every X, and then Yoneda lemma tells O(n+1) is homotopy eq to Diff(S^n)

And that was Mike's statement.

@Anubhav Bye.

Danu

12:35 PM

Hi @Balarka

Balarka Sen

Heya

Danu

@BalarkaSen That sounds real nice.

Balarka, maybe you wanna have a small look at something for me and see if you agree

(Huybrechts, again...)

Balarka Sen

If I can help, I certainly will.

Danu

11 hours ago, by Danu

@TedShifrin So what I said about the proof: (i) he calculates the cocycle of $\det \mathcal T_{\hat X}$, not of $K_{\hat X}$. (ii) His rule for getting the cocycle from a divisor is upside down---this is confirmed by the corollary he refers to. Also, once you flipped the cocycle to get that of $K_{\hat X}$, one had better find an excuse to flip the other one too, right? :) Otherwise they won't match.

It's just checking something I'm pretty certain about

This ^^ About proof of 2.5.5

Forget about the first part of the proof

Start at "this is a local calculation"

Also, if you don't feel like checking the proof then maybe look at this line of reasoning instead:

Balarka Sen

I never muddled with canonical bundles over blowups, but I'll have a look.

Danu

12:38 PM

12 hours ago, by Danu

Oh, never mind! I know what happened---but it seems like Huybrechts makes a rather "big" mistake in his proof.

@BalarkaSen You don't need to know anything about it.

I explain the whole line of reasoning there ^^

Balarka Sen

Alright.

Danu

You don't need the details of the proof, mostly

Just the general idea, with some specifc details

Balarka Sen

12:51 PM

@Danu Yeah, the choice of the cocycle is odd. I think your explanation makes sense.

Danu

@BalarkaSen To me, the one with the canonical one was really clear

(but in fact I disovered the problem with $\mathcal O(E)$ first)

I'll explain in my email to him... I've got like 30 typos

Just from the first 2 chapters

Balarka Sen

yikes

Danu

Okay, 25 now actually

But yeah

Not too precise...

Balarka Sen

1:14 PM

OK, gotta go

Martin Sleziak

This sounds as if Yoda joined math.SE: I'm looking for a function that surjective but not injective is and this fulfills. (From this question. It is rather unclear what the OP wants, so I guess it's going to be closed and deleted sonn.)

Danu

@MartinSleziak Lol

Martin Sleziak

Just to make clear, I just wanted to share the unintentional joke the OP made. I am not calling for close or delete votes. (It already has some downvotes. And closing it as unclear would certainly be deserved at the current state. But it might still be good to give OP the chance to clarify the question.)

user228700

1:33 PM

Hello everyone :-)

user228700

I've got a bit of a homework-tsy question regarding logarithms. Is anybody interested to help me clear my doubt..?

user228700

Can anybody tell me if there is any flaw w/ the following set of formulae/rules:(because I'm running into trouble while applying these for problems)

user228700

user228700

I generalized all the above equations as:

user228700

$log_b(a)≥\alpha$ →
$a≥b^{\alpha}$, $b>1$ AND
$0<a≤b^{\alpha}, $0<b<1$

user228700

1:47 PM

And

user228700

$log_b(a)≤\alpha$ →
$a≥b^{\alpha}$, $0<b<1$ AND
$0<a≤b^{\alpha}, $b>1$

user228700

This seems correct and all, but it doesn't work out very well when solving problems like:

user228700

(I forgot to mention that $a$, $b$ and $\alpha$ may be any real numbers or any real-valued function)

user228700

Anybody..? No? OK :-(

Martin Sleziak

2:03 PM

@KaumudiHarikumar Can you say for which $t$ is $\log_{1/5} t\ge 0$?

Is it $t\ge1$ or $t\le1$?

If you can answer this, all that remains is to solve either $2x^2+7x+7\ge1$ or $2x^2+7x+7\ge1$. (Depending on the answer to the previous question.)

Similarly for the second one.

0celo7

$E\subset\Bbb R^n$ linear subspace, $\Bbb R^n-E$ connected $\Leftrightarrow\dim E\le n-2$.

*Proof.* ($\Leftarrow$) We will show $\Bbb R^n-E$ is in fact path connected. It is sufficient to show this for $\dim E=n-2$, for any vector subspace of dimension $n-3$ or less can be realized as a subspace of an $n-2$ dimensional subspace. Let $p,q\in \Bbb R^n-E$. Let $L\subset\Bbb R^n$ be the straight line from $p$ to $q$. (We also use the notation $L=\overline{pq}$.) We set up coordinates $x^1,\dotsc, x^n$ with the origin on $E$ and with $E$ spanning the first $n-2$ coordinates, that is, $E$ is the…

@BalarkaSen In case you wanted to see the final product of the connectedness proof ^^

My prof said it was "much better than what he had in mind" (i.e. the hint he gave).

arctic tern

If E is a subspace of an inner product space V, and its orthogonal complement is C, then orthogonal projection can be used for a deformation retraction V\E -> C\0.

0celo7

@arctictern This is for an analysis, not topology, class.

arctic tern

still

0celo7

I don't know what a deformation retraction is.

arctic tern

2:13 PM

given v,w in V, we can connect them to v',w' in C\0 by a straight line. then it suffices to show C\0 is connected.

Danu

nice

That's really great

Such efficiency

@arctictern I guess you don't even need to use explicitly that it's a defo retr, to argue that it doesn't change the issue of connectedness.

user228700

@MartinSleziak Uh, $t$?

Martin Sleziak

2:30 PM

@KaumudiHarikumar $t$ is just a name of variable.

Like x,y,z,w,u.

So if you prefer, when is $\log_{1/5} x \ge 0$? For which values of $x$?

I still think that it is nicer to write t, since then we can write $t=2x^2+7x+7$. Writing $x=2x^2+7x+7$ would be something completely different.)

user228700

@MartinSleziak Well, since the base, 1/5 is positive, this is true for any $x>0$

Martin Sleziak

I don't think that is correct.

What is $\log_{1/5} 1/5$? What is $\log_{1/5} 5$?

user228700

1 and -1

Martin Sleziak

Great!

So you see that $\log_{1/5}x$ is not always positive.

So I will ask again, and since I prefer this notation, I will denote variable by $t$.

For which values of $t$ we have $\log_{1/5} t\ge 0$?

You already saw that $1/5$ is such value, but $5$ is not such value.

user228700

@MartinSleziak Right, yes. If the base and the number are on the same side of unity, the logarithm will be positive. If on the opposite side, it will be negative, correct?

Martin Sleziak

2:36 PM

So $\log_{1/5} t\ge 0$ if and only if $0<t\le 1$.

Here $t$ denotes a real number?

Do we both agree on this?

user228700

@MartinSleziak Yes, we do.

Martin Sleziak

And your original problem was to solve $\log_{1/5}(2x^2+7x+7)\ge0$.

Which (because of we just said), is the same as solving $0<2x^2+7x+7\le 1$.

user228700

@MartinSleziak Yes...

user228700

@MartinSleziak OK, I get it.

Martin Sleziak

In fact, the part with $0<2x^2+7x+7$ is necessary for the expression in your inequality to be defined at all.

And the part $2x^2+7x+7\le 1$ is the part which comes from the conditions that the logarithm should be non-negative.

user228700

2:39 PM

Yes, yes, OK.

Martin Sleziak

Of course, this is how it works if the base of the logarithm is smaller than one (like 1/5 or 1/3).

user228700

Okay, I think I can proceed with the problem set now. Thanks so much sir/ma'am! :-)

Martin Sleziak

If the question is $\log_{5}(2x^2+7x+7)\ge0$, then it is precisely in the opposite way. I.e., $2x^2+7x+7\ge1$.

user228700

@MartinSleziak Yes, I got that. Thank you!

Martin Sleziak

But from your replies it seems that it's already clear how to go about similar problems.

@KaumudiHarikumar No problem. And good luck with your studying.

user228700

2:41 PM

@MartinSleziak Thanks! Have a nice day!

Mike Miller

@Balarka Your argument is correct, but the converse is also easy to see - if the map O(n+1) -> Diff(S^n) doesn't induce an iso on homotopy groups, you can construct bundles over spheres that either can't be given a linear structure or can be given multiple.

Danu

o/ Mike

Balarka Sen

3:26 PM

@0celo7 Good.

@Danu @arctictern You can prove more generally that chucking out a codimension 2 submanifold doesn't disconnect R^n - take a path between two points in R^n, make it transverse to the submanifold. That's disjoint.

Just sayin'

0celo7

@BalarkaSen Yeah.

Another problem on that problem set was to show $S^1$ is not homeomorphic to $S^n$ for $n\ge 2$. Fundamental group shows this immediately, but that's cheating.

I think transversality homotopy is similarly cheating.

Balarka Sen

Chuck out a point

But I guess you already know that

0celo7

@BalarkaSen Chuck out two points.

Balarka Sen

Ah, yes, thanks.

0celo7

You need to disconnect $S^1$

Then use stereographic projection

Tbh, connectedness proofs like that are very unsatisfying.

Balarka Sen

3:33 PM

Why?

0celo7

@BalarkaSen I'll have to think about why exactly. brb

Danu

@BalarkaSen But that needs a lot of machinery

Still nice.

Balarka Sen

@MikeMiller Thanks for checking (I don't actually know why maps to BDiff(S^n) classify sphere bundles, but maybe I'll write down a proof). I just got back, I'll ponder on what you said in a minute.

@Danu Of course it does, we're proving it for codimension 2 submanifolds in general.

Lozansky

does anyone know a good lecture series for complex analysis?

introductory

Balarka Sen

Not just hyperplanes.

Danu

3:35 PM

Yeah, that's true

So that also works for other manifolds right

Balarka Sen

For sure.

Mike Miller

@BalarkaSen If G is a topological group that acts effectively on X, X-bundles with structure group G (transition functions are in G) are classified by maps into BG - you turn an X-bundle into a G-bundle and vice versa.

Danu

All pretty nice---Huybrechts uses this at some point in his book :P

Anubhav

hey @BalarkaSen did you get any answer?

Danu

So $\Omega^p_X$ and $\bigwedge^{p,0}X$ are identified as complex, but not as holomorphic vector bundles, is that right?

Balarka Sen

3:38 PM

@Anubhav Answer to what?

I did write down a proof, modulo the fact that maps to BDiff(S^n) classify sphere bundles.

Anubhav

that why homotopy equivalence of O(n+1) and diff(S^n) is enough for that question

can you show me the proof?

@BalarkaSen ?

Balarka Sen

I wrote it down when we discussed. Suppose every sphere bundle is a linear (i.e. appears as unit bundle of a sphere bundle). Then there is a 1-1 correspondence between [X, BO(n+1)] and [X, BDiff(S^n)]. Yoneda lemma then says BO(n+1) is homotopy eq. to BDiff(S^n). That in turn implies O(n+1) is homotopy eq to Diff(S^n).

Anubhav

Why is there a 1-1 correspondence?

Is it some kind of property of classifying bundles BG?

Mike Miller

@Anubhav If Y -> Z is a homotopy equivalence the map [X,Y] -> [X,Z] is a bijection for all X.

Just write down what it means to be a homotopy equivalence, more or less.

Anubhav

ohhh...sorry, I misread your comment

Yes, but is there any easy proof available?

Balarka Sen

3:43 PM

@MikeMiller Um, I'm using the 1-1 correspondence of sphere bundle with [X, B(stuff)] here though. I think you're thinking about the contrapositive you said.

Which is easier, because one doesn't have to invoke Yoneda, yeah.

@Anubhav I don't think one can avoid the classification of bundles by maps to BG.

Mike Miller

You don't need Yoneda. If that's a bijection for all X, it's a bijection for S^n, which by definition means it's a weak homotopy equivalence.

Balarka Sen

Ah, true.

Anubhav

hmm...

Mike Miller

You could simplify to showing that G-bundles over S^{n+1} are in natural bijection with pi_n G without developing classifying spaces.

Balarka Sen

@Danu What's the difference between complex and holomorphic vector bundles?

(I don't really know the two bundles you wrote down, so I can't help. Just asking out of curiosity)

Danu

3:52 PM

Trivializations being biholomorphic or just smooth

(both should restrict to $\Bbb C$-linear maps on the fibers

Balarka Sen

Oh, so your base manifold has a complex structure?

Danu

Yeah, everything on complex manifolds of course

Else there is no notion of holomorphic vector bundles

Balarka Sen

Got it.

Danu

Sorry, I thought it would be implicit in most things I talk about nowadays :P

Balarka Sen

Yeah, I should have figured.

@MikeMiller Thanks, that makes a lot of sense. I still need to learn principal bundles - my knowledge on that is inconsistent.

Anubhav

4:02 PM

bye...now I'll go and study...

Martin Sleziak

4:35 PM

In case somebody is interested:

in Calculus and analysis, 43 mins ago, by Saeid Nourian

I'm the author of a graphing software that is mostly used by calculus III professors and students. Would anyone here like to volunteer to try it out and give me feedback?

Mike Miller

Dang it, I have to learn infinity categories now.

Secret

5

Q: What happens if the image of a set is empty?

Can someone give me hint how to prove that for a function $f:X\rightarrow Y$ and $B\subseteq X$ we have $$f(B)=\emptyset \Rightarrow B=\emptyset \ ?$$ Is my idea that, supposing $x\in B\neq \emptyset$, we would have to have, since $f$ is a function, a $y$ such that $f(x)=y$ and therefore $y\in ...

elementary-set-theory

Question on an answer:

\begin{align}
& f[B] = \varnothing \\
\equiv & \;\;\;\;\;\text{"basic property of empty set"} \\
& \langle \forall y :: y \not\in f[B] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\cdot[\cdot]\;$"} \\
& \langle \forall y :: \lnot \langle \exists x : x \in B : f(x) = y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"simplify: apply DeMorgan to $\;\exists x\;$"} \\
& \langle \forall y :: \langle \forall x : x \in B : f(x) \not= y \rangle \rangle \\
\equiv & \;\;\;\;\;\text{"rearrange quantifications -- to prepare for one-point rule"} \\

Balarka Sen

@MikeMiller Oh no

rip

Secret

Are we implicitly assumed f[B] and B to be universal sets of the domain and range of f, therefore if all x and y are not in those sets, then they are automatically empty?

Mike Miller

arxiv.org/abs/1609.09132