#real sets

[aurianneor]

The Director and the Jedi

The process of filming as it should always be:

Filming on schedule, no pressure

A casting looking for talents rather than looks, with costumes designed according to the individuals. 

The revival of real sets, real props, puppeteers.

A peaceful director with a strong vision but letting the actors do their jobs

The Director and The Jedi (2018) - Imdb: http://www.imdb.com/title/tt8080556/

Sugar, my favourite artist: https://aurianneor.tumblr.com/post/163823899260/sugar-my-favourite-artist-today-there-is-no

Hips: https://aurianneor.tumblr.com/post/172003434170/hipsmore-sugar-my-favourite

About this film, have a look at : I can inspire hope: https://aurianneor.tumblr.com/post/172587941070/i-can-inspire-hope

[tilde-he]

trying to understand the “Real sets“ paper by Janelidze and Street

Part 1

@twocubes​  referred me to this paper, and I said I’d read it some time that I don’t have a midterm the next day, so I’m trying to read it now.

Page 3 they define a function delta which, Oh wait nvm they say “given 0 in A”. Ok, page 4, the definition of a series magma, uh..? So, when it says that Sigma is a function from A^Nat to A, and that it is defined as taking a to sigma_{i in Nat} a_i , is that just defining the notation sigma_{i in Nat} (something)_i , or is this refering to some previous or expected-to-be-understood idea of sigma_{i in Nat} (something)_i ? Like, so far I just have that A has an element called “0″, but I don’t have any structure on the set yet. I’ll keep reading. Ok, so that that diagram (still on definition 2.1 on page 4) commutes is saying that, well, that delta composed with Sigma^Nat is equal to A! , and so, that for all a in A, Sigma^Nat (delta(a)) = A! (a) , but I’m not sure what A! is? I get the idea that it has something to do with uniqueness. I imagine that it means, like,

but I’m not sure.

Oh, hold on,

I think I maybe get it now?

It is probably saying, like, ok, well, the delta is sending a to a sequence of sequences, with the nth sequence having a in the nth spot, and the Sigma^Nat is probably like, applying Sigma to each of those sequences, resulting in a sequence of the sums, and the image of the A! is like, yeah, ok, so it is saying that the sum of a series where all the terms are 0 except for one term, is that one term, regardless of where that term is in the series. Ok. That seems sensible. Not sure why they didn’t just say that.

I also note that this seems to require that all series in a series magma are sum-able, which is a little surprising to me, but alright.

It defines a category of things like this. It justifies this as “Since series magmas are models of an algebraic theory, thereis a corresponding notion of morphism” which confuses me a bit, because I don’t see why one can’t just, oh, is it in order to see that using that choice for what the morphisms are, uh, actually makes a valid category? In any case, yes, I do buy that the result is a category.

I don’t know what it means for a category to be co-complete, or remember what it means for a category to be complete. I certainly don’t know what it means for a category to be “Barr-Tierney exact”. Ok, complete means all small limits exist, co-complete means all small colimits exist. Limits and colimits are saying that stuff defined by universal properties exist, like products and whatnot. And “small” here just means “like the size of a set, not proper-class sized”. ok. So, why must this category be complete and co-complete?

Oh, well, this doesn’t answer that question, but I see that {0} would be “the” initial object of the category. Ok. Ah, it would also be the terminal object. Cool. So this category has a zero object.

Ok, well, can I see why arbitrary products would exist in this category? Well, if you just take the set product of the Ais, and have the 0 be the element of that which has the 0 from each of the Ai s, and has the Sigma on this thing just do the different parts separately, and does that satisfy the universal law for products? I think so, yes. Wikipedia tells me that a category is complete if it has all small products, and all equalizers. I can see the small products part, but I don’t have a good intuition for equalizers (yet). Ah, ok. uh.

So, for any f,g:A->B , want an E, eq, s.t. eq;f=eq;g , and with E, eq, universal for this. Uh huh. Ok. It seems to me that this is essentially saying that need the subset E of A such that f and g send it to the same thing, and also need the subset E to also be one of these “series magmas”, and that the inclusion map is a morphism in the category.

But, to define Sigma for Eℕ , isn’t an issue. Just define it to be the restriction of the Sigma for Aℕ to Eℕ . And, E will be an object of the category for the same reason that A is, as the thing that the commutative diagram requires, is just some thing about each of the elements of the object, and so, ah, wait, hold on. This is not so! What if have some element of Aℕ which Sigma sends to an element of A which is not an element of the chosen element of Aℕ ? So, then, what can we do? Can we just take the closure of the intersection, taking the smallest subset of A which contains E, and which is closed under the series sums? No, that can’t work, because then it would contain things which f and g send to different elements. Perhaps there is a largest subset of E which is closed under this thing? Or perhaps we can define E’s Sigma differently when the output would lie outside of Sigma? Like, if we send it to 0 instead? That would, I think, produce a valid object of the category, but would the inclusion function be a morphism of the category? It would not! Take some sequence of elements in E whose sum (using A’s Sigma) is not in E, if this was mapped by E’s sigma to 0, then the inclusion morphism from E to A would have to send 0 to whatever A’s sigma sends it to, in order for the diagram 2.3 in the paper to commute, but that element has to be something other than 0 (as 0 is in E), but also it must be 0, because morphisms in this category must send 0 to 0.

So, I guess, take the largest subset of the subset of A which equalizes f and g, which is closed under these series. But, why must there be such a largest such subset?

Suppose that E = { x in A | f(x)=g(x) }, and E1 , E2 are two subsets of E such that they are closed under a series of them under A’s Sigma . What of a series of elements of the union of E1 and E2 ? Must it also be in that union? Or, at least, must it be in E? Well, hm, if we split the sum up into the part in E1 , and the part which is in E2 and not in E1 (using 0s for the other terms), then we get two terms, one from E1 and one from E2 , but, that doesn’t seem particularly useful, because I don’t even know that 0 + x + 0 + 0 + y + 0 + 0 + 0 + ... = 0 + x + 0 + 0 + 0 + y + 0 + 0 + ... with these sigmas.

So, I don’t think I have a way of adding 2 elements.

....

aaaand the only google search result for “Barr-Tierney exact” is this very paper? and none of the papers cited by this paper are by anyone named Barr or Tierney? What. How am I to figure out what “Barr-Tierney exact” means? Ah, well, https://ncatlab.org/nlab/show/exact+category comes up when I search for “Barr exact”, and it says “An exact category (in the sense of Barr)”, so, I guess that’s probably what is meant? Still don’t know what that means though, because idk what a kernel pair is, but uh, ok.

Oh hey! There is a page on n-cat-cafe about this paper. Maybe reading that will help me!

hmm. Not really. Not for this part of the paper anyway. It might help me understand later parts of it.

Ok, let’s, uh, continue despite my not understanding why SerMg is complete (not to mention co-complete and exact). It now defines a “series monoid” as being a series magma where double sums of series of series, uh, have the two summations commute.

I don’t understand the statement of Proposition 2.2 . Specifically, it says

If Σ′: Aℕ →A is a morphism for the Σ structure on A

and I don’t know what it means by that. Does it mean, like, If Σ′ acts as a morphism from the object ( Aℕ , Σℕ ) to the object ( A, Σ) maybe?

Guessing it is something like that.

Ok, SerMn , the subcategory of SerMg consisting of series monoids, uh, is still a category because it just takes a subset of the objects. Alroight.

Now, this category should, I think, be easier to show that the equalizers and stuff exist in it. Oh, do arbitrary products still work in SerMn ? I think so? Well, of course can still have the product as an object in SerMg, the question is whether the product is also an object of SerMn. And... yeah, if the sums of the different parts of the product are separate, then, yeah, if each of the parts commute, than the whole thing should as well.

Ok.

Now, with the whole, thing I was saying about the E1 and the E2 , If we split the series into the two parts, and have those two parts be the first entries in a series of series, and the rest of them all being the constant 0 series, then because of the commutativity of the additions thing, doing the sums one way results in getting just the original series back and adding it up, and doing it the other way results in the sums of the two subseries being there, and like, adding those with infinitely many copies of 0, and the commutativity shows that these have to be the same. And, because this argument doesn’t depend on where the 2 are in the series, therefore, uh, well, the sum of those two things doesn’t depend on where it is in the series?

I think can extend this to say that any finite sum is well defined, and doesn’t depend on the order?

Yeah, so, take some (an)n in ℕ with finite support, and another with the same elements, so like, the same thing but under some permutation of the natural numbers, and then, like, let an,m be defined to be , an,p(n)=an , and for all other pairs (n,m), an,m=0 . As the vertical and then horizontal sum must equal the horizontal and then vertical sum, and that these are the sums of the original an and the permutation of it picked, therefore the sums of those two things are the same, and so the sum of the elements doesn’t depend on the order. Cool. Actually, that didn’t seem to depend on the fact that it had finite support. Alright, cool, so we have that the order of the terms in the sum doesn’t matter. That’s nice. More than what I expected.

examples 2.4 and 2.5 make sense to me. Err, their definitions do anyway. And I see why ex 2.4 is a series monoid.. But why does ex 2.5 have the commutative property thing? Well, I guess that’s just a fact from analysis. Yeah ok.

Oh, right, I wanted to see if I could show that SerMn is complete.

So, have the sum of terms from E1 and from E2 which are elements of E1 and E2 respectively, and, because both of those are subsets of E, these elements of them are elements of E, and so f and g send them to the same thing in B. and, furthermore, f and g must send their sum to the sum of what they send them to individually, and therefore f and g send their sum to the same thing, and therefore, the sum is in E.

Ok. But that doesn’t seem to quite get me to showing that there is a subset of E which contains E1 and E2 and which is closed under the sums? Oh, wait, what if I just take the subset of E of all things which can be expressed as a finite sum of things from the union of E1 with E2 ? Then, uh, every element of this set (call it E3) would either be:

.. does tumblr not have a bullet points feature? Ok whatever. 1) an element of E1 , 2) an element of E2 , 3) a sum of an element of E1 and an element of E2 3) a sum of

oh wait, uh, because the addition is now known to be commutative, then I guess “a finite sum of elements from the union” is just the same as “a sum of an element from one and an element of the other”. Ok.

So, taking a series where terms are in E3 , can take each element of the series and split it (not necessarily uniquely) into an element of E1 and an element of E2 (err, assume that 0 is in both E1 and in E2), and then sum the parts in E1 and the parts in E2 separately, and then sum those together, the result will be in E3 . So, E3 is closed under these series sums!

So, for any two subsets of E which are closed under the sum-of-series function (and contain 0), there is a subset of E which contains both of them, and is also closed under that function.

But, does that let me establish that there is a maximal one?

My idea is to approach this by transfinite induction, take the set of all subsets of E which contain 0 and are closed under the sum-of-series function, and take some well-ordering of them. At each element of this (possibly transfinite) sequence, if we have that there is a subset of E which contains 0 and is closed, and has all the previous ones as subset, we can use this way of combining 2 of them in order to get to the next one. But, what to do for the limit ordinals? Well, if we use the same thing about “if we split each term into parts from the two subsets”, if we instead split it into parts from countably many subsets, then the same general proof should still hold. And, that lets us do induction for everything up to the first uncountable ordinal.

Ok, so what if we have a collection of arbitrarily many subsets of E, which each include 0, and are closed under the sum of series function, and then we have a subset of E which is comprised of finite or countable sums of finitely many or countably many of those subsets, then a series of elements from that set, could be split into series, and ... hm, not sure. But, I think this might work. and then given that, one could use the transfinite induction to find a maximal subset of E like that. and then, because the arbitrary products would work, this would show that SerMn would be a complete category.

But for now, I’m going to leave this off there, because I should go to bed. Wow, didn’t even finish reading the first 5 pages of the 33 pages of this paper tonight. Oh well.

[hyperbolicechochamber]

Star Wars The Force Awakens Comic-Con Trailer (We Went To the Desert Edit) 

remember when movies did not have to be advertised to tell you how it was made before it even comes out? I’ll take the prequel trilogy over this nonsense any day