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Lee Mosher

 Discussion on answer by Robert Shore:…

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Lee Mosher
Jan 24 22:13
You yourself gave the definition of $\omega$ --- the set of natural numbers --- in your post, using a definition that is clearly a set in ZF.
 

 Discussion on question by Hearth: Wha…

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Lee Mosher
Dec 10, 2024 10:29
Regarding vectors, how are you on linear transformations? Although you write that a tensor is "...some sort of generalization of a vector...", you'll really also need to understand that it is some sort of generalization of a linear transformation.
 

 Discussion on question by Miss_Unders…

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Lee Mosher
Nov 3, 2024 03:41
It is very far from a coincidence. You are simply reversing Hubble's original calculation, where he derived the age of the universe from $c$ and the (observed) Hubble constant.
 

 Discussion on question by urhie: defi…

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Lee Mosher
Oct 11, 2024 17:03
If you want to reach an intuitive understanding of the "standard" definition of limits, and of your own attempted definitions, or to develop an intuition for them, then put those definitions to work. Try to use them to prove theorems. For example, try to use them to prove: the limit of a sum of real valued functions is the sum of the limits; the same for products; if $f(x)$ is differentiable at $x=a$ then $f$ is continuous at $x=a$. As you go along, compare your developing proofs with the standard proofs you find in the book.
 

 Discussion on question by Paradoxy: D…

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Lee Mosher
Aug 21, 2024 14:26
@Paradoxy: You wrote in your comment I defined $M$ by using axiom of schema. Perhaps what you did not realize is all you get from the axiom of schema is a bare, naked set. You did not, yet, get a manifold, nor even a topological space. What else do you need to get a topological space? You need a collection of open sets. What else do you need to get a manifold, in addition to a topology? You need charts. Can you get a manifold without a topology, using only charts? Yes, but then you have to fill in additional details, including properties of the overlap maps between charts.
Lee Mosher
Aug 21, 2024 14:26
The given coordinate charts of the manifold automatically give you a basis. For example if $\psi : U \to \mathbb R^n$ is one particular coordinate chart of an $n$-dimensional manifold $M$ with $U \subset M$ then you get a whole pile of basis elements of the form $\psi^{-1}(V) \subset U \subset M$ where $V$ ranges over all open subsets of $\mathbb R^n$. Take all of these piles of basis elements together, one for each coordinate chart. Voila, you have a basis for $M$.
 

 Discussion on question by Frode Alfso…

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Lee Mosher
Aug 8, 2024 23:22
No, this is a fine place for it.
Lee Mosher
Aug 8, 2024 23:22
I suppose one could sift through the LaTeX files of set theory papers on the arXiv, although that will only go back to the early-mid 90's. LaTeX fonts were in existence 10-ish years before that.
Lee Mosher
Aug 8, 2024 23:22
Just to clarify, you are not asking about when the capital P was first used for the power set (which I suspect was well before LaTeX fonts were designed). Correct?
 

 Discussion on question by Princess Mi…

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Lee Mosher
May 25, 2024 20:39
By the way, you might be interested in reading the recursion theorem. You'll find that recursion is, at its root, not a circular process either.
Lee Mosher
May 25, 2024 20:39
Yes, the development of mathematics is like a textbook, it occurs in order. When one states a definition, or an axiom, or a theorem, that statement may only use concepts that have been defined or axiomatized somewhere earlier in the textbook.
Lee Mosher
May 25, 2024 20:39
Just take it as an axiom of logic: any concepts used in a definition must be either previously defined or previously axiomatized.
Lee Mosher
May 25, 2024 20:39
That was the genius of the Peano axioms: something seemingly so simple, and yet you could use them to define all of arithmetic.
Lee Mosher
May 25, 2024 20:39
If you can define addition and derive its properties starting from simpler axioms, then you may do so.
Lee Mosher
May 25, 2024 20:39
I should say that there are other ways to develop analysis axiomatically than to start from Peano's Axioms. Another traditional way, which you can see carried out in Fitzpatrick's "Advanced Calculus", is to start from axioms of the real numbers themselves.
Lee Mosher
May 25, 2024 20:39
There is no morality behind it; nothing is "supposed to" or "not supposed to" happen. What mathematicians discovered, starting with Euclid, is that the axiomatic method is a systematically practical way of developing mathematical theory.
Lee Mosher
May 25, 2024 20:39
Well, in your post you gave $S$ in terms of addition, but that's not how it is done in Tao's book. Instead, $S$ is assumed to be given, and it is assumed to satisfy some simple axioms, and that's it. Everything else is derived from that. This is the essence of the axiomatic method: something has to be given from which everything else is derived; mathematics does not rise unborn from a giant scallop shell. Hopefully what is given is as simple as possible.
Lee Mosher
May 25, 2024 20:39
Well, there it is. That definition assumes that $S$ is given. That's all there is to it.
Lee Mosher
May 25, 2024 20:39
"the one which was given...* Which one is that? What is the definition of addition that was given?
Lee Mosher
May 25, 2024 20:39
You wrote Say we did define $S(n)=n+1$... What definition of $+$ are you using? Am I correct in guessing that you are implicitly assuming that the natural numbers and their operations of addition and multiplication have already been defined and so you are making use of them?
 

 Discussion on question by some_math_g…

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Lee Mosher
Jan 8, 2024 12:49
Well, without some further clarification, all that can be said is that (c) does not make sense. Smoothness can be a consequence of more powerful theorems, but that does not seem appropriate here.
Lee Mosher
Jan 8, 2024 12:49
What is the source of this question -- with this exact wording? Part (c) indeed does not make sense "without using partial derivatives".
 

 Discussion on question by The Shepard…

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Lee Mosher
Nov 6, 2023 02:04
Duplicate title on math.stackexchange
 

 Discussion on question by Uzai: Why w…

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Lee Mosher
Oct 19, 2023 17:57
Unfortunately, the Hilbert hotel went bankrupt; where are those infinite monkeys going to work?
 

 Discussion on question by Rohit Pande…

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Lee Mosher
Aug 7, 2023 03:18
Rather than dancing around with terminology, perhaps a direct quote from the answer of Keith Thompson is in order, and I add some emphasis: The criterion for an exoplanet to have solar eclipses like the spectacular ones we have here on Earth is a moon that happens to have an angular size large enough to cover the photosphere, but not so large that it also hides the corona.
 

 Discussion on question by Denis: Isn'…

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Lee Mosher
Apr 29, 2023 08:38
You wrote: "If there was an objective criteria that could define `what is art and what is not', then art wouldn't be any different from any other field, let's say engineering... It would become absolutely artificial and meaningless term." I imagine, though, that you, just like me and everyone else, make artistic judgements all the time, based on your own subjective criteria: you like a song or you don't, you like a movie or you don't, you like a painting or you don't. And however unobjective your artistic judgements are, they are not meaningless at all.
 

 Discussion on answer by FShrike: How …

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Lee Mosher
Mar 7, 2023 14:04
Regarding your remark what's the "semi" all about?, the difference between a conjugacy and a semiconjugacy is that a conjugacy must be an isomorphism of the category. So in the (apparent) topological setting of this post, a conjugacy must be a homeomorphism.
 

 Discussion on question by Charles: Al…

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Lee Mosher
Feb 24, 2023 15:53
Your statement So $(a,b)$ must be one of the two forms... seems not related to the given pair of sets. Every open interval $(a,b)$ has both of those forms, for every value $z \in (a,b)$.
 

 Discussion on question by Franklin: F…

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Lee Mosher
Nov 17, 2022 13:23
Because $8^n$ does not mean $8n$, it is as simple as that.
Lee Mosher
Nov 17, 2022 13:23
Regarding correctness/validity of your solution, I'll echo some of the earlier comments: as it stands that $8^n$ makes your solution invalid. Correcting that would make it a valid solution. But it is like a plant that has been allowed to grow too much and become rangy, it needs a lot of trimming. So for all of those reasons, 5 points out of 10 seems about right.
Lee Mosher
Nov 17, 2022 13:23
It's just a bad idea to mix additive and multiplicative notation for a group operation, all you'll do is to confuse anyone trying to understand you. If you choose $m+n$ to denote your group operation, then you should stick with $-n$ as the inverse operation.
 

 Discussion on question by Kevin Njoko…

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Lee Mosher
Jul 29, 2022 02:59
Regarding your statement I can't possibly see why this question was downvoted: To see why, you can look at our guidelines for how to ask a good question, with emphasis on providing context. Your post is a bare problem statement with no context, and such questions are often downvoted and closed on this site.
 

 Discussion on question by Ethan Dlugi…

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Lee Mosher
Jul 10, 2022 13:41
I presume that your fourth Cartesian factor $\mathbb Z^{\mathbb Z^2}$ is where you record "vertical stripes". With that in mind, associated to each square (i.e. to each element of $\mathbb Z^2$) is a certain coordinate chosen from $\mathbb Z$. What coordinates are assigned to that square for three situations described in your earlier comment, namely: a "negative vertical stripe going down", versus a "positive vertical stripe going down", versus a "positive vertical stripe going up"?
 

 Discussion on question by Novice: Am …

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Lee Mosher
Mar 1, 2022 12:37
What happened when you asked your professor the questions you are asking us, e.g. Why is he referring to polynomials?
 

 Discussion on question by n2k: Is a f…

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Lee Mosher
Jan 22, 2022 07:08
Now that your question has been clarified in the comments, let me suggest that you edit your post to write out exactly what quantity you want to prove is algebraic. You should express that quantity as a mathematical formula typeset with mathjax. Without that, your question may soon be closed.
 

 Discussion on answer by manassehkatz-…

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Lee Mosher
Jan 11, 2022 11:38
Aside from the descriptive start and predictive continuation of Moore's law as said in the previous comment of @MSalters, it will never be dictative as the very first comment to this answer seems to assume.
 

 Discussion on question by Caleb Brigg…

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Lee Mosher
Dec 22, 2021 21:04
The argument you give in your first paragraph is not a "continue ad nauseam" argument. It is a straightforward contradiction argument. Given a real number $x$, by assuming the existence of a next real number $y$ after the real number $x$, one derives a contradiction. Therefore the assumption is false, so there does not exist a next real number after $x$.
 

 Discussion on answer by Lee Mosher: W…

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Lee Mosher
Dec 19, 2021 23:39
This question of "more scientific" notation seems to be beyond the scope of your original question. Instead of continuing to expand your question in the comments, you should decide whether this answer is acceptable, or await another answer which is more acceptable.
Lee Mosher
Dec 19, 2021 23:39
I will repeat: there are many notations. Two different authors do not always use the exact same notation. If you understand the mathematical concepts, then I am confident that you can figure out how one author uses notation to represent the mathematical concepts, and how a different author uses different notation to represent the same mathematical concepts.
Lee Mosher
Dec 19, 2021 23:39
Your terminology "projection plane" is not standard. From what you wrote, I take it that you are referring to the plane $\{(x,y,1) \in \mathbb R^3 \mid x,y \in \mathbb R\}$. I explained the difference between $(x,y,z)$ and $[x:y:1]$ in my answer, starting with the first bullet point.
Lee Mosher
Dec 19, 2021 23:39
There are many notations. You must read each author carefully to learn what that author means by their notation.
Lee Mosher
Dec 19, 2021 23:39
I have tried. You'll see at the end that I explain the exact mathematical difference between the point $[x : y : z] \in P^2$ that is represented in homogeneous coordinates by the point $(x,y,z) \in \mathbb R^3$.
Lee Mosher
Dec 19, 2021 23:39
I've rewritten my answer to try to address your confusion.
Lee Mosher
Dec 19, 2021 23:39
Another notation you may encounter is $[x:y:z]$ for the point in the projective plane that I am denoting $[x,y,z]$.
Lee Mosher
Dec 19, 2021 23:39
No. A point of $\mathbb R^3$ is an ordered triple of real numbers $(x,y,z)$. Notice that $[x,y,z]$ or $[x : y : z]$ is a different notation than $(x,y,z)$.
Lee Mosher
Dec 19, 2021 23:39
They are different things, as my answer explains. A single point $[x,y,z]$ in the projective plane that is represented by the point $(x,y,z) \in \mathbb R^3$ has many other representatives in $\mathbb R^3$, namely $(2x,2y,2z)$, or $(.001x,.001y,.001z)$, or $(-x,-y,-z)$, or indeed any point on the entire line $(rx,ry,rz)$ (except the point where $r=0$).
Lee Mosher
Dec 19, 2021 23:39
In some formal sense, the point $[x,y,z]$ in the projective plane is the line $\{(rx,ry,rz) \in \mathbb R^3 \mid r \in \mathbb R\}$.
 

 Discussion on question by Bhawana Dey…

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Lee Mosher
Aug 30, 2021 20:07
You have defined $\tilde X$ as a set, however you have not specified a topology on that set. Without that, asking whether $X$ and $\tilde X$ are homeomorphic does not make sense.
 

 Discussion on question by PiKindOfGuy…

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Lee Mosher
Jun 23, 2021 10:56
See the related discussion here. Roughly speaking, the point made there is logical connectives such as "if--then" need to formalized by having rigorous, consistent truth tables, in order to be useful in mathematics.
 

 Discussion on question by Ultimate Ha…

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Lee Mosher
Jun 17, 2021 13:02
By the way, I think you misunderstood the comment of @JoseAvilez. Yes, after you have picked $B_n$, and if the function is not yet surjective (it won't be, because $\mathbb R$ is not finite), you can then pick $B_{n+1}$. But that comment meant: after you are done picking $B_n$ for every $n \in \mathbb N$, and therefore the definition of your function $B_n$ is complete and there are no more terms left of that function to be picked, can you prove that the function is surjective?