Consider the proof of the statement that every set's cardinality is strictly less than that of its power set (let the set $A$ be nonempty below). We know there exists an injection, now assume there exists also a bijection $f:A\to\mathcal{P}(A)$ (we're going for a contradiction). Then my text claims we can form the set $$\{a\in A:a\notin f(a)\}.$$This set is confusing me. Why can we form it? Could this set be empty?
ok, so I guess the formation of this set has nothing to do with the fact that we're assuming $f$ is a bijection. As $f(a)=\{a\}$ shows, we'll just get the empty set.
The Jews won't recognize me, as I've never been bar mitzvah'd, and now the mathematical community is going to shun me for not being confused by the empty set?! Who are my PEOPLE?!
in trying to prove that every open set $G$ in $\mathbb{R}$ is the countable union of disjoint open intervals, I started with saying $R$ is separable. and I took all rational points of $G$ and enumerated them $\{r_1,r_2 \cdots\}, $and took each $\varepsilon_j$ ball around $r_j$ to be my countable basis. We have $$\cup_{j=1}^{\infty}B_{\varepsilon_j}(r_j)=G $$. Now my idea was that, if two intervals overlapped, they are a part of a bigger interval. Loosely speaking,
I wish to keep doing this until I stop at a disjoint collection
how do I mathematically enforce this though
when I say "two intervals overlap, they are part of a bigger interval", i mean that they are part of the union of them both, which would be an interval in itself
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.
This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as poss...
you're noticing aspects of a key difficulty or feature of this problem, which is that you do need something that's kind of specific to intervals as intervals, and not just as (for example) examples of "a particularly nice countable basis for a topological space"
e.g. in any R^n, the open balls with rational centers and rational radii are a countable basis for the usual topology, but it is not generally true that an open subset of R^n is a disjoint union of a countable number of such balls
nick yeah i think what that question/answer illustrates is the enormous number of ways deal with what is basically the same idea, not just via notation/symbols (although there is a lot of that) but also what notions they use to get at it
the concept that sticks in my mind is that in R, "open connected subset" = "open interval" (suitably interpreted - you have to include R and "half-infinite" intervals as "intervals") so that as some answers/comments point out, the maximality stuff is giving you the decomposition of your G into its connected components
which is something that you can also do for open subsets of more general connected spaces like R^n, without the expectation that the components themselves will look any particular way (e.g. one obviously does not have "open connected subset" = "open disc" in R^2)
i think there's at least one answer relating this phenomenon more generally to order topologies (of which R is an example, but e.g. R^2 is not) but yeah
Here it is, a purported proof that every even integer is the difference of two odd primes: https://math.stackexchange.com/questions/4966037/proof-of-the-fact-that-delta-bbbp-geq-3-sqcup-pm-1-2-bbbz
Can anyone find counter-claim to any argument used? It's quite simple, and elegant. I should though in retrospect maybe have used that $\lim$ should commute with $\cap$. That would be an alternative approach.
Because $\cap$ is supposedly an inverse-limit of something, and inverse-limits commute with themselves, but I don't know if that means you can do $\lim (A \cap B) = \lim A \cap \lim B$ whenever $\lim A, \lim B$ exist.
Actually, the proof is of every even integer is the difference of either two primes or a prime and $\pm 1$.
You are approximating the limit with (hopefully simpler) objects. I find inverse limit much clearer than direct ones, at least as far as topological spaces are concerned
@AlessandroCodenotti do you have any (motivating) concrete example in topological space?
@Thorgott In AT, homology class is (relatively) intuitive and has a geometrical representative but cohomology class does not have a geometrical description though one can give a more concrete description using differential forms.
If $C$ is the circle $|z| = 2$ taken with positive orientation, then
$
\int_C \frac{2z \, dz}{z^2 + 2} = 4\pi i.
$
The solution given in the book goes like this:
Solution Using partial fractions, the integral given can be written as
$
\int_C \frac{2z \, dz}{z^2 + 2} = \int_C \frac{dz}{z + i\sqrt{...
Folland writes in his text that $X$ is countable if $\mathrm{card}(X)\leq \mathrm{card}(\mathbb N)$ and then says it is countably infinite if $X$ is countable but not finite. Then, in the following proposition, he proves that if $X$ is countably infinite, $\mathrm{card}(X)= \mathrm{card}(\mathbb N)$. He says it suffices to consider $X$ as an infinite subset of $\mathbb N$. Why can we do that?
Above, $\leq$ means there's an injection and $=$ means there's a bijection.
that injection induces a bijection between $X$ and its image, which is a subset of $\mathbb{N}$
so if you want to find a bijection between $X$ and $\mathbb{N}$, you can equivalently find a bijection between that subset of $\mathbb{N}$ and $\mathbb{N}$ itself
(having equal cardinality is a transitive relation)
@nickbros123 the property you are looking for is local connectedness. It's not true that every open set has open connected components, unless your space is locally connected. Then you can decompose every open set as disjoint union of open connected sets. If you have a basis of size $\kappa$, you can then show that the union must be of size $\leq\kappa$. For $\mathbb{R}$ it just so happens that convex and connected sets are one and the same, feature of its one-dimensionality
By axiom of choice for two sets $X, Y$ either $\text{card}(X)\leq \text{card}(Y)$ or $\text{card}(X)\geq\text{card}(Y)$. But then if we define infinite and finite then what you want to prove might just be a definition
@Xander I asked a question on MSE a couple months, cross-posted a slight variation of it on MO a couple days ago and it was answered there. I then also came up with another answer to the original version of the question I posted on MSE (the other answers I got on MO do not answer the original version of the question posted on MSE) and posted this as an answer on MO. I now want to also re-post it on MSE to get the question off the Unanswered queue. Should I make the answer CW?
@nickbros123 well yes, technically the definition of an interval is that of a convex set, which just happen to be of the form $[a, b]$ or $(a, b)$ or $[a, b)$ or $(a, b]$ for some $a, b\in \mathbb{R}$ or $a = -\infty, b = \infty$ when appropriate
but not everyone defines intervals this way, I suppose
convex set in the sense that if $a, b\in C$ and $a < c < b$, then $c\in C$
it just so happens that the order-theoretic convex and geometric convex coincide
maybe an interesting question would be, in a vector space $V$ over a field $k$, what would it mean for $C\subseteq V$ to be convex
I'd think that we'd need an order structure on $k$, and then define it just like for $\mathbb{R}$, that if $x, y\in C$ and $t\in [0, 1]_k := \{s\in k : 0 \leq s \leq 1\}$, then $tx+(1-t)y\in C$
@Thorgott The general policy on the SE network is that questions should not be duplicated across the network. If it is the same question (or if the same answer addresses it), the unanswered version should be deleted, since it already has an answer on the network.
that would remove the possibility for people looking for that or a similar question on MSE to find it there, which undermines the site's "repository" philosophy
Well, my intention was to get it off of the unanswered list by supplying an answer. I think it's a good question, so I don't love the idea of deleting it. (And, of course, I followed all the policy recommendations I could find when I cross-posted it in the first place.)
@Thorgott None of what you are saying contradicts what I have said. Have I said it is a poor question? (No---I don't even know what question you are talking about). Have I said that getting it off of the unanswered list is not an admirable goal? (No---I don't have a strong opinion here).
According to your own description, the question has already been answered, just on MO. So the answer exists.
@Thorgott To be honest, there are duplicates of the same questions all over stack exchange, so if you were to post an answer it wouldn't really matter that much
Note, also, that I am just giving you the standard SE line: if a question has already been answered on the network, then copy-pasting the same answer is, technically, against the rules.
You asked what the policy was, and I have told you what it is.
@Jakobian That was certainly the impression given. And note that "copy-pasting" needn't mean literally copying and pasting (though, again, that was the impression I got).
The official SE policy is that significantly similar questions should not be cross-posted on multiple sites. Period.
If two nearly identical questions are asked on multiple SE sites, and one of them has an answer while the rest do not, the appropriate action is to delete the unanswered one. This happens all the time.
It is possible to migrate a question from one Stack Exchange site to another by closing, but if I have a question that I think is on-topic for multiple Stack Exchange sites, is it OK to post it on both (multipost)?
For example, I have a question that's earned me the tumbleweed badge on SO and I...
I already broke rules (technically) by cross-posting to begin with and I don't intend to delete my old question either, but I also won't go further and post an answer when I'm not supposed to as I've now been informed. The MO version of the question is of course linked on the MSE question, so a pointer to the answer exists, which is good enough for me.
@Jakobian It is one thing to break the rules when you are ignorant of the rules. It is another thing entirely to ask what the rules are, be told that you are planning to do something which is against the rules, and then do it anyway.
The reality is that I, personally, have other things that I would rather focus my attention on, and that I, personally, would be unlikely to take any action towards a cross-network duplicate being answered. But the policy remains the same, and encouraging other people to violate the policy is not a good look.
I have no idea what a "good luck" is in this context. If you need help with the idiom, I can be more clear: don't tell people to break the rules. This is a good way to get suspended.
@Xander For future reference, if I have a question on MSE that has been left unanswered for some times and I wanna post it on MO (cause I think it is appropriate in substance and that site potentially has more users knowledgeable in the niche that my question falls into), then the most appropriate course of action would be to delete my old question and simply post a new one?
@Jakobian I think this is because I read text primarily using my "inner voice" so to speak. I think writing text is similar. And then I translate what it says to text, and this sometimes fails when words have similar sounds
Let $\varepsilon$ be an open set of $\mathbb R$, and $U:\varepsilon\to\mathbb R$ be a function. If $\delta\in\varepsilon$, we say that $U$ approaches the limit $x$ near $\delta$ if for every $f>0$ there is an $a>0$ such that, for all $V\in\varepsilon$, if $0<|V-\delta|<a$ then $|U(V)-x|<f$.
Yes, indeed, but I wouldn't like to give only this subject, even if it requires a lot of knowledge even at a theoretical level
The oral exam is not mandatory but I would still like to do it
I mean, now I'll have to prepare the oral part too, I already know some things, but I still have to see the demonstrations
I remember the exercises part of algebra and geometry quite well, I should do the theory well here too, then I have physics 2 but I did almost only exercises, what advice would you give me?
If $C$ is the circle $|z|=2,$ taken with positive orientation then $\int\frac{2z}{2+z^2}dz=4\pi i.$
I tried solving the problem as follows:
Let $I=\int\frac{2z}{2+z^2}dz.$ Then, $$I=\int_C \frac{1}{z-i\sqrt 2}dz+\int_C\frac{1}{z+i\sqrt 2}dz=4\pi i.$$ This is because, for any complex number $z_0\i...
I meant that Neiman is going to the houses of chess players to challenge them, earning fide points, but the thing is that he is the one who chooses who to challenge
thomas: your "Log" function there is not continuous on that whole range of t. can't be chosen continuous. so it won't work as an "antiderivative" for FTC formulas
thomas: which is also what sine is getting at at "Log" being a "multivalued function" but maybe a more precise way of looking at it. you could evaluate that integral for smaller intervals of t with specific choices of Log to which the FTC does apply, and then get the right answer by adding those together.
But the result, although wrong, can be interpreted in multiple ways as correct. For example, you can define the value of a series $1+a+a^2+...$ to be $\frac{1}{1-a}$ for not only $|a| < 1$ but any $a\neq 1$. In particular, $-1$ for $a = 2$
See e.g. Divergent series by Hardy
Approach above is by analytic continuation, but there's also other approaches where we could possibly make sense of assigning a non-infinite value to the series $1+2+4+...$
In the above reasoning, all you've used is that we assume that our summation procedure can assign a finite value to $A$, and that it assigns value $c\cdot \sum_n a_n$ to $\sum ca_n$
any summation procedure that does the above two things will assign value $-1$ to $1+2+2^2+...$
the latter condition is reasonable - we want our summation procedures to be linear
oh and also that it assigns value $a_0+\sum_n a_{n+1}$ to $\sum_n a_n$, of course
as Hardy tells us, linearity and property (C) is something you almost always want from your summation procedures, although not all of the common ones satisfy (C) according to Hardy, so perhaps $1+2+2^2+...$ won't have value $-1$ under them (or it can just be infinite)
and of course it'd be natural to expect that $\sum a_n$ is the same as what we usually mean by the symbol when the series is convergent
Those type of summing procedures are called regular by Hardy
and they are called totally regular when series summing to $\pm \infty$ still sum to $\pm\infty$ under that summation method
so here any method of summation that assigns a finite value to $1+2+2^2+...$ wouldn't be totally regular
Consider proving $\operatorname{card}(\mathcal P(\mathbb N))=\operatorname{card}(\mathbb R)$. In Folland's text, he defines the function $$f(A)=\begin{cases}\sum_{n\in A}2^{-n}&\text{if }\mathbb N\setminus A\text{ is infinite}\\ 1+\sum_{n\in A}2^{-n}&\text{if }\mathbb N\setminus A\text{ is finite}.\end{cases}$$I have a hard time understanding this function and how it is injective. First, why are we treating infinite and finite cases separately?
$f$ is supposed to be an injective map $\mathcal P(\mathbb N)\to\mathbb R$.
and, i would say, you are treating the finite and infinite cases separately because that is how folland decided to define f. it isn't helpful to question the definition if the exercise doesn't ask you to.
you might as well ask 'why did folland define this f and not f(A) = [some thing i just made up].' it's really the same level of fighting the premise of the question
I understand that there's a natural bijection between $\mathcal P(\mathbb N)$ and $\{0,1\}^\mathbb{N}$ given by $A\mapsto\chi_A$. Is this of any relevance?
you want to treat elements of $\mathcal{P}(\mathbb{N})$ as binary sequences, if this binary sequence is infinite then put it in $[0, 1]$, if not then put it in $[1, 2]$
this is because numbers can have two binary expansions, one infinite and the other finite
yes, the function which would map a sequence to its binary expansion wouldn't be injective, the map which sends a sequence to its binary expansion has fibers (as in $g^{-1}(y)$ for some $y$ in the image) of size $\leq 2$
but it can be that they are exactly $2$. That's why you need to separate those somehow
"Let $\mathcal{F}\mathrm{in}^{\mathrm{i}}$ denote the category whose objects are finite sets and whose morphisms are injections, and let $\mathcal{F}\mathrm{in}^{\mathrm{i}} \leq n$ denote the full subcategory of $\mathcal{F}\mathrm{in}^\mathrm{i}$ spanned by those finite sets having cardinality $\leq n$."
why wouldn't you put the $\leq n$ in the subscript like everybody else
it's coming from J. "coCartesian" Lurie so I can't convince myself it's a typo
at least he spells it "cocartesian" in Kerodon
but on the other hand he swapped out "having the right lifting property for collection of morphisms $T$" for "being weakly right orthogonal with respect to $T$," which to me is pretty much infinitely less clear
@Jakobian here's one other thing I don't quite understand about the definition of $f$. Why is it defined in terms of $\mathbb N\setminus A$ and not simply $A$? I.e. isn't it harder to tell when $\mathbb N\setminus A$ is finite versus $A$ being finite?
like, he tries generalizing a statement about a map $f$ from the case where it's a cofibration to a more general case where it need not be a cofibration and the argument involves introducing a commutative square $gh=h^{\prime}f$ and then claiming that $gh$ is a cofibration...
@psie forget what I asked here. I was just curious if the following observation is correct; if $A$ is finite, it goes in $[0,1)$ and if it isn't, in $(1,2]$, right?
psie: not quite. it is true that if A is finite then N \ A is infinite and so you would use the first formula to compute f(A) and land in [0,1). but for example if A is the even numbers, then A is infinite, and yet N \ A is also infinite and so you would still use the first formula to compute f(A) and still land somewhere in [0,1).
there is a kind of vague duality between A and N \ A, but "N \ A is infinite" is definitely not the same thing as "A is finite"
i would think of the function as an abstract machine that folland has given you, it happens to distinguish between whether A has finite complement in N or not, so that is just something that folland made up and you will have to deal with.
finite sets do not have finite complement in N, but some infinite sets also do not have finite complement in N.
so i guess that's maybe one first step to understanding the definition. the cases are about the cardinality of the complement of A in N, and are not simply reducible in the most tempting way to cases about the cardinality of A itself.
you do already have understanding of f, which is that 'the first case' of the definition is going to turn out stuff in [0,1) and 'the second case' is going to turn out stuff in (1,2]. that is a very helpful observation. there was just some blurring around what those cases actually were.
so if i have x and y and f(x) = f(y), i know from this understanding and the disjointness of those two intervals that either x and y were both such that they landed me in 'the first case,' or x and y were both such that they landed me in 'the second case.'
i really dunno how folland is expecting you to approach this. where the 1s appear after the 'binary point' in the binary expansion of f(x) tell you which natural numbers are in x, so it's just proof by duh if you can take that for granted.
my own mind's approach would be, if x and y are co-infinite subsets of N and they aren't the same, there's a least element of N that's in one but not in the other, and then f(that one) is going to be larger than f(the other one) because of how summing up powers of 2 works.
where you're only using that they're both co-infinite to ignore the cases in computing what f( ) is going to be
ah. i took a class out of folland but we must have skipped that chapter. i generally like the book but do not recall loving the exercises very much. our instructor mostly wrote his own
@leslietownes I have a question about your argument here, which I cannot convince myself of. If x and y are co-infinite subsets of N, then indeed, if they are not equal, one has a least element of N that's in say x but not in y. But x could have only two elements say, and y many more than two. So would the sum of 2^-n for n in x still be greater than that of y?
yeah the idea is simplest say if 1 si in x but not in y. if 1 is in x then f(x) is going to be at least 1/2, and none of the shit in y will ever get it up to 1/2 because you'd need to put all of the other shit in y to get it there and then N \ y wouldn't be infinite.
if you remove the profanities from that argument it is closer to a formal argument
i guess i misspoke when i said the only use of x and y both being co infinite is in choosing the formula for computing what f( ) is going to be. you're also using it here
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.
This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as poss...
