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May
8
comment The least relation which produces a partial order
(stupid comment removed)
May
7
comment The least relation which produces a partial order
@PedroM. $E^2 = E\circ E$ is the composition of binary relation $E$ with itself
May
2
comment What do the symbols d/dx and dy/dx mean?
It is described in en.wikipedia.org/wiki/Differential_of_a_function
May
1
comment What do the symbols d/dx and dy/dx mean?
@Fixee See my answer. It is nevertheless CAN be defined as a fraction of two functions, rather than an atomic object.
May
1
comment What do the symbols d/dx and dy/dx mean?
Note that $\frac{F}{G}$ is defined for two functions $F$ and $G$ as $\frac{F}{G}(x)=\frac{F(x)}{G(x)}$. Thus $\frac{df}{dx}$ makes sense.
Apr
29
comment Three theorems for the price of one? (like duality)
en.wikipedia.org/wiki/Triality (for vector spaces)
Apr
29
comment Three theorems for the price of one? (like duality)
@HagenvonEitzen Why six?
Mar
26
comment An elementary proof about filters
@AndreasBlass: You are right, the first $\bigcap$ on the right side of the equation was intended to be $\bigcup$. I've corrected the question.
Dec
14
comment About a function ranging filters
Oh, I found an easy solution (but only for people who has read my book): $\operatorname{Back}(f;k)$ is a complete funcoid; from this the conjecture follows.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
Oh, I see: Some people use the word "setoid" differently. So my downvote was a mistake. I can't upvote it back now, sorry.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
This your definition of setoids is not equivalent (however, it is equivalent up to equivalence of categories) to the standard definition of setoids, that is a set with an equivalence relation. Equivalence up to equivalence of categories is not enough for my purposes. I still think that I downvoted correctly,
Dec
5
comment Why “thin groupoids” are not ubiquitous?
Setoid is a set with an equivalence relation on it. (And I know it long ago before I've read your answer.) I understand this. I downvote because switching from a groupoid to a setoid leads to information loss, and this makes it not an answer to my question.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I need to describe a category. A category contains not only objects but also morphisms. As my category happens to be thin, there is a (not necessarily entire defined) function from pairs of objects into a morphism. Having a setoid we cannot define such a function. But I need this function. I need to be able to get the morphisms whenever a pair of objects is specified. Having only a setoid, I cannot do this.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
But having two elements of setoids, I cannot restore particular morphism (such as $f\mapsto f\cap\Gamma$). It is not what I need
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I yet don't understand you and don't see how to express this with setoids. When we switch from a thin groupoid to a setoid, the information about particular morphisms is lost (they are just replaced with a pair of objects), but the whole thing I need is information about what are particular morphisms, depicted as arrows in my diagrams.
Dec
5
comment Why “thin groupoids” are not ubiquitous?
I don't get how this is related with setoids. How to express my diagrams using setoids? I need particular isomorphisms not just the fact that two objects are isomorphic.
Dec
2
comment Another way to express certain filter
See also counter-examples in this thread: groups.google.com/forum/#!topic/sci.math/Plru0S8ePzs
Dec
1
comment Another way to express certain filter
The other direction is surprisingly difficult
Dec
1
comment Another way to express certain filter
In one direction: Let $X \in [\mathcal{A}]_{\mathfrak{B}}$. Then $\exists Y' \in \mathcal{A} : X \geq Y'$. Thus if $Y \geq X$ for $Y \in \mathfrak{A}$ then $Y \geq Y'$ and thus $Y \in \mathcal{A}$. So $\forall Y \in \mathfrak{A}: (Y \geq X \Rightarrow Y \in \mathcal{A})$.
Sep
24
comment Equality of two expressions describing a filter
@AndreasBlass: Thanks, it was my error. Now have been corrected