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2h
comment Bourbaki and set inclusion
Those votes to close make absolutely no sense. If someone thinks this is a bad question, they should simply downvote.
6h
comment How much set theory does the category of sets remember?
Nice question, but this is surely MathOverflow material.
9h
comment Inner product space related to pythagorean theorem.
Use the fact that $\|v\|^2$ is defined as $\langle v,v\rangle$. Use also the fact that, by definition, $u$ and $v$ are orthogonal iff $\langle u,v\rangle = 0$. Start on the LHS and arrive, by a series of equalities, at the RHS.
1d
accepted What are some illustrative examples that demonstrate how $\succ$ can differ in behavior from $>$ and/or $\geq$?
1d
comment Relationships between affine closures and convex closures
@MichaelGrant, a subset $X$ of $V$ is convex iff for all $x,y \in X$ and all $a,b \in \mathbb{R}$, we have: $(a \geq 0) \wedge (b \geq 0) \wedge (a+b=1) \rightarrow ax+by \in X$. But those order relations don't make sense over an arbitrary field. So technically, what we need is for $V$ to be a vector space over an ordered field, or even an $R$-module where $R$ is a partially-ordered commutative ring. Edit. The reason we can talk about convex subsets of $\mathbb{C}$ is because $\mathbb{C}$ can be viewed as a $2$-dimensional real vector space.
2d
revised Submodules $H$ satisfying: “if $ax \in H$ for some non-zero scalar $a$, then $x \in H$.”
added 208 characters in body
Apr
25
asked Submodules $H$ satisfying: “if $ax \in H$ for some non-zero scalar $a$, then $x \in H$.”
Apr
24
comment Restriction of a monotone self-map (“endomorphism of $\mathbf{Pos}$”) to a wide subposet
But it is surprising though. Think vector spaces. Are bilinear maps $X,Y \rightarrow Z$ necessarily linear maps $X \times Y \rightarrow Z$? No. Does the converse hold? No again. There's something very special about posets....
Apr
24
comment Rigorous separation of variables.
Giasou Yiorgos - funny coincidence, I'm Greek too, but unfortunately, not too familiar with the mother tongue (3rd gen), certainly not enough to understand mathematics written in it. Do you know of any English resources where similar issues are addressed? On a related note, it might be worth translating your book into English; perhaps market it as "Ordinary Differential Equations, Done Rigorously" or some such. There's definitely people out there who would really appreciate this kind of thing.
Apr
24
comment Naive Set Theory by Halmos is confusing to a layman like me
@DavidRicherby, your suggestion is a good one if we don't need to mention the variable $x$. If we do, I do think my suggestion is not so bad. It has certain advantages. In particular, the phrasing I give makes it clear that anywhere you see $x \in A$, you can write "therefore blah", and anywhere you see "blah", you can write "therefore $x \in A$." The symmetry is clear. I also like making it very explicit that $A$ is a subset of $X$, which my phrasing does. If you can see how to shorten the whole thing without losing these qualities, I'd be interested to hear your ideas.
Apr
24
awarded  Nice Answer
Apr
23
revised Naive Set Theory by Halmos is confusing to a layman like me
added 126 characters in body
Apr
23
answered Naive Set Theory by Halmos is confusing to a layman like me
Apr
23
comment Restriction of a monotone self-map (“endomorphism of $\mathbf{Pos}$”) to a wide subposet
Regarding the penultimate paragraph; yes, that I what I mean by saying that an $n$-ary function is monotone. Note that this is (rather amazingly) equivalent to requiring that it be monotone separately in each argument.
Apr
23
comment Restriction of a monotone self-map (“endomorphism of $\mathbf{Pos}$”) to a wide subposet
@RespawnedFluff, thanks for the info bud.
Apr
22
revised Restriction of a monotone self-map (“endomorphism of $\mathbf{Pos}$”) to a wide subposet
edited tags
Apr
22
comment Restriction of a monotone self-map (“endomorphism of $\mathbf{Pos}$”) to a wide subposet
@RespawnedFluff, thanks for your input, however I think that's just a terminological thing. Note that a wide-subposet of $P$ is basically just a monotone bijection $f : Q \rightarrow P$. So the concept does live comfortably in the world of order-theory, even if there is a bit of a terminological conflict. You make a good point about graph theory potentially being a good lead.
Apr
22
comment How do I express, algebraically, this comparison of two sets of sets?
@atomh33ls, no worries. Yeah, that's where I'm unsure of your meaning. By $n(A_1+A_2+A_3)$, do you mean $|A_1 \cup A_2 \cup A_3|$, or do you mean $|A_1|+|A_2|+|A_3|$? If they're disjoint, these numbers agree, of course.
Apr
22
revised How do I express, algebraically, this comparison of two sets of sets?
added 402 characters in body
Apr
22
answered How do I express, algebraically, this comparison of two sets of sets?