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comment Give a good reason to define a function from A to B as a triple (F, A, B) rather than a functional set of pairs with domain A and image included in B.
@Raymond My answer to your question shows why the SOP answer to "when are functions equal" is sometimes inappropriate. The SOP answer (equal domains and equal values at all points of the domain) makes the notion of "induced map of homology" ill-defined, in the sense that "equal" functions can lead to unequal homology homomorphisms.
6h
awarded  linear-algebra
13h
comment What would the ring $\mathbb{Z}[x,y]/(x^2-y)$ look like?
In my previous comment, "the isomorphism" should be "an isomorphism" or perhaps "the simplest isomorphism".
13h
comment What would the ring $\mathbb{Z}[x,y]/(x^2-y)$ look like?
The ring in the title of your question is isomorphic to the polynomial ring in one variable, $\mathbb Z[y]$. The isomorphism sends any polynomial $p(y)\in\mathbb Z[y]$ to the equivalence class in $\mathbb Z[x,y]/(x^2-y)$ of the same polynomial $p(y)$ regarded as an element of $\mathbb Z[x,y]$. The ring in the body of your question is more complicated.
13h
comment Let $F$ be a sigma algebra such that every element of $F$ is the union of two disjoint nonempty sets also in $F$. Prove that $F$ is uncountable.
I see that you get an infinite binary tree with nonempty sets from the sigma-algebra at all the nodes, and I know that there are uncountably many infinitely long paths through this tree, but I don't see how that gives you uncountably many sets in your sigma-algebra. The obvious idea would be to associate to each path the intersection of the sets attached to the nodes on the path, but as far as I can tell, those intersections might all be empty and therefore not distinct sets.
16h
awarded  Necromancer
1d
answered Give a good reason to define a function from A to B as a triple (F, A, B) rather than a functional set of pairs with domain A and image included in B.
1d
comment Cohomology space
Statement (a) is wrong if the dimension of $M$ is $1$.
1d
comment How to define map function on “Church lists” using primitive recursion?
I suspect that, in order to get a useful answer, you'll need to say exactly what constructions are allowed in your definition. The two equations you wrote look to me like a perfectly good definition by recursion. I probably wouldn't call it "primitive recursion", but that's because I think of primitive recursion as being defined exclusively for functions on natural numbers. So under my convention, there's no way to define a function on lists by primitive recursion. You presumably are using a different notion of primitive recursion, but what is that notion?
2d
answered Is it true that $A\cong B$ implies $A = B$ when $A$ and $B$ are ordered structures
2d
comment Understanding $\Theta$-notation rigorously
In your context, the rigorous meaning of "equal to within a constant factor" is exactly what you wrote at the end of the question. The reason for the terminology is that the constant factor $c_1$ restricts $f$ from getting much smaller than $g$, while another constant $c_2$ restricts $f$ from getting too mush bigger than $g$. So the values of $f$ must stay fairly near the values of $g$; they're not exactly equal, but they differ by at most a constant factor. (It might have been better to say they are equal to within constant factors (plural) since there are two $c$'s involved.)
2d
comment Is it true that $A\cong B$ implies $A = B$ when $A$ and $B$ are ordered structures
In the context of descriptive complexity, "ordered structure" usually means a finite structure whose underlying set (also called its universe or its domain) is of the form $\{1,2,\dots,n\}$ for some natural number $n$ and whose basic relations include the standard linear ordering relation of natural numbers (restricted to $\{1,2,\dots,n\}$). With this understanding, Immerman's assertion is correct. (With the usual, model-theoretic notion of "ordered structure", Immerman's statement would not be correct, because of counterexamples like yours.)
2d
comment From Primitive Recursive to Recursive by Iterating over more than one Argument?
Presumably the material here attributed to Rósza Péter is from her book "Recursive functions". It's been a long time since I looked at it, but my recollection is that it's mostly about many sorts of extensions of primitive recursion, more than about general recursive functions. So it may give the OP a more detailed picture of what can and cannot be done with various sorts of recursions.
Aug
26
comment Couple of questions on the Axiom of Extensionality
Halmos's example gives an interpretation of $\epsilon$ that doesn't satisfy the axiom of extensionality. It does, however, satisfy the converse implication because, as Asaf Karagilaexplained in his answer, that converse is logically valid (as long as one regards equality as part of logic).
Aug
26
comment Closeness of measures on a cardinal
@AsafKaragila $D$ is normal because $g$ represents $\kappa$ in the ultrapower by $U$. In more detail, if $f$ is regressive on a set in $D$, then $f\circ g<_Ug$ and therefore $f\circ g$ is constant on a set in $U$, which makes $f$ constant on a set in $D$.
Aug
23
comment Implicit Differentiation. Please help me understand why!
"Why don't I ... take the derivative of $y$ and get $1$?" You do take the derivative of $y$ but you don't get $1$. If you were differentiating with respect to $y$, then you'd get $dy/dy=1$, but you're differentiating with respect to $x$. (If you really wanted to differentiate the left side of the equation with respect to $y$, then you'd have to also differentiate the right side with respect to $y$. You can't differentiate one side with respect to $x$ and the other with respect to $y$. You have to do the same thing to both sides of the equation.)
Aug
23
answered Question about definition of bounded linear functionals
Aug
23
comment Motivation for different mathematics foundations
Surreals don't need classes any more than ordinal numbers or sets do.
Aug
23
comment Proof that transpose of Hadamard Matrix is also a Hadamard matrix
@user26857 Yes, $A^\top$ is the transpose of $A$.
Aug
23
comment Motivation for different mathematics foundations
@AsafKaragila I don't think Paul was claiming that hyperreals and surreals are a foundation of mathematics but rather that they transcend ZFC in the sense of not fitting (comfortably) into the ZFC foundation. This weaker claim, though, is also wrong.