@Jakobian Some of us appreciate mathematics in much the same way that we appreciate other forms of art / intellectual exploration.

@leslietownes hard to pursue that when publish/perish logic applies tho

sounds like the mindset of someone whose paper on "semiclassical's constant" got rejected from the proceedings of the transnistrian hydrological society

@leslietownes Oh, g-d... you're one of the Transnistrian Separatists, aren't you?

It's part of Moldova! Get over it!

more like the mindset of someone who has sorta given up on actual research

in favor of teaching

I like teaching, most days. It is relatively fulfilling.

Though I had grading. :(

Grading sucks.

that it does

also i'm not really that happy with the lectures i've delivered

but it's tough to teach the second semester of algebra-based intro physics

What is "algebra-based intro physics"? (please keep in mind that I took one semester of physics in high school, and one quarter of QM in grad school, and otherwise have zero physics background)

for first semester it's not that different

but doing wave physics without calculus is...tough

just from the description it sounds like what my hs physics was, i.e., do basic newton stuff without assuming calculus

oh god wave stuff too?

yup

fluids + oscillations + electromagnetism (with a loooot more emphasis on the electro than the magnetic)

more acoustics than EM waves, to be clear

@Semiclassical "Wave physics"? Is that, like, the Dirichlet problem?

How do you do that without calculus?

And funky anal?

there was a physics sequence for biology majors at my undergrad that did a cook's tour of those things in its second semester, although i think at least nominally it did require calculus (albeit not the calculus sequence that the actual physics class required)

more like all the implications of "a travelling wave has the form $u=A\cos(kx-\omega t)$"

so mostly just wavelength/frequency/phase velocity

but that includes Doppler effect, standing wave resonance, sound intensity vs sound level...

now we're doing electric force and electric field

Bar bar bar bar bar bar...

no group velocity / dispersion tho

which is fine with me, there's too many velocities to keep track of with waves

phase velocity, group velocity, signal velocity, particle velocity (for sound waves)

i'm probably forgetting some

Hey! I've watched the 3Blue1Brown series on linear algebra very carefully, and I'd like to know your advice on how to continue my learning path. I'm seeking resources with a similar approach, whether they are videos, books, or any other format

What are you trying to learn? More linear algebra or something else?

Like, 3B1B has similar videos on other math subjects (Differential Equations, Probability, Neural Networks, etc.), and the channel owner Grant Sanderson did a great series for Khan Academy on the differential half of Multivariate/Vector Calculus. He also hosts a yearly math video competition which has spawned tons of other content creators making similar style videos.

It's kind of revolutionizing math on Youtube. It's almost hard not to come across them searching for arbitrary topics.

X4J the limit comparison approach is in some sense "the same thing as" using the taylor expansion of sin at 0 (not so much the taylor expansion of f at 0, f is too ugly to have one of those at 0). more generally if f(0) = 0 and f(x) = 0 + a_1 x + O(x^2) and g(x) is a nonzero function that goes to 0 as x goes to infinity, then lim x to infinity f(g(x))/g(x) [the comparison/l'hopital thing] will coincide with a_1 [the coefficient of x in the taylor polynomial for f]

@user10478 Thank you very much! I'll check that out

so in some sense i'm not sure if it makes sense to distinguish the approaches

If $\pi_1(M)$ of a manifold $M$ contains a surface subgroup $\Gamma$, then $M$ contains an immersed surface whose $\pi_1$ is $\Gamma$?

@leslietownes This is great, thanks

I recently learned about typewriter sequence. Can someone tell me the name of some more sequences that one should keep in their inventory of counter examples?

most things like that don't really have well established names or fixed definitions (i had to google that, for example, and while the results were related, they were more examples of a broad idea than one single fixed thing). you often find things like that given as examples and/or exercises in analysis books, around where they introduce different notions of convergence. i am not sure they really come up so often that they need to be part of anybody's "inventory"

whenever you have two kinds of convergence, there's always the question, does one imply the other. if the answer is "no" there is usually a counterexample kind of like that lurking somewhere.

like what terry tao is doing here: terrytao.wordpress.com/2010/10/02/…

in fact, that web page is a better summary of that sort of thing (in its context) than you find in most books.

i wish i could remember what we called an example like that in my analysis class. we had some other name for ours, something like "the wandering step" but not exactly that. i don't actually like the term "typewriter sequence" because kids these days will need a video explainer for why anyone would call it that.

@leslietownes Thank you for the helpful response.

> Theorem Let $f:X\to Y$ be a continuous map between two topological spaces $X,Y$ and let $X$ be connected. Then $f(X)$ is also connected.

> Proof Suppose that $f(X)$ is disconnected. Then there exist nonempty, disjoint open sets $U, V$ in $Y$ such that $U \cup V \supset f(X)$. Since $f$ is continuous, $f^{-1}(U), f^{-1}(V)$ are open, nonempty, disjoint sets such that $$X=f^{-1}(U) \cup f^{-1}(V),$$ so $X$ is disconnected. It follows that $f(X)$ is connected if $X$ is connected.

Is it correct to write $U \cup V \supset f(X)$? Why?

You want $U\cup V=f(X)$

that's what I thought

yeah, you have to work inside of $f(X)$, not inside of $Y$

it is not in general true that if you have a disconnected subspace of a space that it can be covered by two disjoint open subsets of the ambient space

@Thorgott are you also saying that $U,V$ should be in $f(X)$ and not in $Y$?

yes

right, ok, this is where my confusion stemmed from; how do we know $f(X)$ is open in $Y$? (rhetorical question)

we don't

@psie Does it need to be?

The assumption that there exist disjoint open sets $U, V$ with $f(X)\subseteq U\cup V$ and $f(X)\cap U, f(X)\cap V\neq \emptyset$ is stronger than saying $f(X)$ is disconnected

@Thorgott Oh yes I missed that, I thought they were picked in $f(X)$

This is all wrong

to have that you'd need stronger assumption on space $Y$ i.e. that its $T_5$

instead, you want $U, V$ to be open subsets of $f(X)$ where $f(X)$ is equipped with subspace topology

so that $U = U_0\cap f(X), V = V_0\cap f(X)$ where $U_0, V_0$ are open in $Y$, but not necessarily disjoint!

Or just say, without loss of generality $f(X) = Y$

The confusion stems from two inequivalent ways of saying that a subspace of a topological space is connected

In general the definition of connected doesn't translate well to the ambient space, unless your space is nice enough to begin with

At least in a way that would be "natural" to a student learning topology

that requires a proof as well. So even if you're a fan of metric spaces, you need to argue on this further

so that it actually doesn't depend on the ambient space

what I accidentally removed: in general we want the notion of "connected subspace" to be the same as "connected when given subspace topology" and the same holds for many other properties P with connected replaced by P

well hopefully that was good enough... I'm just going to go now

When learning say sequences and series, there are various "tricks" like summing contractive, recurrence related sequences to get a final and initial term, there are tricks for monotone inc serieses like considering 2^n termed subsequences and bringing inequalities, etc. and a myriad of tests not to mention, like cauchys test, root, ratio, condensation etc. how to keep oneself sharp with the details of each of these?

There are only 3-4 problems that concern themselves with application of these, and they're frankly boring, direct application

@nickbros123 In the "real world", you try a few standard things, then use Google.

And you get a sense of what things are "standard" by working though a bunch of "boring, direct applications".

Recently someone asked me how to show the p-harmonic series ($\sum n^{-p}$) for real p>1 converges, and I blanked: I had studied like atleast 2 methods for this series. I found cauchy condensation test, but I couldn't recall the consideration of 2^n subsequence trick

I'd say there's a difference between ideas/patterns that are so relevant that they persist throughout a number of proofs/applications and hence worth remembering and ad hoc tricks only used at one point or another and hence not worth remembering, but only experience can tell the difference between those.

@Thorgott ad hoc tricks look so ingenious though

I can't help but get tricked into believing they're worth remembering

What exactly does "connect together a nonzero component" mean here? And why is $\alpha_6+\alpha_3+\alpha_2+\alpha_1$ deducted for 2nd highest weight?

Also can somebody recommend a textbook (preferably with examples) or atleast with the rules for obtaining irreducible rep. decomposition of tensor product of irreducible representations of simple Lie algebras clearly stated?

I am interested in a generic rule for all the simple Lie algebras than for rules merely valid for classical algebras for which I am more comfortable with Young tableaux

@BalarkaSen

I like Ruberman's answer

What is the smallest prime factor of $$\sum_{j=1}^{14} j^{j!}$$ ?

What is the smallest odd prime factor of $$\sum_{j=1}^{19} j^{j!}$$ ?

@Peter 4

@Peter 15

(I would normally answer "3", but three is both odd and prime, so that can't be right).

@XanderHenderson Did you notice that the exponent is "$j!$" ?

@Peter Did you notice that I wasn't being serious?

what r some lesser known interesting math fields other than geometry, algebra and number theory

@RyderRude $p$-adic analysis. Fractal geometry. Analysis on fractals. Planar algebras.

thanks!!

the last one seems completely new knowledge !!!!

i find planar algebra similar to simplical complexes and homology. is it a related idea

I saw it come up with Khovanov homology in knot theory, so it does have tendrils to homological ideas.

@RyderRude The people I know who are interested in planar algebras are C*-algebra and noncommutative geometry folk. I am not sure how closely tied to simplicial complexes and homology they really are.

But sure, there may be an interesting connection.

We can ask leslie

Cool @onepotatotwopotato

@XanderHenderson p-adic analysis seems not so lesser known these days

@anak It also looks extremely close to the contact category of the 2-disk

Where the objects are "germ of a contact structure on the 2-disk" and morphisms are "attach bypass"

@BalarkaSen What's the contact category of something? I am unfamiliar with the term.

I don't know in detail, but I described above what I vaguely understand it to be for the 2-disk

Oh I see. Sorry, I completely did not read on.

Each of the disk diagrams here seem to be like a convex neighborhood of a disk in a contact 3-manifold, with the curves being dividing curves

Let's call the first three pictures 1, 2, and 3. Think of the arcs as like dividing curves.

Attaching a bypass gets you from 1 to 2, and then attaching another gets you from 2 to 3, and then finally attaching another gets you from 3 back to 1

Very weird tbh, I had never encountered planar algebras before.

I saw them in a knot theory course.

The white and black regions alternate also, which is exactly what dividing curves do in a convex surface - decomposes the surface into + and - regions

@BalarkaSen, you might have an idea on this, but do you know of any like normal form for a top degree form $\omega$ which has a zero/critical set being some codim. k submanifold? Like if the critical set $\Sigma$ is a hypersurface, locally described by the vanishing of some coordinate function $\rho$, I want to say that $\omega$ locally looks like $\rho\mu$ for some volume form $\mu$, but I haven't been able to nail it down for some reason.

For the more general case where $\Sigma$ is the vanishing of several coordinate functions, I am not sure what I would want it to look like.

@anak For any oriented manifold $M$, the top exterior power $\Lambda^n T^*M$ is a trivial line bundle. A positive volume form $\mu$ is a section of this, and any other form is therefore a function multiple of $\mu$.

Yeah, but I am wanting to relate the function multiple part to the local description of the critical set as the vanishing of some coordinate functions.

$\omega = \rho \mu$, you're demanding $\rho$ to vanish on $\Sigma$. In the space of smooth functions on $M$ vanishing on $\Sigma$, the "generic" ones are Morse-Bott. That is to say, they are locally of the form $\rho(x_1, \cdots, x_n) = x_1^2 + \cdots + x_k^2$ where $\Sigma = \{x_1 = \cdots = x_k = 0\}$ in the local coordinate.

It's a purely local problem. You want to ask what the generic functions $\rho : \Bbb R^n \to \Bbb R$ vanishing on $\Bbb R^k \subset \Bbb R^n$ look like.

Maybe my description above is not accurate (eg, not so for $k = n - 1$).

But either case this has not much to do with forms

Ah OK: For $k = n-1$, it is $\rho(x_1, \cdots, x_n) = x_n$. This is because the sub-class of transversal functions is dense.

For lower $k$ (aka, higher codimension), it is going to be one of those, because $d\rho = 0$ along $\Bbb R^k \subset \Bbb R^n$. Then the sub-class of functions with nonzero partial Hessian is dense, forcing it to be Morse-Bott.

@BalarkaSen Darn that Tate guy!

Well I am starting with a form which has a critical set which is a submanifold. Then I want to come up with something like a local description $\omega = \rho\mu$ in a neighbourhood $U$ where $U\cap\Sigma = \{x\in U \mid \rho(x) = 0\}$. Maybe you understood my problem backwards, or maybe I don't see clearly how one can get rid of $\omega$.

I don't see why that is relevant, @anak.

$\mu$ is fixed. All you need to figure out is what $\rho$ looks like.

That is given by Morse-Bott theory

Do you want to not impose any genericity restrictions?

@BalarkaSen Just to confirm: you are saying no matter what my submanifold $\Sigma$ is, if I have a top-degree form $\omega$ with critical set $\Sigma$, then $\omega = \rho\mu$ for some nowhere vanishing volume form $\mu$ and $\rho$ such that $U\cap\Sigma = \{x\in U \mid \rho(x) = 0\}$ in some local chart $U$, this $\rho$ being given by Morse-Bott?

Yes. First of all, do you agree that any top-degree form $\omega$ on an orientable manifold with a fixed positive volume form $\mu$ satisfies $\omega = \rho \mu$?

For some function $\rho$.

For some $\rho$, yeah.

OK, so now you have told me that $\omega$ has vanishing set (or maybe even critical set) some submanifold $\Sigma$. What does that say about $\rho$?

Oh my am I really just stupid

And so we can just use $\rho$ as our like "defining" function for $\Sigma$.

For some reason I was like "start with the defining function and find the $\rho$", but it's really just trivially easy the other way.

Well, all you know is $\rho^{-1}(0) = \Sigma$. Now one has to come up with a local form for a function $\rho : M \to \Bbb R$ such that $\rho^{-1}(0) = \Sigma$.

And that's just using a Morse chart kind of thing, is what you are saying?

@anak If $\rho$ is generic enough the answer is either "the defining coordinate for $\Sigma$" if $\mathrm{codim}(\Sigma) = 1$, or a certain Morse-Bott quadratic form.

But even if it is not generic enough, I think you can write down a normal form by following Milnor's book where he proves Morse lemma.

Yeah, I see. I didn't think of using Morse-Bott. Thanks so much!

In full generality, it might look like $\rho(x_1, \cdots, x_n) = x_{k+1}^2 g_1 + x_{k+2}^2 g_2 + \cdots + x_{n}^2 g_{n}$ where $g_i \geq 0$ are non-negative functions such that $g_i^{-1}(0) \subset \Sigma$ (possibly empty).

For the vanishing case, not the critical case. In the critical case you get cubes, not squares (again with a codimension restriction).

Yeah, that makes sense. Thanks! What have you been working on recently?

I'm finally doing contact topology. :P

Yay!

But also sad because I have strayed so far :(((

What do you do nowadays?

Still geometry, but kind of floating around with dynamics stuff, optimal transport, etc.

Are you doing more of the Eliashberg/Murphy direction for contact topology?

Hey, that's kind of cool!

I have a ridiculous idea I want to tell you along that direction.

@anak I wrote my Master's thesis on Murphy's theorem, and was actively thinking about high-dimensional contact topology. Then I temporarily got lured by dimension 3. I will get around to dim >= 5 once I have written up some observations in dim 3 that I made.

I think Yasha is coming this September to India, so that's a good opportunity to come back around to that :P

When I was doing a little contact topology, I kind of noticed you had like the "french" school of contact topologists who were more into the lower dimensional, geometric stuff, and then everyone outside of that school that were into much more high-powered stuff. Do you find that to be the case? I never did get too deep, so my impression could be way off.

I think the last decade or so has been bad for "geometric" geometric topology, yeah. Lots of invariants from differential geometry, less cut and paste. But there's a lot of unsolved cut-paste problems even in low dimensions.

It's a matter of the general movement of the tide. You know how these invariants that people seem to like are uncomputable for hyperbolic 3-manifolds, right?

That's why no one knows anything about contact topology on hyperbolic 3-manifolds.

Oh, I wasn't aware there were issues with hyperbolic manifolds.

It's ridiculous lol

I wouldn't have guessed studying hyperbolic manifolds would have been a common pursuit, though. Is there a reason they are looking at them besides that in itself?

It's because if you picked a closed 3-manifold randomly, then with probability 1 it is going to be hyperbolic.

Absolutely dominating class

Everything else is peripheral, falls on the wayside

Wait, maybe I don't know what a hyperbolic manifold is. Are you referring to the subset of Riemannian manifolds?

By a hyperbolic 3-manifold I mean a 3-manifold which is of the form $\Bbb H^3/\Gamma$ where $\Gamma$ is a discrete subgroup of isometries of $\Bbb H^3$. Equivalently, those which admit a metric of everywhere constant sectional curvature -1.

Then my claim is, look at the class of all closed 3-manifolds. Pick one randomly, whatever that means. It is hyperbolic.

Oh I see! I thought you were saying that the actual metrics will be generically hyperbolic.

Ah no

Got the confusion

Should have been a little clear about that

Makes sense, though. Well makes sense in that my mind doesn't break immediately considering it. Is there any easy way to see why they generically admit a hyperbolic metric?

@anak There's a precise theorem, which I don't want to state. Let me try to convince you instead.

Heh, sounds good.

Would you agree that a typical 3-manifold has infinite fundamental group?

Not necessarily closed 3-manifold?

Closed, sorry.

Infinite order, yeah.

Great.

Next, would you agree that a typical 3-manifold is prime? Doesn't break as M # N

M, N \neq S^3

Yes, though I kind of rely on experience with knot theory there.

That's exactly what I had in mind

But also: if it's not prime, you cut, break it as M - ball, N - ball, and fill both spherical boundaries, and then pick one of the two. After some finite stage you'll run out of spheres. If your random law on manifolds is "uniform enough", performing this operation on a given random 3-manifold should also result in a manifold with "uniform law"

That seems reasonable.

Finally, hardest bargain: would you agree that a typical 3-manifold is atoroidal? I.e., there is no embedded torus which is injective on pi_1

I can definitely bargain that one on the premise that I have no intuition for the property.

Here is one intuition. Take a diffeomorphism $f : \Sigma \to \Sigma$ of a surface of genus g. The mapping torus $T(f) = (\Sigma \times [0, 1])/(x, 0) \sim (f(x), 1)$ is a 3-manifold. If $f$ preserved a circle (i.e., $f(\gamma) \sim \gamma$), that would give rise to an embedded torus inside $T(f)$.

And these are basically the only ways you can find embedded tori in $T(f)$ (the fiber is genus g >= 2, so the fiber cannot be my torus -- it has to be in the "horizontal direction")

Do we expect generic 3-manifolds to "look like" mapping tori?

But a typical diffeomorphism of a surface of genus g >= 2 preserves no disjoint collection of curves

It's a closed condition. If it did you'd be able to cut along these circles and reduce the diffeomorphism to simpler ones.

@anak In some sense yes, in some sense no.

Anyway if you've reached until this point: Thm (Thurston & Perelman). A closed prime atoroidal 3-manifold with infinite pi_1 is either S^1 x S^2, a nonorientable S^2-bundle over S^1, or hyperbolic.

That's pretty cute. That's the Perelman, too?

Yep

It was solved in a special case by Thurston. In full generality it was only established after Perelman's work

Pretty nice. Thanks! Does this all fall under the pursuit of the geometrization conjecture?

Yep. This one's the hyperbolization conjecture, Poincare conjecture is part of the elliptization conjecture.

I really ought to look into Perelman's work more closely. I read a short history book of the Poincare conjecture but didn't really look more closely at the specific details of the work mentioned.

O'Shea's book. More pop math history sort of thing. Hardly any detailed math.

If you start out with a metric on any closed 3-manifold, evolve it under Ricci flow, the manifold starts looking like a bunch of large bubbles (the hyperbolic parts) connected by extremely thin tubes to a bunch of "flattened", almost 2D, manifolds, which could themselves be interconnected by extremely thin-tubes in a web-like way

The thin tubes are $T^2 \times [-L, L]$, $L$ large. The flattened pieces are all the rest of the geometries

I believe, $C(X;\mathbb{C}) = C(X;\mathbb{R})\otimes_\mathbb{R} \mathbb{C}$?

The whole web-like thing is called a graph manifold

If you didn't have any pi_1 injective tori to start with, all you'd have is the bubble -- because there's no tubes to join it with other parts. That's hyperbolization

That's a nice visualization!

no, that doesn't sound right

@Jakobian it's probably a "completed tensor product" or some other imaginary thing the anal people pretend exists

@Thorgott what do ideals in $C(X; \mathbb{C})$ look like in terms of $C(X;\mathbb{R})$?

For an ideal $I\subseteq C(X;\mathbb{R})$ let $I_c = \{f+ig : f, g\in I\}$. I found a source that claims $I\mapsto I_c$ is a bijection between prime ideals of both rings

this probably means the ideals are not in such correspondence

whats an example of ideal $J\subseteq C(X;\mathbb{C})$ not of the form $J = I_c$ for some ideal $I$?

probably $J = (f)$ for some $f\in C(X;\mathbb{C})$ gives an example

If $X = \mathbb{C}$ and $J = (\text{Id}_\mathbb{C})$, then if it were $J = I_c$ then $(z\mapsto \text{Re}(z))\in I$ so that there would be continuous $g:\mathbb{C}\to\mathbb{C}$ with $g(z) = \frac{\text{Re}(z)}{z}$ for $z\neq 0$ which is impossible

so it seems like while the map $I\mapsto I_c$ is an order embedding of the ideal structure of $C(X;\mathbb{R})$ into ideals of $C(X;\mathbb{C})$, its not surjective, however many special types of ideals are still bijectively mapped between those two rings.

So swapping $\mathbb{R}$ to $\mathbb{C}$ doesn't add any important structure, although it does add some structure.

@Thorgott @leslietownes

also the article: arxiv.org/pdf/2001.09659.pdf

was about to ask what the relevance of nationality has to do with math

hopefully, nothing

with scientific articles you mean

all national languages "should" merge into the language of mathematics

last time I've been looking for articles about Henstock-Kurzweil integrals and one article I found promising ended up being full of errors and incomplete arguments, something I'd call a bullshit article

researchgate.net/figure/…

uh, there's certainly a storied tradition of folks with no institutional affiliation, or affiliation only with some regional university that is not well known for its mathematics, posting junk on the arxiv. and some places have more universities than others by virtue of population alone. i wouldn't go much further than that very nonspecific level of generality

also, without getting too deep into credentialism, whatever your definition of 'obscure regional university' is, CSU long beach probably is one :D

Let's see if I'm making sense (I've been tumbling in logical abstraction, trying to understand the logics of a theorem). Suppose one has a statement of the form $$A \implies (B\vee C).$$ The proof goes as follows; the author assumes $A$ holds, but also not $C$, and then derives that $B$ holds. Is this logically OK? Does this prove the statement?

@leslietownes I'm about to get canceled here

in recent years csulb has rebranded several of its athletic teams (which used to be mostly "49ers" except the baseball team) as "beach" teams, which is about as annoying as it sounds

@psie its okay

phew, that's a relief to hear...I will give you some more context...

i'm not going to refer to a basketball team as "the beach"

nobody ever says "i'm going to give you less context"

This is from Magnus's Metric Spaces. The author tries to prove the contrapositive of the statement. Leslie suggested the contrapositive is "If $f(X)$ is not connected, then either $f$ is not continuous or $X$ is not connected." Now, using continuity of the function is an essential ingredient in the proof. This confuses me, since from the contrapositive statement it seems like this is something we can not assume. All we can assume is that $f(X)$ is not connected.

please don't say that i suggested that "the contrapositive" of anything written in English was anything else. this was actually a big part of the point i was trying to make

i said that, whatever the person on MSE was posting, was OK, as a logical rearrangement of what they were trying to prove or show

@psie Repeat after me: Math is not English. Math is not English. Math is not English.

to revisit my point very briefly, there is no one to one correspondence between English statements and statements in anybody's formalism, and so language like "the contrapositive of __" is vaguer than it sounds

We often say things in spoken or written language which are not 100% rigorous, or which don't precisely fit into some framework of well-formed logical propositions.

psie this gets back to whether you choose to include background hypotheses as part of the "P" in something like "if P, then Q." we might even have discussed this example

If you want to find the "contrapositive" of something written in plain language, your first step is to write out the original plain-language statement using more formal notation, and then take the contrapositive of that (which you can then translate back into plain language, if you like).

if you want to impose a tiny bit of formal structure on this, they are using "X is connected" as P and "f(X) is connected" as Q (suppressing dependence of any of this on X and f) and the continuity assumption is lurking in the background as part of the universe your statements are talking about

@leslietownes I mean, it is a CS. They do real work at CSes. Not as prestigious as a UC, but not bad...

I suggest not worrying about semantics

logic is all about vibes

@Thorgott word, man.

*snaps*

@Thorgott citation needed

@AlessandroCodenotti Oh, man! I got yer citation right here! Like, take a hit, man!

Crackle, pop.

Imma go home and make a pie. I want pie.

psie: looking back it seems the example i used was "if a is even and b is even then a+b is even"where a and b being integers is understood, and is not part of what you negate when you cast that as "if P, then Q" and form "if not Q, then not P"

I think I understand the logic better now. If we accept "a" contrapositive to be what I stated above as Leslie's contrapositive (sorry!), then we suppose that $f(X)$ is disconnected. The function $f$ is either continuous or it is not continuous. If $f$ is not continuous, then the conclusion "$f$ is not continuous or $X$ is not connected" is true, and the proof is done. Thus only the case of $f$ being continuous remains, and that is what deserves a special little treatment; the image above.

@XanderHenderson well of course (although for many years its most famous and probably most-cited faculty member was an antisemitic crank)

@leslietownes True.

:-/

@XanderHenderson meat pie?

good o' fashion apple?

@user85795 No, my daughter cannot eat apples.

It is going to be strawberry, because that is what she picked out.

coolio

@user85795 He died. :(

:( rip

psie: if you follow that rabbit hole a little further, if you have as a background assumption that f is continuous, you can not put it into your symbology at all, so your P and Q are just "X is connected" and "f(X) is connected", and your if not Q then not P is just "if f(X) is not connected, then X is not connected," with the background assumption still being everywhere that f is continuous, and you are not stopping to consider discontinuous f as an alternative case

i term this a "rabbit hole" because in context, i think it wouldn't naturally arise to look at the case of f discontinuous if you didn't insist on (1) detouring through symbology and (2) encoding the continuity hypothesis as part of that symbology

September 28, 2022

I think leslie is probably a good person to go for when it comes to logic since he studied law

if you don't like the answer you get, just appeal to a higher court

SCOTUS

why does this acronym sound so nasty

it has a reputation of nastiness

phonetically it resembles a slang for scottish people, I guess

@Jakobian Because it resembles the word scrotum.

What is an example of a "monotonicity property" of the real numbers? I'm reading the following sentence in my notes:

> Our use of the extended real numbers is closely tied to the order and monotonicity properties of $\mathbb R$.

Hence I wonder...

hashtag measure theory