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12:00 AM
these are the definitions i am working with
 
12:14 AM
@Thorgott ok, am finally getting off of work so I’m able to think about this. I see in the case of f=id that I’d be defining h as the constant map to c. So certainly c not in the range is a necessary condition for h to be injective. I guess I’m still stumped in terms of seeing how to argue that f injective and c not in the range get me to the conclusion.
I must somehow be able to show that f^k(c) = f^j(c) means k = j but I still can’t quite see how
 
12:28 AM
suppose $j\ge k$ wlog
can you use injectivity of $f$ to argue that $f^k(c)=f^j(c)$ implies $c=f^{j-k}(c)$?
 
12:54 AM
If $f:M\rightarrow \mathbb{R}^k$ is an immersion an $M$ is a compact manifold with boundary
then does that mean $f$ is an embedding?
If $M$ is without boundary then result is obviously true
 
I would say obviously false
 
I forgot to say $f$ is injective
 
then yeah
 
 
1 hour later…
2:17 AM
@Thorgott OK I think this might get me there but hoping to follow up. Let me consider the corestriction of $f$ to its range and then apply to both sides, so that I get as you say $c=f^{j-k}(c)$ (incidentally, how would you make this precise? induction?). Now we conclude that $j = k$ for otherwise we would have $c \in \ran f$
 
I'm not sure what purpose corestricting $f$ is supposed to serve here?
 
probably nothing. I guess I'll think about making the claim precise by induction. It's a set theory book after all so I really should do that.
Dang, still realizing that I don't have a notion of order
i.e. $j \ge k$ as yet not well0defined
going to try to adapt
 
2:41 AM
you know that the natural numbers are well-ordered, don't you
so you know what order is
 
I do, it's true, but I am hoping to solve exercises with only what I've been given up to that point
I figure that's good practice in general?
 
it might be good self study practice in general, but it severely limits your ability to get useful feedback from third party sources
particularly if it amounts to (as it seems to here) not "how do i do this proof by induction" but "how do i do proof by induction in this particular way"
 
That's definitely true and well taken, so apologies then for that confusion
 
i took a class out of enderton though, and remember doing this stuff the way his text liked to do it
even while/after i was being less fussy in other classes
 
im finding it a remarkably good book at the moment, not sure how you felt/feel about it
 
2:47 AM
yeah, i like it
but, 99% of real life difficulty in induction proofs is figuring out how to structure the argument so that induction is possible (roughly speaking, "figuring out what to induct on"), and i don't know of any textbook that gives a particularly good sense of this
in crummy 'intro to proof' books it is literally always drop dead obvious what you are supposed to induct on, because the problems are basically made up to fit into some recipe
and in something like enderton it's like "wait, do we have the order on natural numbers 'yet' or are we still phrasing everything in terms of the successor operator"
which is certainly a kind of difficulty worth overcoming in induction proofs, but maybe not the most representative kind
 
Ya that's fair
I'm actually finding following the proof I linked very instructive fortunately
 
one thing that might not be too clear from enderton is that it's actually a much deeper book than a random book on naive set theory
 
I would not have (though I should have) thought of defining that as the set to be shown as being inductive
 
in particular, there's stuff in there that basically people who aren't PhD students in set theory probably shouldn't bother learning
 
insofar as it treats the axioms? or what do you mean there?
 
2:52 AM
and its not well delineated away from the stuff thats like "here is a way to encode ordered pairs as sets" and "here is what an inductive set is"
insofar as it treats the axioms in a depth greater than most textbooks, and consequently teaches more of set theory than any average math PhD would need to know
a lot of people expect a book with 'set theory' or equivalent as its title to be a kind of 'just the facts' handbook, and enderton is more than that
 
hmm, if you recall specific instances of what you mean i'd definitely be curious. i'm basically done Ch 4 and the "deepest" things I've taken I think relate to proving that certain sets exist
by using axioms like pairing, union, etc to show that some set contains them, and then axiom schema of specification to specify down to that set
 
okay, chapter 4? you are doing the beginning of the textbook
you're doing the basics now
 
but I don't know if that's deep, and maybe what you're alluding to is later
 
haha correct :)
Ch 5 IIRC will be about constructing the reals formally (my favorite pastime given my analysis book has already done that)
 
2:56 AM
yes, again, that is the basics
 
then 6 on cardinality, and i peeked ahead to 7 and saw some talk about regularity which was cool because it's not in halmos
 
i am talking more about chapters 6, 7, 8, and 9
which mix some stuff that basically anybody shoudl know with deeper stuff
and my whole point above is that it isn't clearly delineated which is which
within the textbook, anyway
 
i will perhaps come back to you on this point l;ater on then if that's OK :)
 
at some point, i think you need to put the set theory down
maybe not quite yet, but sometime soon. we can revisit :)
 
I'll just agree with leslie
 
2:58 AM
the set theory is awfully fun though, i hate to say it
enderton is an undergrad book in the end though, no? surely not the end of the world to power through it
 
its like quicksand for people. they are on their way to something else and they land in set theory and the world just stops
 
I suspect that's because us rookies are drawn to maths given that it promises complete precision, and then you see that you have to go down, down, down... to actually realize that promise. I am content for now going no further than Enderton though
 
it's not the end of the world to power through anything, and i don't want to make it out like it's scary or inaccessible, i just wonder whether everyone bothers to examine why they want to do set theory
 
books are not meant to be read from start to finish
2
that's my general advice
 
it's like if you were learning a language, at some point in acquiring the rules of grammar, and their big exceptions, and the histories of how words or figures of speech developed from other words or languages, eventually you reach a point where you're learning stuff that > 99.9% of native speakers have never heard of and never use
and with something like set theory, you hit that point sooner than a lot of people think
and yeah i actually wonder how many books are actually worth reading start to finish
a lot of standard textbook advice fits the format "read through chapter x and then stop," where x is often only a little more than halfway through
 
3:05 AM
the only book I ever read from start to finish was Halmos
and that was all the set theory I ever needed, too
did it during a spring break once, it's the mathematicians paying your dues to read set theory at some point, I guess
 
But surely something to be said for just enjoying a good book :) I agree some are slogs, but some are just really pleasant. (Baby) Rudin fell in the first category (at least for me), but Enderton I am really enjoying. LADR was another beautiful one
LADR was long ago and I'm so out of practice that i gotta do another LA book sadly, hence hoffman and kunze
 
@leslietownes Baby Rudin: stop after chapter 7.
 
3:26 AM
@EE18 Try LADW
 
have heard great things about that one Soumik
Incidentally, has anyone read the books intro to Statistical Learning or elements of statistical learning?
I know for mathematicians these things are probably looked at as applied statistics/stuff we've known about forever
but figured i'd ask
 
3:43 AM
@Thorgott I think so. I can't help it. I do it anyways. I start from preface and all the way to the end.
 
 
3 hours later…
6:46 AM
even with baby rudin? i don't believe it.
i believe that some people might have hit chapter 8, because i did, despite xander's advice (which is the best advice and the advice taken by my first analysis prof). but nobody has ever gone through chapters 9, 10, and 11 of rudin. you can't. your brain shuts off.
 
7:01 AM
if you finish chapter 11 of rudin you wake up and someone congratulates you for exiting the simulation
 
7:38 AM
🤣
 
8:01 AM
I was evaluating this limit and got answer as 1. but the answer key claims to be 3/2
What is wrong in my method ?
Can anyone please correct me ?
 
8:42 AM
@KavinIshwaran everything
Your method "proves" that every limit is equal to 0 as well
$\lim_{x\to a} f(x) = \lim_{x\to a} \frac{f(x)}{x-a}\cdot (x-a) = \lim_{x\to a}(\frac{f(x)}{x-a}\cdot 0) = \lim_{x\to a} 0 = 0$
@leslietownes because I'd like to get insight into independence proofs and methods such as forcing
 
@leslietownes not with Rudin :)
 
9:03 AM
@Jakobian Ah. I get it. I have seen some questions where they apply limits individually. But didn't know where it is applicable and where it isn't.
 
@KavinIshwaran I don't know what that means
 
@Jakobian suppose if there is a sinx term in an expression and the x tends to zero, they multiply and divide by x and take that particular term to be 1
 
@KavinIshwaran you can't do that unless it makes sense with respect to the rules that limits obey e.g. products, sums, quotients of limits
You would have to be more specific and only then we can discuss if in a given limit it makes sense to make such substitution. In general the answer is no, it doesn't make sense
 
9:18 AM
to be more specific, in a question which goes like this :
limit x tends to zero (tanx - xsinx)/x they multiplied sin x by x and divided it with x (that particular term alone and then LH rule is used)
@Jakobian Oh Ok. I will make a memory note on this.
@Jakobian Thank you for the clarifications :-)
 
The above is supposed to be an example of a sequence that converges in mean but not pointwise a.e. I simply do not understand it, since I believe there are some typos. How can n increase from $2^m$ to $2^m-1$? Also, is $f_4$ correct? Mightily confused.
Trying to figure out what the correct example should be...
 
@KavinIshwaran sounds like wrong reasoning
 
@Jakobian Oh Ok, I will restrict myself from using that
 
9:34 AM
@psie typewriter sequence
f_4 is incorrect as well
26
Q: The Typewriter Sequence

FabianThe typewriter sequence is an example of a sequence which converges to zero in measure but does not converge to zero a.e. Could someone explain why it does not converge to zero a.e.? $f_n(x) = \mathbb 1_{\left[\frac{n-2^k}{2^k}, \frac{n-2^k+1}{2^k}\right]} \text{, where } 2^k \leqslant n < 2^{k+...

 
ok, then $f_4$ should probably be $1$ on $0\leq x\leq 1/4$ instead
and $0$ elsewhere
 
 
3 hours later…
Mad
12:39 PM
https://math.stackexchange.com/a/2335986/695930
How can i show that the sets are equal?
if i pick a discontinuity point, then it is not hard to show it is in the the union of cuts, but how do i go in the other way.
 
1:17 PM
Hi everyone, wondering if someone can point me in the right direction before I resort to asking a question. I'm analysing numerical stability of an operation and can incur in an error that is (number of elements of the vector smaller than machine precision)×ε, which obvs. ≤ (size of the vector)×ε, but this seems super loose?
 
2:13 PM
I'm awake!!!!!!!!!! But at what cost?
 
i hav a 1D manifold which is the real line, R. i define two atlases : Atlas 1 : ψ(x)=x. Atlas 2: ψ(x)=2x,x≥0;ψ(x)=−3x,x<0
why arent these two atlases two different smooth structures?
the first one is a smooth structure becuz it's a homeomorphism onto $R^1$ and there's only one chart so the compatibility of charts holds. the second one is also smooth for the same reason
sorry i meant $\psi (x)=2x ; x\geq0$ and $=3x , x<0$ for the second atlas
 
2:30 PM
Pickin up speed, runnin outta time. Math test aproachin, scary way of life :O
 
@RyderRude they are. But if you equip $\mathbb{R}$ with either of them you get diffeomorphic manifolds
perhaps thats what you meant
 
@Jakobian but the functions that are smooth in one atlas are no longer smooth in the other
im asking this because i read the real line admits only one smooth structure
 
2:45 PM
@RyderRude what are you objecting. Which part of what I said
@RyderRude there's multiple smooth structure on $\mathbb{R}$ but no matter which one you equip $\mathbb{R}$ with, those will be, up to diffeomorphism, your standard $\mathbb{R}$
The important word here is up to diffeomorphism
 
sorry i thought "up to diffeomorphism" meant that the atlases have a smooth transition map between each other
 
Diffeomorphism between manifolds $f:M\to N$ is a smooth map with smooth inverse
You're confusing it with the situation where the identity map on $\mathbb{R}$ would be a diffeomorphism
 
What is editing a tag wiki? I looked it up before asking but found nothing O.o
 
Another example. You can have two distinct topologies on a set, but both spaces to be homeomorphic
It doesn't contradict anything. They would only be the same topologies if we required the homeomorphism to be the identity map
what about another thing, for example, group structures on a two element set
they are all isomorphic, but there is exactly two, depending on which point we require to be the identity element of the group
 
2:56 PM
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. The proof starts by a reducing the proof to simpler special cases. We can assume $f$ has compact support by DCT and moreover that it's positive. I understand that. However, then it is claimed, since $f$ is measurable, there is an increasing sequence of simple functions of, note, compact support, that converge to $f$.
I understand why there's such a sequence, but I don't understand why the sequence can be compactly supported. Why?
 
@YourLordJoyBoy Tags can be given descriptions. You generally shouldn't touch that since if you add a useless tag and give it a description then it will stay forever until it gets moderated
Unless you're trying to give description to an already existing tag
creating tags is at 1000 rep so you couldn't even if you tried, but yeah
@psie they approximate $f$ from below, so if $f$ has a zero then so do they
 
ah, makes sense actually, didn't think that far, thanks!
 
 
3 hours later…
6:07 PM
@ZaWarudo See above, though this applies for pretty hard problems
 
7:02 PM
is there an easy way to prove that the sum of the ideals $V(x^2-y)$ and $V(y^2-x)$ is not the entire ring?
 
7:14 PM
@Derivative ideals in what ring
 
$\mathbb C[x,y]$
 
Is $V(x^2-y)$ the principal ideal generated by $x^2-y$
 
yes
 
Suppose $1 = f(x, y)(x^2-y)+g(x, y)(y^2-x)$. Do you see how this leads to a contradiction?
 
no, I did try
 
7:21 PM
Polynomial on the right has no constant term
 
ah sorry I think I misunderstood
I'm trying to find two complex analytic varieties such that the sum of their ideals is strictly included in the ideal of the intersection
where the ring I'm working with is the ring of holomorphic germs in $\mathbb C^n$ localized at the identity
and your example doesn't do what I want because I want a function that vanishes at the origin
 
its your example
 
how?
 
@Derivative what do you mean how. Its you who gave the example, here. I only proved it to you
 
the example proves the claim about complex analytic varieties and not just the claim about polynomial rings, correct?
I can much more easily take the planes {xy=0} and {xz=0} duh
nevermind
 
8:32 PM
@AlessandroCodenotti What's the relationship between a Dieudonne complete space and Cech complete space?
 
> Theorem 2.25. A subset $A \subset \mathbb{R}^{n}$ is Lebesgue measurable if and only if for every $\epsilon>0$ there is an open set $G$ and a closed set $F$ such that $G \supset A \supset F$ and $$\mu(G \backslash F)<\epsilon.$$ If $\mu(A)<\infty$, then $F$ may be chosen to be compact.
I'm reading a proof of the fact that the space $C_c(\mathbb R^n)$ of compactly supported continuous functions on $\mathbb R^n$ is dense in $L^1(\mathbb R^n)$. Let $A$ be a bounded set. Somewhat out of the blue, it is claimed that from the Borel regularity of Lebesgue measure, there exists a bounded open set $G$ and a compact set $F$ such that $G \supset A \supset F$ and $\mu(G \setminus F)<\epsilon$.
I know the theorem above, but it doesn't say that we can choose $G$ to be bounded. Does anyone have any clue why $G$ can be bounded?
 
the simpler question might be, what is the quickest path in your document to seeing that G can be bounded. i might look at the proof of "theorem 2.25" to see where the existence of its "G" comes from. it might be as simple as tracing through that argument
 
ok, good idea, I'll have a look
 
as a background vibe, you could think about covering of A by "simpler" open sets (e.g. dyadic rectangles, or other sets whose diameter you can control), and convince yourself that if A is bounded and the sets in your cover have controlled enough diameters, the union of the cover will be bounded too
but that might not be the simplest route in your reference
and that is more of a vibe than an argument
 
9:03 PM
I think I found an answer. If $A\subset\mathbb R^n$, then $\mu^\ast(A)=\inf\{\mu(G):A\subset G, G \text{ open }\}$. If $A$ is measurable, we can find an open set $G$ such that $\mu(G)<\mu(A)+\epsilon$ for some $\epsilon>0$, and if $\mu(A)<\infty$, we'll have $\mu(G)<\infty$.
 
@psie If you have an unbounded $G$, intersect it with a ball containing $A$
 
... and that's the cleanest answer
 
ah yeah :) thanks
 
9:19 PM
@robjohn did you get any good pictures of the eclipse?
 
9:36 PM
@AlessandroCodenotti I figured there exist spaces which are Dieudonne complete but not Cech complete, and there are spaces which are Cech complete but not Dieudonne complete
So seems there is no relationship
 
9:59 PM
I see
I've never encountered Dieudonne complete spaces before
 
 
1 hour later…
11:08 PM
@AlessandroCodenotti Do you know about realcompact spaces?
 
11:32 PM
In this question, shouldn't $f_F(x) = d(x,F) = \inf\{d(x,y) \ : \ y \in M\}$ read $f_F(x) = d(x,F) = \inf\{d(x,y) \ : \ y \in \color{red}{F} \}$? This confuses me.
 
psie: yes
 
great, thanks for the quick answer :)
 
thanks for asking something with such a quick answer :)
 
11:47 PM
@psie yes.
 

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