Mathematics - 2024-09-25 (page 1 of 2)

VulpesInculta

00:02

Okay I will give that a try.

Well that worked but the answers are differen.

leslie townes

degrees vs. radians?

VulpesInculta

Used degree setting for both sides input

@leslietownes would you like the question in full>

Ben Steffan

Is $\mathrm{FinSet}_*$ isomorphic to $\mathrm{FinSet}_*^\mathrm{op}$?

Apparently about half of people like to define Segal's category as the skeleton of the former (notably this is the case in HA) and half as the skeleton of the latter, and then both sides claim that Segal's category can be described as having finite sets as objects and partial functions as morphisms.

If this is true, then the two versions of Segal's category are isomorphic, hence $\mathrm{FinSet}_*$ and its opposite category must be equivalent (which I should have asked about in place of "isomorphic").

This seems... unlikely to me, because if it were true then why bother with $\mathrm{FinSet}_*^\mathrm{op}$ to begin with?

"...he yells into the void" :^)

Jakobian

00:50

@BenSteffan there is no initial object in $\text{FinSet}_*$ right

Ah no sorry.

Ben Steffan

any one-point set is a zero object

Jakobian

@BenSteffan I think this would be true because a finite-dimensional vector space is isomorphic to its dual.

Jakobian

01:15

while I have no idea what I am reading, I've found this for example math.stackexchange.com/questions/3772422/…

14

Q: Where does Segal's category come from?

Segal's category $\Gamma$ is the skeleton of the category $\text{FinSet}_{\ast}$ of pointed finite sets. It is used to write down $\Gamma$-spaces, which are functors $\Gamma \to \text{Top}$ satisfying some conditions, and which model infinite loop spaces. I would like to be able to tell myself a ...

convention?

Thorgott

02:24

@BenSteffan no, right? smash product distributes over wedge, but not the other way round

Obliv

02:45

What does bounded mean again, for sequences?

I know convergence implies boundedness. I think it means that it's a closed or open interval with a finite number on each end.

this seems impossible

maybe if $(b_n)$ was divergent.

Jakobian

@Obliv A sequence $(a_n)$ of real, complex numbers, vectors in a normed space, is called bounded if there is a natural number $M$ with $|a_n|\leq M$ for all $n$

@Obliv it is impossible

Obliv

How would I justify it? Are the bounds of $(a_n-b_n)$ simply the bounds of $(a_n)$ minus the bounds of $(b_n)$?

Jakobian

I guess more generally, a set $A\subseteq X$ of a metric space is called bounded if there is $x\in X$ and $r > 0$ such that $A\subseteq B(x, r)$ i.e. $A$ is contained in a ball

@Obliv For each natural number $M$ you can find $n$ so that $|a_n|\geq M$. But $b_n$ being convergent, there exists $M'$ with $|b_n|\leq M'$ for all $n$. So $|a_n-b_n|\geq |a_n|-|b_n| \geq |a_n|-M'$ can get as large as you want

Obliv

Strange how $\frac{1}{n}$ goes to $0$ yet the harmonic series doesn't converge. Too bad we won't go over series in this class.

@Jakobian ah, makes sense. Thanks.

Jakobian

I suppose boundedness can be also defined in any uniform space. If a uniform space is given by a family of pseudometrics $\mathcal{D}$ then $A\subseteq X$ is bounded if for every $d\in\mathcal{D}$, the $d$-diameter of $A$ is finite.

Or in another direction, given a topological vector space $X$, $A\subseteq X$ is bounded when for every open neighbourhood of $0$, $U$, there exists $a > 0$ with $A\subseteq a\cdot U$.

And, I think, the two approaches are equivalent for locally convex topological vector spaces (I suspect this is the only instance when a topological vector space can be given a uniformity)

Pizza