12:01 AM
force of habit, i guess
old habits are the hardest to break :-)

You're about the only regular here whose math knowledge I know basically zero about.

12:15 AM
Is it common for people in grad school to attend undergraduate courses to flesh out their background?

12:37 AM
@SubhasisBiswas Thanks

1:01 AM
Hello!
How can we deduce that if $\gcd(n,7)=1$ then $n^{6}\equiv1\pmod{7}$?
Is it because of Fermat Little's Theorem?
Thanks!

1 hour later…
2:09 AM
That's precisely Fermat's little theorem, @manooooh.

2 hours later…
3:53 AM
@TedShifrin good to know. Thank you Ted!

2 hours later…
5:32 AM
plato.stanford.edu/entries/quine-nf/#4 says "extensionality cannot be defined in a language without equality", but doesn't this contradict en.wikipedia.org/wiki/…?

2 hours later…
7:03 AM
Hi guys
I have a doubt regarding matrices
r′=r−s^(r.e^3)/s3
How will I write that as a vector?
By the way its not s to the power of or e to the power of , they are s cap and e cap respectively
e cap is a basis vector

7:33 AM
hey chat, if someone of you knows a bit homogeneous space and want to earn a bounty, please give a look at this question, it would help me alot
https://math.stackexchange.com/questions/3234925/understanding-projection-of-vector-field-in-homgeneous-spaces

3 hours later…
10:24 AM
May 11 at 16:52, by Tobias Kildetoft
If you lecture long into the void, the void starts lecturing back.
The mystery of the lecturing void continues
May 11 at 16:55, by Ted Shifrin
@anakhro: Generally lack of responsiveness, lack of interaction. With too small a class, people are quieter, in general.
Hmm...
I want to create a mathematical object that behave somewhat like this
However, it is not clear what context needs to be abstractised to produce said object
A "void lecture" ,assuming Ted does capture most of the gist, is one where you transmit information into it, and getting only total silence as a response
Thus one way to abstract this is the zero map, which is trivial
What if, you can produce "a zero with intricate structure", abstracted from how in most practical case, what seemed silent is full of implied information in nonverbal form
and said information may be recovered given the correct tools
One of the easiest nontrivial way to imbue a structure on an element that otherwise behave like zero is to have a semigroup where there is a subsemilattice. Then you have elements that behave like an identity except for a few elements.
Likewise, nth order absorbers are also straightforward to implement
Apr 4 at 0:30, by Rithaniel
So, an "absorber" is defined an element $a$ of a magma $M$ satisfying that $am=ma=a$ for all $m\in M$. Obviously, you can only have one absorber per operation on the object, but is there a concept of a "second order absorber?" Ie an element $b$ satisfying that $bm=mb=b$ for all $m\neq a$?
In addition, suppose we have a strange zero function $\mathscr{o}$ such that:
$$\int_a^b \mathscr{o} dx > 0$$
but $\mathscr{o} + f = f$ for some functions $f$, then we will often end up with divergence issues e.g.:
$$\int_a^b f dx = \int_a^b \mathscr{o} + \mathscr{o} + \cdots + f dx$$

3 hours later…
1:55 PM
Hey folks, how do you remember that $f(A\cap B) \subseteq f(A)\cap f(B)$ is the only operation where in general we don't have equality? I always know there's one thing prohibiting the image from being a lattice homomorphism, but I keep forgetting which one.

1 hour later…
2:59 PM
@Luke: one way to think of the equality is as a generalization of the definition of injective functions. Indeed, the definition of injective functions is literally the same as $f(\{a\}\cap\{b\})=f(\{a\})\cap f(\{b\})$.

think of two horizontal disjoint disks at different heights in 3d getting floopsquished into a venn diagram in 2d by projection
taking that as f makes those set theoretic equalities/inequalities obvious for me at least

3:33 PM
@KarlKronenfeld ah, nice, this is even more minimal than the example I came up with. Thank you!

@Luke Why not simply use the definition of $f(A)\cap f(B)$?
@Luke In any case, another way is $\emptyset =f(A\cap A^c)\subseteq f(A)\cap f(A^c)$, so the reverse inclusion does not hold in general.

@user170039 The problem is not „understanding it on a deeper level“, I just needed a minimal example or mnemonic so I get the correct fact with minimal effort / distraction when proving other things.

4:20 PM
fun game

@LeakyNun Is it correct that if X is baire space then for $A \subseteq X$ set of second category then $X \setminus A$ is of first category?

no idea

ok

4:57 PM
Three girls $G_1,G_2,G_3$ and three boys $B_1,B_2,B_3$ are made to sit in a row randomly. The probability that at least two girls are together is ....
My try: Probability of no girl together: B_B_B here in dashed places girls be seated so probability is $\dfrac{4*3*2*3*2}{6!}=\frac15$ and hence probability of at least two girls are together is $1-\frac15=\frac45$. Am I correct?

@TedShifrin hey

hi Leaky

you would love that game

Huh?

5:17 PM
Hey everyone, I am going over some information theory stuff, and I came across the sentence "Let $\mu_n$ and $\mu'_n$ be two probability distributions on the state space of a Markov chain at time $n$, and let $\mu_{n + 1}$ and $\mu'_{n + 1}$ be the corresponding distributions at time $n + 1$". I am kind of confused at what this means exactly. How can you have two different distributions on the same markov chain?

@Eran No

@WilliamOliver are the probability distributions independent of each other or do they infulence each other's evolution on the markov state space?
are the distributions stationary?

I think they are meant to be stationary
@Ultradark Here is the full proof I am trying to understand i.gyazo.com/f21af4a93fea879b73c3ad9162e08b63.png

hi demonic @Alessandro

@Ultradark Maybe they aren't meant to be stationary, and that is what I am misunderstanding. I think what i am confused by is this: given a distribution on the state space of a Markov Chain, how exactly does that relate to the actual probability of an event in the Markov Chain?

5:35 PM
@AlessandroCodenotti What would be a counter example?
(0,1) as subset of reals?

6:07 PM
Hi @TedShifrin @AlessandroCodenotti @LeakyNun (and possibly others)

6:18 PM
hi @Tobias — how goes the new career?

Good. So far I am enjoying it
it's not math, but still plenty of challenges to be had

Well, that's a good thing!

It does mean getting used to dressing a lot better than I used to :)

Well, mathematicians needn't always be slobs :P

sure, but we also didn't have a dress code, so I tended to be on the casual side of things
and now I need to wear a freshly ironed shirt and neat pants (but at least the tie is optional)

6:24 PM
@Eran: Why can't the complement of a subset of second category also be of second category?

It can

I do confess that my days of ironing my shirts are long gone, @Tobias.
Wouldn't that be a counterexample, then, @Eran?

"freshly ironed" is actually an explicit part of the dress code for the company

10 messages up

6:26 PM
Last thing I saw was $(0,1)$. How is that a counterexample?
Oh, I was trying to get the complement of an open to be open.

(0,1) as subset of R
(0,1) is second category from baire's theorem

I agree.

the complement is also second category..

Is it? It's not open.
So it takes a little more work.

Does it have to be open?
It's from baires theorem also

6:27 PM
You don't like to wear a tie? @TobiasKildetoft

@skullpetrol I don't even know how to tie one myself

assume it's a union of nowhere dense sets then by baire's theorem the complement is dense in R
(0,1) is certainly not dense in R
Am I wrong?

Time to learn pal :-)

@skullpetrol as I said, the tie is at least optional

Indeed.

6:29 PM
@Tobias: I always liked wearing ties, so I can tie four-in-hand and Windsor knots both.
@Eran: What I recall is that nonempty open sets in a Baire space are second category.

It might also happen that I land on a project where it makes more sense to wear one.
But for now not so much.

Baire's theorem: Let $X$ be a complete metric space then for every countable collection of nowhere dense sets $A_n$ then $X\setminus \bigcup A_n$ is dense in X
obviously $(-\infty,0] \cup [1,\infty )$ is not dense in $\mathbb{R}$
The closure of it is itself.

Huh?

What?
Oh sorry, I proved that (0,1) is of second category, you can apply the same on it's complement.

I know $(0,1)$ is second category.
So you're saying that the closure of an open set is always of second category?
I guess that might be right.

6:37 PM
Assume by contradiction $(-\infty,0] \cup [1,\infty)$ is first category. That is, there exists $(A_n)_{n=1}^\infty$ nowhere dense sets such that $(-\infty,0] \cup [1,\infty) = \bigcup_{n\in \mathbb{N}} A_n$, By Baire's category theorem we get that $\mathbb{R} \setminus (-\infty,0] \cup [1,\infty)$ is dense in $\mathbb{R}$, but (0,1) is not dense in R

OK. I was just going to take $X = (0,1) \cup (2,3)$ and be done.

7:26 PM
Hello from London

hi
@Semiclassical
Can I ask you some questions in Topology?

You can ask, but if it’s point-set topology I can’t help
And judging from what you’re talking about above...yeah, no help here

@Semiclassical what are you up to in London?

Wel, I was in Italy for a conference and I wanted to spend a few days after in England to see some sights

Oh cool :)
Palace of Westminster is undergoing refurbishment atm so that's a real disappointing sight

7:33 PM
In topology, the long line (or Alexandroff line) is a topological space somewhat similar to the real line, but in a certain way "longer". It behaves locally just like the real line, but has different large-scale properties (e.g., it is neither Lindelöf nor separable). Therefore, it serves as one of the basic counterexamples of topology. Intuitively, the usual real-number line consists of a countable number of line segments [0, 1) laid end-to-end, whereas the long line is constructed from an uncountable number of such segments. == Definition == The closed long ray L is defined as the cartesian...
Why would gluing together more than ω1 copies of [0,1) make the space no longer locally homeomorphic to R?

7:45 PM
@Eran yes

@AlessandroCodenotti Let $X,Z$ be metric spaces such that $X$ is complete, let $f:X \rightarrow Z$ s.t the set of continuity points of $f$ is dense in $X$, How can I prove that the set of discontinuities is of first category?

8:20 PM
@user76284 who claims that?

8:30 PM
Hello
If there are two elements of order two in an abelian group, then there must be a subgroup of order $4$.
I am trying to do this as follows:

Let $a$ and $b$ two such elements. Consider the set $A=\{a,b,ab, e\}$. Clearly, $pq^{-1}=pq \in A$ for all $p, q \in A$
Here $q^{-1}=q$ since every element is its own inverse

@user76284 Look at neighbourhoods of $\{\omega_1\}\times\{0\}$

@AlessandroCodenotti , can you verify my work here
just a little problem
I am starting out in group theory

That's correct

9:03 PM
@AlessandroCodenotti Sorry if this is a stupid question, but can you explain what you mean?

Let's look at $\omega_2\times [0,1)$ for concreteness
With the order topology induced by the lexicographic order
That is $(\alpha,x)<(\beta,y)$ iff $\alpha<\beta$ or $\alpha=\beta$ and $x<y$, ok?
Where $\alpha$ and $\beta$ are compared with the usual order on ordinals and $x,y$ with the usual order on $\Bbb R$
And a basis of the topology is given by the intervals $(a,b)=\{x\mid a<x<b\}$ for $a,b\in\omega_2\times[0,1)$
Now let's look at the point $(\omega_1,0)\in\omega_2\times[0,1)$. We want to find an open set containing it, so in particular it must contain an interval around this point
In particular this interval must contain some point smaller than $(\omega_1,0)$, but those must be of the form $(\alpha,x)$ for some $\alpha<\omega_1$ and $x\in[0,1)$ (They can't be of the form $(\omega_1,x)$ because $0$ is the minimum of $[0,1)$)
But there's uncountably many ordinals between $\alpha$ and $\omega_1$ for every $\alpha<\omega_1$, so in particular every nbhd of $(\omega_1,0)$ must contain uncountably many copies of $[0,1)$, hence it cannot be locally euclidean

I see, thanks!
The same would not be true if they contained only countably many copies of $[0,1)$.

9:21 PM
Right, because every interval in $\Bbb R$ contains countably many copies of $[0,1)$ so that's not an issue
And that's why the long line $\omega_1\times [0,1)$ works
Every open interval contains only countably many ordinals

i.e. non-empty open interval in $\mathbb{R}$ can be "expanded" into a segment of the long line, right?

Uhm what do you mean?

There's a homeomorphism between any interval in $\mathbb{R}$ and $\alpha \times [0,1)$ for countable $\alpha$.

hmm actually I'm not sure wait a moment

1 hour later…
10:41 PM
Hi everyone, got a small problem that's too small for an post, can i post it here?