Activity for celtschk
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #293500 |
Post edited: Fixed a mistake that made the question ambiguous. |
— | 10 months ago |
| Comment | Post #293500 |
By definition, a topological field is a field with a topology such that all field operations are continuous. So the second way it is. **Edit:** I only now notice that unlike in the question title, in the actual question body I wrote "topological vector space" - what I wrote in the title is actual... (more) |
— | 10 months ago |
| Edit | Post #294995 |
Post edited: Added: My concrete idea from the question does not work |
— | 10 months ago |
| Comment | Post #294950 |
A subset and an element are two very different things. The empty set has no elements (that's the defining property), but one subset. The powerset of the empty seet has one element (the empty set), but two subsets (the empty set end itself). While (at least in the standard set theory) a set is nev... (more) |
— | 10 months ago |
| Edit | Post #294995 | Initial revision | — | 10 months ago |
| Answer | — |
A: Are there other topologies on $\mathbb R$ that make it a topological field? The answer is yes. The key to this is found in the Wikipedia article about p-adic numbers: > $\mathbb {C} {p}$ and $\mathbb {C}$ are isomorphic as rings, so we may regard $\mathbb {C} {p}$ as $\mathbb {C}$ endowed with an exotic metric. The proof of existence of such a field isomorphism relies... (more) |
— | 10 months ago |
| Edit | Post #294832 | Initial revision | — | 11 months ago |
| Question | — |
In this representation of surreal numbers and gaps with classes of ordinal numbers, are all gaps represented? One way to represent the surreal numbers is the sign representation, that is, as function from a (von Neumann) ordinal to the set $\{+,-\}$. Now given that there are two values, I got the idea to represent this as set of ordinals, where $+$ means the corresponding ordinal is in the set, and $... (more) |
— | 11 months ago |
| Comment | Post #294778 |
The continued fraction of a rational number has an exact maximal denominator. For example, $3/5 = 0 + 1/(1 + 1/(1 + 1/2)$. Therefore this doesn't work as irrationality measure.
Normal numbers are surely interesting, but there are definitely irrational numbers that are not normal. So normality... (more) |
— | 11 months ago |
| Comment | Post #294778 |
Note: I'm not really happy with the tag but that's the best I could find. If you find a better one, feel free to edit. (more) |
— | 11 months ago |
| Edit | Post #294778 | Initial revision | — | 11 months ago |
| Question | — |
Is this a proper measure for irrationality? It is often said that the golden mean $\phi$ is the "most irrational" number. However what I've never seen is an actual measure of irrationality which would make such a statement meaningful in a quantitative way. Now I thought about what such a measure would look like. Obviously it would need ... (more) |
— | 11 months ago |
| Suggested Edit | Post #294748 |
Suggested edit: (more) |
helpful | 11 months ago |
| Comment | Post #293754 |
Just imagine what happens with a different distribution for the components of $z$, say an equal distribution in the interval $[0,1]$ in $3$ dimensions. With this distribution, the full distribution of $z$ is an equal distribution inside a cube centred at the origin, clearly not a rotationally inv... (more) |
— | about 1 year ago |
| Edit | Post #293500 | Initial revision | — | over 1 year ago |
| Question | — |
Are there other topologies on $\mathbb R$ that make it a topological field? As is well known, $\mathbb R$ with the standard topology is a topological field. It is also not hard to check that the discrete and the indiscrete topology on $\mathbb R$ result in a topological field, simply from the fact that all functions from a discrete topology are continuous, as are all fun... (more) |
— | over 1 year ago |
| Comment | Post #292671 |
That is indeed a very important information. With thousands of dimensions, reject sampling probably would not be a good strategy. Another important question is the number of vectors. there are of course at least (dimension+1) of them (or else you'd not have a finite volume), but it should make a ... (more) |
— | almost 2 years ago |
| Comment | Post #292671 |
My guess would be that the most efficient way to sample would be to random sample from a well-chosen parallelogram-shaped bounding box, and then discard the point and repeat if the point lies outside of the shape. While you have the additional effort of generating more points than you ultimately ... (more) |
— | almost 2 years ago |
| Comment | Post #292225 |
Are you sure you didn't miss a complex conjugation in the inner product formula? (more) |
— | about 2 years ago |
| Comment | Post #292215 |
If you know in advance that the thing you search is in one of those four smaller rectangles (rather than anywhere in the 9 rectangles that make up the three squares), that would be exactly the kind of information which would change the probability. Except that if you knew it before even searching... (more) |
— | about 2 years ago |
| Comment | Post #292207 |
Of course it does. If the top right square now has a probability of $2/3$, then this means it is twice as likely in the upper right square than not. Which is a very strong hint that this square is the one which should be looked at next. Indeed, the probability distribution is exactly that: The in... (more) |
— | about 2 years ago |
| Comment | Post #292207 |
Well, I guess you could find a piece of paper with additional information in the visited square. *Somewhere* the extra information must come from, after all. (more) |
— | about 2 years ago |
| Comment | Post #292189 |
Thank you for the answer and the reference. A small note: When you write $\lg$, you obviously mean the base 2 logarithm (the paper uses just comparisons with powers of 2). You should be more explicit about that, either by writing $\log_2$, or by using one of the standard notations for the base 2 ... (more) |
— | about 2 years ago |
| Edit | Post #292178 | Initial revision | — | about 2 years ago |
| Question | — |
Is there a closed formula for multiplication of imaginary units in the direct limit of the Cayley-Dickson construction? The Cayley-Dickson construction is a way to systematically construct, starting from the real numbers, a sequence of ever higher-dimensional real algebras $Ak$ which starts with complex numbers and quaternions. The rules are as follows: Each algebra comes with an operation called conjugat... (more) |
— | about 2 years ago |
| Edit | Post #292099 |
Post edited: Fixed a typo |
— | about 2 years ago |
| Edit | Post #292099 | Initial revision | — | about 2 years ago |
| Answer | — |
A: What does it mean by saying that $C([0,1])$ is a subset of $L^\infty([0,1])$? This is actually an instance of a broader concept called identification. To understand the concept of identification, it is possibly better to first look at a simpler example, and only then see how this applies to $L^\infty([0,1])$ versus $C([0,1])$. What is identification? Example: $\mathbb ... (more) |
— | about 2 years ago |
| Edit | Post #291882 |
Post edited: fixed a typo |
— | about 2 years ago |
| Edit | Post #291882 |
Post edited: |
— | about 2 years ago |
| Edit | Post #291882 | Initial revision | — | about 2 years ago |
| Question | — |
For which spaces are all continuous functions either constant or the identity? The Sierpinski space has a particular property: All continuous functions to itself are either the identity or constant. Obviously the empty space and the singleton space share this property. My question is now: Are there other such spaces, and how would you find/construct them? I already fo... (more) |
— | about 2 years ago |
| Comment | Post #290765 |
Correct me if I'm wrong, but AFAIK one of the assumptions of the Gödel incompleteness theorem is that you have a *finite* number of axioms or axiom schemes. Which makes sense given the fact that any axiom system we can actually handle will be finite in this way. However it doesn't preclude that t... (more) |
— | over 2 years ago |
| Comment | Post #291687 |
Thank you. Together with the answer of the only-of part below, that completely answers my question.
(more) |
— | over 2 years ago |
| Comment | Post #291691 |
Thank you. That's a quite elegant proof (I actually never considered that the odd numbers form a group modulo a power of two, though it's easy to check; that's a nice fact already by itself). (more) |
— | over 2 years ago |
| Edit | Post #291687 |
Post edited: Fixed some errors |
— | over 2 years ago |
| Comment | Post #291687 |
You're right on all accounts. I'll edit immediately. Sorry for the confusion. (more) |
— | over 2 years ago |
| Comment | Post #287887 |
Thank you (and sorry for the late reaction; I've not been on the site for a long time). Your explanation why my construction fails to do what I intended was very helpful. (more) |
— | over 2 years ago |
| Edit | Post #291687 | Initial revision | — | over 2 years ago |
| Question | — |
All numbers are triangular modulo $N$ iff $N$ is a power of $2$? When thinking about binary representations of triangular numbers, I noticed an interesting property: In the cases I've tested, for the numbers from $0$ to $2^n-1$, each combination of the last $n$ bits occurs exactly once, that is, $k\mapsto k(k+1)/2 \bmod 2^n$ is a bijection on the set $\{0,\... (more) |
— | over 2 years ago |
| Comment | Post #288097 |
Thank you (and sorry for late reply, I wasn't on the site for quite some time).
(more) |
— | over 2 years ago |
| Comment | Post #291562 |
Actually for $x\ne 0$, $\left|x\right|$ is differentiable; it's derivative there agrees with the sign function. So as long as you exclude $0$ from the domain, there is no problem with taking the derivative. Of course that also means that the result also can only be used for $x\ne 0$, but then, it... (more) |
— | over 2 years ago |
| Comment | Post #291588 |
Thank you for your answer.
Either I don't correctly understand your answer, or I see a gap in your argument.
Let's denote the product topology of $V\times V$ when $V$ is equipped with the topology $\mathcal T_k$, with $\mathcal T_k\otimes\mathcal T_k$. Then obviously, the fact that $+$ is c... (more) |
— | over 2 years ago |
| Edit | Post #287880 |
Post edited: Removed further application because I now think I got it wrong |
— | over 3 years ago |
| Edit | Post #287880 | Initial revision | — | over 3 years ago |
| Question | — |
Unification and generalization of limit and colimit The recent post by r reminded me of my own idea to unify and generalise limit and colimit. I also only occasionally dabble in category theory, and thus I wonder if the following is a sound construction, and if so, if it is a known concept. The idea is as follows: We have a category $... (more) |
— | over 3 years ago |
| Edit | Post #287848 |
Post edited: fixed some grammar errors |
— | over 3 years ago |
| Edit | Post #287848 |
Post edited: fixed typo |
— | over 3 years ago |
| Edit | Post #287848 | Initial revision | — | over 3 years ago |
| Answer | — |
A: Criterion in terms of the bases for determining whether one topology is finer than another A counterexample would be the set $S=\{0,1,2\}$ with the topologies $\mathcal T = \{\emptyset, \{0\}, \{0,1\}, S\}$and $\mathcal T' = \{\emptyset,\{0\},S\}$. Clearly $\mathcal T'$ is not finer than $\mathcal T$. Now $\mathcal T$ is generated by the basis $\mathscr B = \{\{0\},\{0,1\},S\}$ and ... (more) |
— | over 3 years ago |