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00:07
i might have been slapped for saying that once
wait, i think i hear a drone outside
ok, i don't want to be suspended again.
basically spent the week in bed after picking up a flu at a super spreader baby shower.
at an influencer's house, no doubt
@copper.hat I spent last weekend in Tucson taking care of my niece while my infant nephew was in the hospital with RSV. Now I am pretty sure that I have the RSV. :(
00:22
did you wear a mask the whole time
a mask?
@XanderHenderson no fun.
i think the last time a virus knocked me out this long (well much longer) was when i had hep-a as a teenager
i am out of any influencer's circle nowadays :-)
@copper.hat Indeed. Though I don't think that I got it as bad as my mother nor my brother-in-law.
hope you all recover asap
And the baby is healthy again, which is a win.
There were a couple of days that were pretty scary.
And I got all of my grades in (20 minutes before the deadline), and my annual professional goals thingy.
excellent
00:28
And I applied for a job in Washington (that I am not going to get, and maybe wouldn't take if it were offered(?), but I needed to update my CV).
good luck!!!
01:10
@BenSteffan after half a dozen iterations of changing the proof strategy due to error or gap, I have managed to prove that $\mathbf{Cat}_{(\infty,n)}$ is cartesian closed for all $n\ge0$
 
4 hours later…
04:52
@SineoftheTime Kramnik or just common people?
@pie No, as Xander said you don't need Fourier to do complex
@Thorgott nice!!!
05:25
Fourier is much cleaner with complex, just fyi.
 
5 hours later…
10:20
Keep math clean: use ℹ️
 
2 hours later…
12:10
@SoumikMukherjee Russian Federation
@copper.hat oh, certainly. But you don't need complex analysis to do it. Just some basics about complex numbers.
@SineoftheTime Sounds like flame baiting by Russian bots to me.
At an introductory level, there really aren't any theorems in complex anal that you need.
Did you introduce imaginary numbers to your algebra 1 students :^)
@think_meaning_buildß nah I don't think so. Btw Kramnik got banned from tournaments with prizes on chesscom
12:24
Yeah I just checked
@think_meaning_buildß What has bots to do with it?
The Russians are notoriously known for their internet bots.
(some conspiracy theorists believe they rigged the US election results :-)
12:50
For those who are interested in professor Hamkins work here is a link to his Reddit question How do you think of the complex numbers?
13:45
The Twitter / X link is deep.
I think this would be a good question to take to MathematicsEducation.SE
14:01
I am trying to prove that $X^{\omega}$ where $X=\{0,1\}$ (i.e, the set of all sequences of $\{0,1\}$) is bijective to the collection of all countable subsets of $X^{\omega}$. I can write an injection from $X^{\omega}$ to a countable subset of itself, by the following method: If $\{x_1,x_2 \cdots\} \in X^{\omega}$, then my corresponding subset will be $\{ (x_1,0,0,0 \cdots), (0,x_2,0, \cdots).... \}$ (This function basically captures where all the 1s occur, so that this is an injection is shown). The direction from the set of all countable subsets of $X^{\omega}$ to $X^{\omega}$ seems tricki
 
1 hour later…
15:23
@nickbros123 the snake like thing that you draw in proving rationals are countable
15:47
@think_meaning_buildß No. The largest difference between MAT 109 and MAT 152 is the introduction of the quadratic formula and complex numbers.
16:26
@SoumikMukherjee im not sure I get the idea
16:47
ohhhhhi i get it now
thanks
@nickbros123 if $A$ is a set of size $\kappa$ and $0 < \lambda < \kappa$, then the set of subsets of $A$ of size $\leq \lambda$, $[A]^{\leq\lambda}$ is of size $\kappa$
@Jakobian unfortunately I don t have enough set theory knowledge to understand this
actually what I wrote is not true
but what is true is in above screenshot
and $|[\kappa]^{\leq\lambda}| \leq \sum_{\xi\leq \lambda} \kappa^\xi \leq \lambda\cdot \kappa^\lambda = \kappa^\lambda$
In this case $A = X^\omega$ is of size $\mathfrak{c}$ and $\lambda = \aleph_0$ so that $|[A]^{\leq\aleph_0}| = \mathfrak{c}^{\aleph_0} = \mathfrak{c} = |A|$
17:21
well itll take me a few days to understand that, havent gotten to that part in munkres yet
just working munkres right now fyi, not doing set theory proper
18:04
@nickbros123 you don't write a sequence like $\{x_1, x_2, ...\}$
you write it like $(x_1, x_2, ...)$
An injection of $X^\omega$ to countable subsets of $X^\omega$ is easy, you just send $x$ to $\{x\}$
For the other direction, you might take $Y = X^\omega$ and $Z = Y^\omega\cup \bigcup_{n\geq 1} Y^n$ and using a well-order on countable subsets of $X^\omega$, obtain an injection of them into $Z$.
now it might or might not be easy for you to see by an elementary argument that $|Z| = |Y|$
18:19
@XanderHenderson Thanks for the reply.
note that you probably do need well-ordering at some point, and that the argument is probably not very different from the proof above about embedding into a product
Soumik's argument, taking inspiration from $\mathbb{Z^+}\times \mathbb{Z^+}$ seems to be intuitive. I can do lexicographic ordering on these sequences no?
@nickbros123 and what is Soumik's argument
basically taking any countable subset of $X^{\omega}$, and giving it the lexicographic ordering (based on 1>0), and listing them out as such (row i is the i-th sequence): $$\begin{matrix}(x_{11} \ x_{12} \ x_{13} \cdots ) \\ (x_{21} \ x_{22} \ x_{23} \cdots) \\ \vdots \end{matrix}$$. Now I make the sequence $(x_{11},x_{12},x_{21}, x_{13},x_{22},x_{31}, \cdots)$.
Thats not an injection
do you mean countably infinite subsets or countable subsets
in the latter case, this argument could be replaced by first observing that $|X^\omega| = |X^\omega\times (\{\infty\}\cup \mathbb{N})|$
18:36
right, I was thinking about countably infinite subsets
for this you'd need that the ordering on $X^\omega$ well-orders the countable subsets
but this is false
so this argument still fails
even if you well-order $X^\omega$, the well-ordering on countable subsets won't be of the order-type of $\omega$
so that it won't be a sequence
so this argument just can't be formalized, its wrong
You could try give each countably infinite subset a distinct ordering, but then how can you guarantee injectivity
not exactly sure if this is what Soumik had in mind, maybe, who knows, the suggestion was so vague that it could have been anything
either way its clear that what you are trying to do doesn't work
19:47
@Jakobian if we have a well order on $X^\omega$ and consider a countable subset of it, then can't we order the elements of the subset in lexicographic ordering using the well order of $X^\omega$?
@SoumikMukherjee what do you mean by that
20:04
I have the set of all sequences of 1s and 0s. I well order that set. Now if I have a countable subset of the original set then don't I get an induced order on that subset?
yes, and?
did you read what I was saying
1 hour ago, by Jakobian
even if you well-order $X^\omega$, the well-ordering on countable subsets won't be of the order-type of $\omega$
@Jakobian I was wrong. It won't be injective.
@Jakobian I just read
 
3 hours later…
22:49
ew
overleaf going the way of all things
has been for a while, too
23:43
just lost my mind over a proof for 15 minutes until I realized the author just erroneously flipped an inequality sign in a central definition (and a bunch of auxiliary indices as a consequence)

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