« first day (2085 days earlier)      last day (2929 days later) » 

5:00 PM
Use \widecheck
 
This doesn't work.
 
Usage : \widecheck{f}
 
5:28 PM
0
Q: Show that a sequence of uniformly bounded continuous functions with Lipschitz condition is pre-compact in the space of bounded continuous functions

Jessy CatI am attempting to solve the following problem: Let the sequence of continuous functions $\{x_{n}(t) \}_{n=1}^{\infty}$, $0 \leq t < \infty$ be uniformly bounded on $t \in [0, \infty)$ and on each bounded interval $[0,A]$ also satisfy the Lipschitz condition: $\forall t_{1}, t_{2} \in [0,A]$...

 
5:41 PM
Anyone into linear algebra here?
 
user147690
@ValerySaharov Depends
 
user147690
Actually I really shouldn't get distracted sorry :P
 
I'm impressed by SE. They changed their ToS without becoming more evil. That's extremely rare for companies/websites in general.
@ValerySaharov I know what a vector is.
 
6:10 PM
Hey guys, what's the QPD over here on Math.SE? Seems incredibly high just looking at the question feed.
 
it's high
 
user147690
It's probably not constant valued, but we can give an average?
 
yesterday's looks to be approximately 800
 
Wow, that is high. Thanks, I was just curious.
 
@Owatch source?
 
6:14 PM
I know that Area51 shows basic stats like QPD for beta sites. Is there anything like that for graduated sites that you know of?
 
user147690
Yep, you can even run your own scripts on the data.
 
@mike going to a symplectic topology seminar today. fingers crossed that i get something out of it :P
 
what's it about
 
user147690
Data on all of the graduated exchanges probably @JPhi1618 data.stackexchange.com
 
user147690
6:15 PM
Oh there you go
 
user147690
Looks like all exchanges in general, graduated or not
 
something on dehn twists
 
user147690
Number of questions per day
 
@AlexClark, Thanks! I'll have to check out some queries to see how that works. Looks very interesting.
 
6:18 PM
"Let $[G,G]$ denote the commutator subgroup of a group $G$. For $x$ in $[G, G]$, we define the commutator length $cl(x)$ of $x$ to be the smallest number of commutators in $G$ whose product is equal to $x$. The stable commutator length $scl(x)$ of $x$ is the limit $scl(x) = \lim_{n \to \infty} cl(x^n)/n$. By providing explicit factorizations of powers of Dehn twists as products of commutators, we give a new upper bound on stable commutator lengths of Dehn twists."
 
if they didn't say something like "generalized dehn twists", it's probably about surfaces
yeah, that's goign to be about mapping class groups of surfaces
sounds interesting tho
 
i probably won't understand much of it
 
Can anybody think of an unordered basis $\{ x_1, \dots, x_m \}$ where $(x_1, \dots, x_m)$ isn't a basis?
 
do you know what a dehn twist is?
 
I've been trying to get my head around these but I have no idea what's going on.
 
6:20 PM
@Kari An unordered basis is precisely a set of vectors that form an ordered basis if you order them
 
topologically, yeah
 
that's all this talk will be
 
That's what I thought but my set of lecture notes suggests that we find an example of such an ordered basis, @TobiasKildetoft. :-/
 
mmkay. i know about them from how riemann surfaces transform under monodromy
 
It's impossible, right?
 
6:21 PM
@Kari Yes, unless they define things in weird ways
 
'mapping class group' isn't a term i'm familiar with, though, so :/
 
If $X$ is independent and spans $G$ (where $X \subset G$) then $X$ is an unordered basis.
That's how it's been defined for me, @TobiasKildetoft.
 
What is it with people and their need for questions to have "context"?
This doesn't make sense to me.
 
@TheGreatDuck There are good reason to require context
@Kari Right, then it is not possible
 
i'd chalk it up to "need for questions that aren't homework"
 
user147690
6:23 PM
@TheGreatDuck Example question that you disagree with concerns over context?
 
or, more charitably, questions where, while motivated by homework, the poster has clearly done a certain amount of due diligence
 
Like what? I mean one could try to use it to prove somethings not homework, but context doesn't show anything about homework issues.
 
user147690
I don't see how it is hard to provide context
 
@AlexClark There's tons of questions with 5 or 6 answers where the post has been closed.
 
@TheGreatDuck context also provides the information necessary for people to figure out what sort of answer will actually be helpful
 
user147690
6:25 PM
How could one be unable to write any context?
 
But context says
""This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – RecklessReckoner, T. Bongers, Harish Chandra Rajpoot, Shailesh, Claude Leibovic"
"any attempts you have made to solve it"
 
@Semiclassical diffeomorphisms form a group, right?
 
@TheGreatDuck Yep, that is a good explanation of what should be there
 
so, everyone is required to have tried solving before asking?
that makes no sense
 
user147690
Well I would certainly hope so
 
6:26 PM
i mean sure, there are obvious silly cases.
like arithmetic
 
user147690
Not just for our sake, but the students sake
 
"but for the students sake"
 
@Semiclassical mapping class group is (as far as I recall) just diffeomorphisms modulo isotopiy
 
how is that homework under /any/ circumstances?
 
user147690
It's not about being homework. The question is pretty bad
 
6:28 PM
@TheGreatDuck How could you ever think that would be anything but homework?
 
Floor function isn't used in classes anywhere
 
lol
 
@TheGreatDuck Of course it is.
 
and it certainly isn't anylized graphically
 
@TheGreatDuck Once again, yes, of course it is.
 
6:29 PM
OP is probably truly and honestly inspired by the beauty of the floor function
 
^
besides
it's got 5 answers
 
user147690
LMAO @MikeMiller
 
@TheGreatDuck That is a terrible reason to keep a question open
 
user147690
Honestly I doubt it isn't a near duplicate
 
it's already been answered and accepted. It's clearly a decent question. Otherwise it wouldn't be answered.
 
user147690
6:30 PM
On top of being bad
 
how is it "bad"
it asks a question
short and simple
 
user147690
I mean even the title was without thought
 
user147690
"floor functions"
 
Ok I suppose that wasn't the best choice.
but the question itself is clearly understandable.
 
@TheGreatDuck That is not what makes a good question
 
6:32 PM
But you have yet to say what makes it a bad question.
 
it's unclear to me why either side of this argument is engaging with the other
 
@MikeMiller Good point
 
Excuse me?
Are you trying to say I'm out of line.
 
no, i'm saying neither of you are going to convince the other
 
user147690
He's saying neither party will change their mind
 
6:33 PM
oh
it wasn't meant to be an argument
i was just asking why questions that have already been answered are getting closed
that doesn't make sense
the purpose of closing is to reject a question from th site
 
user147690
Bad title, doesn't show any attempt, doesn't provide any information of what they actually don't understand, doesn't give us any motivation to even help them, doesn't give a good example for what a question on the site should look like
 
"doesn't give us any motivation to even help them" since when does an asker have to justify getting their question answered?
 
since the introduction of closure and deletion buttons :)
2
 
user147690
If you are going to single out one point, and ignore the rest, that isn't really reasonable
 
the purpose of closure and delete buttons is to remove content that is innapropriate, not just some question you don't think is worth answering.
 
user147690
6:36 PM
Anyway I have to get back to work, read the rest of my points
 
user147690
Gives us motivation to even help them is probably sufficient not to be closed was how you should of read that
 
"Doesn't show any attempt" this is a Q&A site. There's no reason a person needs to give an attempt. They ask and get answered. Nothing more, nothing less.
 
@TheGreatDuck Except this is not just some Q&A site (to see why this is a good thing, spend a bit of time on Quora)
 
"doesn't provide info on what they don't understand" Ok, that I can see, but that depends. One shouldn't need to include that in an answer.
@TobiasKildetoft Fair enough. It's just weird to see people targeting well trafficked and up-voted answers with closures for no context. After all, the best metric of a question is whether or not people were able to effectively answer it. If it was, then there's no real reason to mess with it.
 
@TheGreatDuck No, that is not a good metric for quality at all.
 
6:40 PM
I mean, why even waste their time patrolling it? It'll die on its own if it isn't popular. (And if it is popular then it must be a god question)
*good
(Or at least thought-provoking enough to not necessarily need context.)
 
Someone to help out a little?
0
Q: Approximate spectral decomposition

Valery SaharovI am interested in effective computations in finding approximate spectral decompositions in some suitable format. Let $A: H \rightarrow H$ be a Hermitian operator on an $n-$dimensional Hilbert space $H$ with the spectrum $\{\lambda_1, ... \lambda_m\}, m \leq n$. Then, $A$ can be decomposed as: ...

It's embarassing how this question is being ignored
 
My brain just liquified.
 
user147690
@ValerySaharov Embarrassing for who and why?
 
Why is it embarrassing?
questions are ignored all the time
that's why we have the crusade of answers
 
Well, since this problem seems to have been addressed somewhere for sure
 
6:45 PM
@TobiasKildetoft Thanks!
 
I just couldn't find a good reference
 
nobody says that answers have to be about new research by the user
if it were, this site would be eons smaller
 
new research?
why
 
?
Nobody says the site needs to only contain questions that are previous unanswered.
 
6:52 PM
I didn't say it either
 
7:25 PM
@MikeMiller Thinking. No dice yet.
What I'd want to do is to lift the two loops up to the universal cover, make the bound a disk by Schoenflies. But the intersections are butting in. I need to get a way to remove them.
 
lift the two loops?
 
Yeah, lift them up to paths in the universal cover $\Bbb R^2$.
 
Don't really understand the idea, but godspeed.
 
I guess I am really thinking of homotopy rel some basepoint in this case.
There are just way too many assumption I am making, and way too many problems there could be.
Hi @Krijn.
 
@BalarkaSen Hey! Symplectic Geometry today
 
7:30 PM
Turns out it's not trivial why blowing up successively resolves singularities after all.
 
@BalarkaSen Isn't that like a big theorem or something?
 
There is this invariant called the "arithmetic genus", which is the inductive thing you use to push through the argument. Blowing up reduces arithmetic genus.
@TobiasKildetoft Oh I'm talking of curves.
 
@BalarkaSen Ahh, how'd you find out?
 
Asked an algebraic geometer.
He referred me to Fulton's algebraic curves. Apparently that has an elementary proof (using blowups). I'll read it when I get back home.
I mean, there are much simpler ways one could get away with using normalization. But the geometric content of that construction is not clear.
It's a very algebraic thing.
 
7:34 PM
Ah well, don't let it bother you too much
 
And, well, at some point I'd maybe want to read resolution of singularities. So knowing what happens for curves is probably better before getting into higher dimensional varieties.
@Krijn Why? The point of doing mathematics is to let things bother you and when you figure out why it's bothering you, whine about nothing bothering you!
At least it works like that for me.
 
I'm more of the bigger picture than the details
 
@BalarkaSen Hm? I don't agree. Arithmetic genus is just a (signed) count of the dimension of spaces of holomorphic forms.
I guess maybe I don't know what it is for singular things?
 
@MikeMiller For non-smooth things, whoops.
Of course for smooth things blowup is an isomrphism, so genus is the same.
(arithmetic genus = topological genus for smooth things, IIRC)
 
I was saying that I don't agree there's no geometric content to it. It counts holomorphic forms.
Sure, it's a bit formal in that you take a signed count of dimension, but it's still a count of something geometric.
 
7:39 PM
Oh I was merely saying that the geometric content of normalization is not clear ;)
I don't know what arithmetic genus is yet.
 
You know, take a singular curve $X$, take $k[X]$, take integral closure inside $k(X)$ and then take the corresponding variety. That is still birational to $X$ (same functin field) but doesn't have any singularities anymore. It's not clear what happens geometrically.
I'd like to think more about this at some point.
@Krijn Me too, but the big picture of blowing up curves is not clear to me ;)
 
@BalarkaSen Let a line arrangement be a finite set of lines L and incidence data ( a finite set of points belonging to each line in each line some non-trivial finite subset of L) so that any two lines intersect in a point. Is every such arrangement realizable as a union of lines in $\Bbb CP^2$?
 
Not sure if I parse the definition of a line arrangement.
 
A set of lines, each of which are marked with finitely many points.
 
7:45 PM
And what is the relation between those marked points and the lines?
 
Well take a finite set of actual lines inside CP^2
 
OK.
 
These can intersect in triple points, double points quadruple points etc but each line intersects another at a unique point.
 
I agree. (I guess n-uple point means n lines intersecting at that point)
 
So the incidence data is just the combinatorial definition of these intersections
 
7:47 PM
Ah, starting to make sense.
 
So for instance the data (however you define the actual combinatorial construction) could be that $L_1,L_2$ and $L_3$ intersect at a point $p_1$ and $L_1$ and $L_4$ intersect at a point $p_2$ for lines $L_1,L_2,L_3,L_4$ and $L_2$ and $L_4$ intersect at a point $p_3$ etc.
 
Yep, I get it. So I have these abstract things called "lines" $\ell_1, \ell_2, \cdots$ and a finite set of them $L$. And I have these abstract things called points $p_1, p_2, \cdots$ and a finite set of them $S$. An incidence data is the power set $P(L)$ equipped with a set $S$ of points so that each line in an element of $P(L)$ (which is a set of lines) passes through a point in $S$ (so it's an assignment $P(L) \to S$). Is that it?
TL;DR I am basically saying it's a bijection $P(L) \to S$, I think.
Is that a valid definition of what you're trying to say?
Hmm, I think my previous edit had it right. Not a bijection $P(L) \to S$, but a bijection $A \subset P(L) \to S$. From a subset of the power set.
 
It should just be a map $f$ from $L$ to $S$ so that $f(\{L_i,L_j\})=\{pt\}$
 
Well, I'm off, watching The Wire
 
Where $f$ just means $s$ is in $l_i$
 
8:01 PM
@BalarkaSen You do different things from math sometimes, right?
 
@Krijn Maybe.
Like chatting here.
@PVAL Hmm. I see.
So a map $L \times L \to S$. OK. I can work with that.
Very interesting question.
My money is on no. I'd have to think seriously.
 
I don't know where the $\times$ is coming from
 
You pick two lines $\ell, \ell'$ from $L$ and they intersect due to hypothesis. So call the intersection point $p$. You're assigning $(\ell, \ell') \mapsto p$, no?
 
no.
 
Hmm, I see.
 
8:06 PM
Because 3 lines could also intersect at one point
The condition is that any two lines intersect in a unique point (the same condition for lines in $\Bbb CP^2$).
 
Yes. Now I am confused. It seems my power set definition should be the right definition. How can you define it to be a map $L \to S$? $L$ can have several marked points in it (several lines could intersect with $L$ at many different points).
 
Well how bout its just a subser of $L \times S$.
 
why do you need such a painfully formal definition
 
You don't
I don't
 
@MikeMiller I am not sure how I can prove something without a definition. I understand the idea, but I am not sure if I can work without a definition.
 
8:09 PM
what a pain
 
@PVAL OK. Makes sense.
So shall I just start thinking about it without caring for the definition?
 
It's a finite set "of lines" $L$ with a finite set "of points" $P$ with a relation $\epsilon$ with the condition that for every $l_1,l_2 \in L$ there exists a unique $p \in P$ with $p\epsilon l_1$ and $p \epsilon l_2$.
That should be perfect and plenty formal.
 
K. So it's a graph with vertices in $L$ and edges in $S$ such that there is a unique edge passing through two vertices. Gotcha.
 
No.
You need to represent triple points of intersection
 
Ah, well. Forget about the graph then. Your definition is good.
 
8:17 PM
@MikeMiller At some point you at least need to convince yourself there is a formal definition. Then you can immediately forget about it.
 
Let me think about the problem now that I have the definition. Thanks, @PVAL.
 
@robjohn is not around?
 
8:39 PM
Suppose my incidence data $(L, S, \in)$ is so that for any $\ell_1, \ell_2, \ell_3$ in $L$ there is no $p$ such that $p \in \ell_1, p \in \ell_2, p \in \ell_3$. Then I am sure I can realize this data: start by taking two lines. we have to avoid 1 point if we are to choose a third. this can be done. after that we have $1 + 2 = 3$ points to avoid hitting.
by induction after choosing the $n$th line we have $n(n+1)/2$ points to avoid hitting if we are to choose an $(n+1)-th$. this can be done, because space of lines passing through each of the $n(n+1)/2$ points form a closed set in the moduli space of all lines in $\Bbb P^2$ (namely, $\Bbb P^3$), so generically I can pick one from the complement, which is quite a large open set.
So this does work (keeping picking lines until you reach $|L|$ lines, which is your quota. if you have more points than $\binom{|L|}{2}$, then just chuck them out somewhere). Now onto examples with triple points.
 
user147690
What is the upper triangular lie group called?
 
user147690
Heisenberg 3 group is a special case I imagine, if that helps
 
If my incidence data is so that there is exactly one triple point (i.e., exactly one triplet $\ell_1, \ell_2, \ell_3$ so that a point $p$ passes through each of them), after choosing $|L| - 1$ many of them through my process described above, I can pick it to pass through one of the $(|L| - 1)|L|/2$ double points (there always exists one line passing through a given point, so this can certainly be done). Then I have realized it just fine.
@PVAL I am starting to believe this can always be done. I think some modification of my argument should definitely do it. I am a bit worried about whether there are trivial cardinality obstructions I am not thinking through though. Maybe I'll ponder a bit more.
 
9:04 PM
@Balarka Avoiding hitting a point is an open condition but if you want to hit a point it won't be.
 
I agree. But I am not sure what you are referring to. My construction above uses the fact that avoiding $n(n+1)/2$ points is an open condition.
 
Well if you want triple points...
 
But see for a single triple point it's clear 'cause after doing my construction with $|L| - 1$ lines you just have to pick a line which passes through one of the $(|L| - 1)|L|/2$ many double points. That can be done because there's a line through a given point, so just choose a double point ;)
Oh dear no there's a subtlety here. You're absolutely right.
I have to pick the last line to pass through exactly one of the $(|L| - 1)|L|/2$ double points.
Fix: so choose a $p$ point from the $(|L| - 1)|L|/2$ many double points. Take the collection of all lines passing through $p$. That's a closed set $X$ of the moduli space. Now consider the subcollection of lines passing through $p$ and some point from the rest of the double points (there are $(|L| - 1)|L|/2 - 1$ many of them now). That's a closed subset $Y$ of $X$. So $X - Y$ is open in $X$ and you pick from there.
Whew. These things are hard to visualize.
 
Well what is a line in $\Bbb CP^2$?
 
It's a projectivization of a linear subspace? Copy of $\Bbb P^1$?
 
9:12 PM
Sure, but what data do you need to define it?
as a projective algebraic variety for instance?
 
An equation $aX +bY + cZ = 0$. A tuple of coefficients $(a : b : c) \in \Bbb P^3$ if you wish. Is that what you want?
 
Yes, and now this is the projectivized/complexified version of what (I'm looking for a map from line arrangements in $\Bbb R^2$ to $\Bbb CP^2$.
But even in that sense that you mentioned maybe you can prove something about how a few lines intersect.
 
@PVAL I see. It seems like a nice problem, and I am prepared to believe there are no counterexamples as of now. I do think my argument can be appropriately modified, I am thinking about it.
Is there an easy/neat way to do it? Mine (well, not a proof, but an approach) is probably more of a brute force thing.
 
@BalarkaSen There are likely numerous neat ways to do it.
it refering to either the negation or affirmation of the question
 
I see. I'll think about it for a bit though, so please don't reveal :) Thanks again for the problem.
I am thinking as follows: suppose my incidence data $(L, S, \epsilon)$ has exactly $k_2$ double points (so $k_2 - \binom{|S|}{2}$ many points are unused), $k_3$ triple points, $k_4$ 4-uple points and so on until $k_n$ $n$-uple points where I have used up all the $k_2$ points I could use.
Then suppose I start by $k_2 - k_3$ points (these are the double but not triple points), and use up $|L|$ wisely to draw lines with no triple points with $k_2 - k_3$ points covered. I have already described an algorithm for how to do it.
Now let's take the triple but no 4-uple points $k_3 - k_4$. I use up more of $|L|$ wisely to draw lines so that each line passes through exactly one of the $k_2 - k_3$ double points. This can also be done by the process I described, I think.
And... I can't expect that to work. The new lines I drew intersected, so my # of double points got way more than $k_2 - k_3$. Crap.
 
9:43 PM
Oh, so I think I messed up the numbers in the previous thing too. This situation itself cannot happen: if there were $4$ lines and $1$ triple point and nothing else, then three of the lines $\ell_1, \ell_2, \ell_3$ must intersect at a point $p$ whereas another line (which may not pass through $p$ by assumption of not having a 4-uple point) intersects all 3 of those lines so I get $3 + 1 = 4$ points.
Not as easy as I thought. Need to think carefully.
 
9:55 PM
@MikeMiller Heard a story about the Malgrange preparation theorem today, not sure if you care though. Apparently Thom was giving a lecture in TIFR and Malgrange was in the talk. So he wrote down the analogue of the Weierstrass preparation theorem and said "this should be true" or something of that sort, to which Malgrange objected by "then you should prove it" or so. :)
 
He's right that it's not a triviality.
 
I admittedly don't know too many applications of the Weierstrass preparation theorem though. I have seen it used to prove that local ring of a variety at a smooth point is a UFD in Shafarevich. Is there a general reason it is interesting?
 
that's a pretty big deal...
but idk read griffiths and harris and seehow many times it or that fact appears in the first 20 pages.
 
I believe you that it's useful. I was just wondering if there is any obvious reason why it's so useful - the result itself doesn't surprise me much. Perhaps I am overthinking it and it's not more than a tool.
 
10:22 PM
@Semiclassical Source is here. The line in question is on the second page. And is where he solves for $u_{3}$.
I'm going to sleep now.
 
10:44 PM
@PVAL Do you know if Bob is going to be at the Rice thing for any amount of time?
 
@MikeMiller No I don't.
 
Darn.
 
Think if you have a question for him the easiest thing to do is email him.
or maybe ask me
and if I dont know
then email him
 
It's the puncture thing.
 
I asked him about it. He said he had some notes written down but didnt remember off hand if that result was part of those notes.
You could also email Kyle
 
10:46 PM
OK, I'll probably do so tomorrow ro something.
 
I don't think my construction works
because I'm not sure where the image of half of the cyllinder goes
or half of the solid torus (i.e. the solid cyllinder goes)
If I take out a cyllinder and reglue it by the image of a cyllinder in a 3-ball I'm not exactly sure what you get.
It isn't a 3-ball though because the meridean doesn't bound a disk (he pointed this out)/
 
I didn't understand it when you first sent it and didn't make an effort to because you told me it didn't work.
Out of curiosity did you tell him I asked you?
 
I'll ask Ahmad if he knows.
no
I just asked about this construction
I may have said someone asked me about this, but I don't remember
 
I don't think I know who Ahmad is.
 
He's a grad student here thinking about embedding 3-manifolds
and maybe HF now.
 
10:52 PM
Oh that's exciting.
 
I don't really understand how you do that stuff. Like each method is understandable but it seems you have to be a super-genius to figure out how to do it.
The constructions appear out of thin air to me.
Did you see the Leg. knot complement paper?
 
I did but haven't read it.
 
I think the proof is kind of trivial
 
I also want to read Vivek's paper on the conormal to a knot at some point.
 
I'm wondering if a similar idea works in the transverse setting
 
10:57 PM
My dream is an obstruction to embeddings that doesn't also obstruct embeddings into homology spheres. I would be deeply weirded out if everything that bounded a homology ball embedded in $S^4$.
That just seems like too much freedom in a dimension like 4.
 
@Owatch there's two errors, one on your part and one on his. first, you said you got $(-2/3,2/3,-2)$, but the last component should actually be 2/3 (check it). normalizing that gives the $e_3$ given correctly in the text; the $u_3$ given is incorrect, and i'd consider it a typo.
 
@MikeMiller This is dangerously close to Poincare.
 
@PVAL Note that I'm not trying to distinguish embeddings into homotopy spheres. But yes I agree it's about as equally much a pipe dream.
 
11:31 PM
hi
 
New question. Help a cat out :(
0
Q: Given that an integral containing $|x_{n}|$ is bounded by a constant, show that $x_{n}\to x$ in $L^{1}(0,\infty)$

Jessy CatThis question is related to another question I asked earlier (feel free to answer that one, too, if you want; I'm not satisfied with what people have said so far). In addition to the conditions mentioned in the other question, suppose that additionally, $\exists \alpha > 0$ such that $\forall n...

 
I am trying to prove the following. Given an equivalence relation $\sim$ on a set $X$, the equivalence classes of $X$ form a partition of $X$. Conversely, if $\mathcal{P} = \{X_i\}$ is a partition of a set $X$, then there is an equivalence relation on $X$ with equivalence classes $X_i$.
my book proves the first direction by saying, suppose there exists an equivalence relation $\sim$ on $X$. For any $x \in X$, the reflexive property shows that $x \in [x]$ and so $[x]$ is nonempty
but what if $X = \emptyset$?
 

« first day (2085 days earlier)      last day (2929 days later) »