5:02 AM
Hi chat! a mathematica and linear algebra query
not[idx_] := Complement[Range[NN], idx]

SC[A_, idx_] := A[[idx, idx]] - A[[idx, not@idx]] . Inverse[A[[not@idx, not@idx]]] . A[[not@idx, idx]]
I am not sure how to interpret the two lines of code
We have matrix $A$ and $SC$
but how are they getting SC from A using the code I dont understand

1 hour later…
6:27 AM
is there an example of convergence which can only be proven using nets but not by sequences?
What is it that nets can do but sequences can't?
In case of nets, we consider elements indexed by any directed set but in case of sequences, we consider elements indexed by a particular directed set -the naturals. So in this sense, nets are 'generalized' sequences.
But what's the point? Is it just for the sake of introducing a definition? Or there is actually a concrete example which demands the need of nets?

what do you mean, 'demands' the need of nets? it's a frame of reference that allows you to use sequence-flavored arguments in spaces where properties like compactness might not be captured by sequential convergence.
there's no 'need' to use them any more than there's a 'need' to use sequence-based arguments however. kind of like how you can do arguments in analysis on R in terms of sequential convergence alone, or in terms of open sets.
the answer to math.stackexchange.com/questions/152447/… lists some examples of spaces where compactness doesn't quite line up with 'sequential compactness,' in both directions.
"X is compact iff every sequence in X has a convergent subsequence" is false, but "X is compact iff every net in X has a convergent subnet" is true. so nets are what you need to do 'sequence-like' compactness arguments in general topological spaces.
makes no difference in something like a metric space.
but again, as in R^1, no need to use 'sequence-like' compactness arguments, you can also just work with open covers and other consequences of the definition.

I mean -" demands the usage of nets" because sequences were not adequate to conclude 'something' and that 'something' could be concluded only using nets.
@leslietownes I see.
@leslietownes I want to conjecture that in Hausdorff, both are the same. I have not yet proven this fact though.

6:50 AM
whats up party people

party? where?

oh I thought here :c

we are the net party people
set builder sounds like a lego kit

damn
never thought of it that way

2 hours later…
8:44 AM

9:27 AM
shouldn't the Jensen's inequality give: $\frac{a}{a+1}\int_0^1a^{(x-1)(a+1)x^a}\,\mathrm{d}x\\ \ge\frac{a}{a+1}a^{\int_0^1(x-1)\,(a+1)x^a\,\mathrm{d}x}$?

No, the measure is $(a+1)x^a\,\mathrm{d}x$
the space has to have measure $1$
we are not using that $\mathrm{d}x$ has measure $1$ on $[0,1]$

I see. I'm not aware of that yet :(. I thought $f(\int_0^1 g) \le \int_0^1 f \mathrm o g$, where f is convex.

Look at this

never mind, I'll get there soon as I've been learning measure theory also.
@robjohn I see. Clearer now. $\int_0^1 (a+1)x^a=1$

$f(x)=(a+1)x^a$ in the Wikipedia article
times the characteristic function of $[0,1]$

9:36 AM
I see. So using characteristic function, we can extend it to $-\infty$, $+\infty$ as integral lower/upper limits.
thanks Robjohn. I never used this version of Jensen's before. :)

yes, to fit it into the format of Wikipedia
I should change the link to Jensen to go directly to that part...

I think it would be better to instead write the inequality in the answer itself.
For, someone may object there is $\pm \infty$ as upper/lower limits in the wikipedia link.

Jensen is often written as $\varphi\left(\int_Xf(x)\,\mathrm{d}\mu\right)\le\int_X\varphi(f(x))\,\mathrm{d}\mu$ where $\int_X\mathrm{d}\mu=1$
@Koro: I have instead linked to a proof of Jensen that proves what I have above.

10:03 AM
:)

10:23 AM
You could write $\varphi(E[X]) \leq E[\varphi(X)]$ if you prefer the probabilistic notation.
@Koro Sequences are nowhere enough general in the context of topological spaces
For example, $x\in\overline{A}$ iff there is a net $x_i\in A$ convergent to $x$.
You can't replace "net" with "sequence" here
Even if we were to only consider nets using well-ordered sets instead of directed sets, that wouldn't be enough.

@Jakobian why not?
If $x\in$ closure (A), then x is either in A or a limit point of A.

because there are examples where it doesn't work

If x is in A, then I take a constant sequence $x_n=x$.
else, I think if convergence can be shown using nets, the same can be done using sequences.

Well it can't
$$\overline{[0, \omega_1)} = [0, \omega_1]$$ but there's no sequence in $[0, \omega_1)$ which would converge to $\omega_1$

10:38 AM
I don't know what $\omega_1$ is.

The first uncountable ordinal

Is there a simpler example?

Simpler? This one is simple

I'm afraid I don't know the 'uncountable ordinal'. :(
I'll take at them. Thanks.
cardinality is familiar to me but the ordinals are new.

ordinals are pretty much, isomorphism classes of well-orders

11:03 AM

12:04 PM
@PM2Ring That looks like my clock. No wonder things seem to take so long.

if $X$ is an uncountable set and $x\in X$, then equip $X$ with the cocountable topology and you have $\overline{X\setminus\{x\}}=X$, but no sequence in $X\setminus\{x\}$ converges to $x$ in $X$
also I've yet to come up with an open set-based rather than net-based argument that the total space of a fiber bundle with compact base and fiber is compact

12:29 PM
@robjohn :) I guess it's a bit reminiscent of The Clock of the Long Now

@Koro Ordinals are really fun. Basically, after every set of ordinals, you have another ordinal. They go: 0,1,2,3,4,5,…,ω,ω+1,ω+2,ω+3,…,
$\omega+\omega=\omega\cdot2,\omega2+1,\dots,\omega3,\dots,\omega^2,\dots,\omega^3,\dots,\omega^\omega$
(for technical reasons we write $\omega2$ rather than $2\omega$)
Unfortunately, there is no $\omega-1$ or $\sqrt\omega$. The fact that these don't exist is actually really useful, since this way the ordinals have the property that there's no infinite descending sequence of ordinals (equivalently, every set or class of ordinals has a minimum value)
(If you want $\omega-1$ and $\sqrt\omega$, you can look into something called the surreals)

@PM2Ring It will run for 10000 years, as long as people are there to wind it??

@robjohn Yes. It's about connecting people to a long concept of time. So it becomes meaningless if there aren't any people interacting with it.

I think it would be more cool to have aliens find the clock running, even if we have joined the dodos.
Or the next civilization after the next ice age.
@Koro: I've added a plot to show the bounds.
The lower bound is very close.

12:45 PM
I guess we could build an atomic powered clock. Some isotopes have huge half-life. OTOH, with a huge half-life the amount of energy released per second is pretty tiny, so you'd need a lot of fuel.
It's hard for humans to comprehend huge ranges of magnitude. I suppose we can sort of contemplate how many millimetres are in 1000 kilometres. But maybe it's a bit easier to contemplate the number of seconds in 30 years.

@PM2Ring almost a billion. Some people have that many dollars.
That would make a stack less than 68 miles high.
the dollar bill is 0.0043 inches thick.

I don't know how thick Australian bills are. They're made of a polymer, and they're pretty thin. We used to have 1 & 2 dollar bills, but we phased them out a few decades ago & replaced them with coins.

@PM2Ring Ooh, I wonder if a stack of one dollar coins would reach the Moon.
Eh, less than 1740 miles.
they are 2.80 mm thick

1:04 PM
Wiki says our notes range from 0.125 to 0.14 mm en.wikipedia.org/wiki/Banknotes_of_the_Australian_dollar
They have various interesting security features like holograms & microprinting. And they have tactile markers to make them easy for blind people to tell their value.
They don't fold nicely like the old paper bills did, but if you let them sit for a day or so under a book they go totally flat.
They stay a lot cleaner than paper money. But if they accidentally go through the clothes dryer, they shrink!

1:53 PM
@PM2Ring Ack! Actual shrinking money
@TedShifrin I'll start on that tomorrow.

2 hours later…
3:36 PM
Suppose that existence of R (set of reals) is given. How do I prove that a uncountable well ordered set W exists with the property that it has a maximum element $\omega_1$ with the property that for any t<$\omega_1$, the set {x in W: x<$\omega_1$} is countable.

you have a typo there at the end

4:25 PM
@Thorgott yes. I mean the set $\{x\in W: x\le t\}$ is countable.

4:46 PM
@robjohn Sorry. Too late!

@Koro take minimum of uncountable ordinals

5:04 PM
@TedShifrin story of my life...

Filed under: easier said than done.

@Jakobian existence of uncountable ordinals is not given.

5:26 PM
Some notes have Braille dots on them.

Are they for the blind?

Supposedly, I presume they are of help.

Do you know how to read Braille?

My linguistic abilities are limited.
At my peak I knew 100 simplified Mandarin characters, but that is long past
thankfully my offspring seem to be better, Hangul & Arabic seem to be among their awareness

5:42 PM
TIL Chosŏn'gŭl in North Korea

한국어surely?

I watched midnight mass tv show.

@copper.hat Don't know.

@Koro When I grew up we went to (Catholic) church every Sunday and religious holidays such as Christmas. On Christmas Eve you could meet your religious obligations by attending midnight mass, since technically the 'important' part of the mass was actually in the early minutes of Christmas itself. The point being that you did not have to attend mass in Christmas day.
Of course, my parents viewed this as cheating and sometimes insisted that we attend two masses on Christmas day, so we always viewed midnight mass with some degree of scepticism.
Also, many of the attendees came straight from drinking at the pub, so the mass could be a little livelier than usual.

6:09 PM
@copper.hat :).
I never went to church. Whenever I planned to go, it rained or anything else happened.

It was an opportunity to see girls :-)

haha

Our schools were single sex and the area was relatively sparsely populated.

@Koro is existence of uncountable sets given?

@Jakobian yes, suppose $\mathbb R$ is given.
The problem is: how should I choose $\omega_1$?

6:24 PM
@Koro then well-order an uncountable set and take an ordinal corresponding to its isomorphism class
Then uncountable ordinals exist

7:23 PM
this has nothing to do with $\mathbb{R}$
$2^{\aleph_0}$ is uncountable

Just by way of evidence that I am not a hopeless curmudgeon, this answer has some really interesting ideas (and links). The question itself is quite interesting, too.

7:39 PM
Why does this preclude your established curmudgeonosity?

@Thorgott what do you mean?

wait, cool graphics in an answer, immediate +1

I'm not sure which part of that message is ambiguous

7:55 PM
Speaking for Thor, I will concur that $2^\omega$ is far easier to construct and understand than $\Bbb R$. But, regardless, well-ordering is required here?

@copper.hat oh, then let me point you to some neat graphics... ;-)

:-)

Oh, well, yes.

8:27 PM
is it so bad if high school students need a calculator to perform -2(2)?
I think so

@TedShifrin for the sake of simplicity, yes
I think you can construct uncountable ordinals without choice even, but who really cares?

@Bob is that evaluating the function $-2$ at $x=2$?

no that is multiplying a positive number by a negative number
something that is often forgotten the year after it is taught

I do not understand your last comment copper.hat

8:29 PM
@TedShifrin Normally, I would suggest that every question ever be closed and deleted, that all of the users be suspended forever, and that y'all get offa mah lawn.
But look! I can like things, too!
@TedShifrin $\mathbb{R}$ is such a huge conceptual leap from the rational numbers (or even the algebraic numbers). It makes me crazy that freshman level calculus texts just take the real numbers for granted.
@Bob That positive eight, right?

@XanderHenderson lol

That's what my calculator says, anyway.

have a nice day
bye

Man, I drove @Bob away. :(

you did not
it is just I need to go
nice chatting
bye

8:35 PM
Also, I bought limes at the grocery today. I intended to drink those, mixed with tequila and homemade orange liqueur. But when I got home, I learned that I had less than an ounce of tequila in the cabinet. What kind of a-hole leaves less than a shot?!
(Hint: it was me.) :'(

@XanderHenderson Even for a Spivak course we assert the LUB axiom and don’t pretend to construct the reals. I’m fine with that, and more in a regular course.
@XanderHenderson the importance of a shopping list

@TedShifrin Sure, you have to get going in calculus, but that doesn't change the fact that there is a big leap from rational to real. Students get really, really hung up on imaginary numbers (probably from the name), but don't seem to understand how hard it is to get to the reals.
The LUB property is not (in my opinion) obvious. It is a reasonable starting place, but it requires a bit of a leap of faith.

It’s only hard if you’re a math pedant.
Most students I’ve taught have no problem with “no holes in the reals”

@TedShifrin I'm not saying that students "have a problem" with the reals.
I am asserting that we sweep a lot of the difficulty of the reals under the rug, and that they get the false impression that working with real numbers is a natural and obvious thing to do, but that imaginary numbers are somehow weird and difficult.

Well, if you do, who cares?

8:42 PM
It is an observation about the psychology of students.

Well, $i$ isn’t on the real number line.

The thing which is mathematically pretty simple causes students a lot of angst, while the thing that is actually quite difficult to come up with is breezed over with nary a comment.

It becomes formalism until we discuss the geometry/algebra of $\Bbb C$ … which may happen in engineering ODE but maybe not even there.

I feel like we are maybe talking past each other. My impression is that the difficulty of calculus (from a historical perspective) is coming up with some notion of the "real numbers". At some level, this is what Newton and Leibniz were groping towards, and what is finally laid to rest by Cauchy and Dedekind. The real numbers are a fairly significant development in the history of mathematics. Like, there is actually something really difficult there in the original development of the theory.
It only seems obvious and easy in retrospect. I surprises me that students seem to have no problem with this, but they get worked up about $\sqrt{-1}$.
This is a cultural or psychological observation, not a complaint.

$\sqrt{1+2+3+4+5+\cdots}$

8:57 PM
@robjohn No, that is $\frac{1}{2\sqrt{-3}}$. Duh.

@XanderHenderson that's definitely true. Another thing of this sort is the number pi, trigonometric functions and angles. I haven't really seen a formal introduction to these outside of complex analysis introduction.

Pi is actually a fun one.
There are subtleties there that we don't generally care about.

9:36 PM
I think the problem is the notation $\sqrt{-1}$.

@copper.hat Honestly, I think that the problem is that children are told from an early age that $\sqrt{2}$ is fine, but that $\sqrt{-1}$ doesn't exist. And then the nomenclature "imaginary numbers" is damaging (I try to use "complex numbers", until I can't).
My daughter had similar problems with negative numbers in first grade: "You CAN'T subtract $5$ from $3$, because $5$ is BIGGER than $3$!!!"

She is correct of course.

Lies.

Negative numbers are a better example of the issue, I think.

Indeed.

9:46 PM
So, that is where the problem lies in my mind.

If you have 3 bananas and I have -2 bananas, how many bananas do we have?

With rotations, it is clear that there is some $r$ such that $r^2 = -1$, but the leap needs to be shown.

@robjohn Too many.
Bananas suck.

Of course, I think of that as $(3,0)$ and $(0,2)$ bananas, so we have $(3,2)$
time for some wine or exercise, i think

is that equivalent to $(1,0)$ bananas?

9:48 PM
I would go make a margarita, but I have no tequila. :(

kids are well versed in the idea of owing something.

@copper.hat Gotta get them to start wracking up that credit card debt EARLY, amirite?

indeed :-)
gimme credit or gimme death
the irony that credit cards end up a debt for many

I really hope that my reimbursement for MathFest arrives before my credit card bill is due.
I am not happy about having to use my own credit card for the hotel (they refused to run the college card, which was used to make the room reservation, because I didn't physically possess that card).
I can pull money from another bank to pay the card, but that is a pain in the ass, and I don't like to touch that account.