$(W,\preceq)$, the surreals and $\mathbb{R}+\omega_{\alpha}$ will be quite useful sets and proper classes to help me build some geometric intuition (if there exist a well defined one) on sets of these sizes, and thus a better intuition of different levels of continuua
@AkivaWeinberger If there is no well ordering on the cardinals due to lack of choice, then $\mathfrak{c}$ may not even be an ordinal, and thus the expression $\mathfrak{c}+\omega_1$ may not obey the ordering of the ordinals, I guess....
@arctictern I know that sounds intuitive in terms of partitions, but then you will expect W to suffer from a similar notion of "rationals are uncountable" contradiction? when some order preserving map embed it to one of its subsets of one level lower in cardinality. Or is it because the countable subsets of W are never dense, thus discrete and hence avoided this problem?
> Observe that $\{r×R | r∈R\}$ is an uncountable collection of pairwise-disjoint non-degenerate ⪯-intervals of $R^2$. Hence, $(R^2,⪯)$ cannot be embedded into (R,≤R) in an order-preserving manner; if this were not the case, then RR would contain uncountably many pairwise-disjoint non-degenerate intervals, so by picking a rational number from each of these intervals, we would end up with uncountably many rational numbers ⎯ contradiction.
How does W avoid this problem as if there exists a order preserving map to embed it into one of its countable subsets, then we can conclude that the countable subset have uncountably many elements, hence contradiction in a similar manner to how if there is order preserving map from some larger set to the reals, will conclude a countable subset (the rationals) will have uncountable elements?
So the only way out is that the countable subset cannot be dense, otherwise uncountably pairwise disjoint intervals of W will contain countable elements hence result in the countabe subset have uncountable elements?
indeed, the existence of an embedding of W into a countable set would be a problem, because it mean an uncountable subset fits inside a countable one. you get a contradiction just from cardinality, no order to speak of. but I never said anything about embedding W into a countable subset.
So is it possible for W to have the same ordering of the reals as long it does not contain the reals as a subset (because the answer said that cannot happen)?
> A common example of this is the cross product of vectors; in this case, the absence of an identity element is related to the fact that the direction of any nonzero cross product is always orthogonal to any element multiplied – so that it is not possible to obtain a non-zero vector in the same direction as the original.
If $a\times b=c$ then there exists an $a'$ such that $a'\times(a\times b)=b$. However $a'$ will depend on both $a$ and $b$. When $b$ is nonzero there does not exist an $e$ such that $e\times b=b$, since $e\times b$ must be orthogonal to $b$.
automorphism depends on what category you're working it. in the category of sets, it just means the group of bijections from a set to itself. (usually this isn't done though.)
Is it possible to have a $> \mathfrak{c}$ set that has the same ordering as the reals, contain the reals as a subset, is not a proper class like the surreals or $\mathbb{R}+\omega_{\alpha}$, and does not contain order embedding that result in the rational uncountable contradiction? I think I have looked hard in MSE and cannot find anything similar?
If you're talking about bijections from $\mathbb{Z} \to \mathbb{Z}$ and you denote the set as $\operatorname{Aut}(\mathbb{Z})$, most people would assume you mean a group automorphism of $(\mathbb{Z},+)$.
@arctictern Not well ordered, has the same linear ordering as the reals in the sense of 1,2,3.... Basically the real line but with more elements, but don't become so large and become a proper class
Ok yes. So the polished question is: I want a $> \mathfrak{c}$ set, that has the standard ordering on the reals, has reals as one of its subset, is not a proper class like surreals or $\mathbb{R}+\omega_{\alpha}$, and does not have order embedding maps to result in the rational uncountable contradiction.
The issue in trying to construct this set, is that the reals are already complete, thus there seemed to be no way to put extra elements in?
In plain english, I want a real line of size $> \mathfrak{c}$ and preserve as many properties of the reals including all the subsets, but is not a proper class. I don't know how to describe that better
@Secret What is a "real line"? What does that mean? If it means the real number line, there is only one up to isomorphism, and does not have size $>\frak c$.
hello, can i deduce $(f_3)$ from (f_1) where $(f_1)$: $f\in C([0,\infty[\times\mathbb{R})$ and for some $2<p<2^*, c_0>0$ $$|f(r,u)|\leq c_0(|u|+|u|^{p-1})$$
$(f_3)$ $f(r,u)=o(|u|),|u|\rightarrow 0,~\text{uniformly on } \mathbb{R}^+$
The cool thing about it is that DHMO taught me Dedekind cuts and conway construction behave similarly to dedekind cuts
Before DHMO taught me that I don't understand surreals
Wrote recipe into notebook
[More recipe requests] What if now I want another set $S$ that has the following properties: 1. $|S| > \mathfrak{c}$ 2. $(\mathbb{R},<)$ can be embedded into it (where < is the standard ordering) 3. The whole set is complete, has least upper bound property, totally ordered 4. Is archimedian 5. Not a proper class 6. No order embeddings to trigger the rational uncountable contradiction
@AkivaWeinberger Not having any maps that exists such that one can prove the uncountable disjoint number of interval will lead to the contradiction that rationals are uncountable
As a bonus problem, our professor of real analysis asked us to prove that the real numbers (a complete ordered field) cannot be extended into an Archimedean field, with no definition of what he meant by extending.
I have tried using proof by contradiction to show that if we have some set, $\math...
@AkivaWeinberger formulation of well-ordering principle in terms of Peano arithmetic: $\forall S[S \subseteq \Bbb N \implies \exists n[n \in \Bbb N \land \forall x[x \in S \implies \exists c[c \in \Bbb N \land n+c=x]]]]$ (have fun decoding this)
Ok, so that means, $\mathbb{R}+\omega_{\alpha}$, $\mathbb{R}^{\mathbb{Z}}$, $\mathbb{R}^{\mathbb{R}}$, $(W,\preceq)$ and the surreals will keep me busy in helping me to develop more algebraic and geometric intuition for higher level infinite sets
Writes in notebook. Infinite sets discussion concluded for now...
In mathematics, specifically in differential equations, an equilibrium point is a constant solution to a differential equation.
== Formal definition ==
The point
x
~
∈
R
n
{\displaystyle {\tilde {\mathbf {x} }}\in \mathbb {R} ^{n}}
is an equilibrium point for the differential equation
...
$N$ is Homeomorphic to the unit sphere, how to explain that $N$ separate the space into two components the interior containnig the origin and the exterior ?
Well, assuming the origin isn't on $N$ itself, you can just define the component containing the origin as the "interior", and the other as the exterior.
@Semiclassical I am trying to solve that exercise. " Given an undirected graph G(V,E) and a subset of the edges F, is there a simple cycle on G that goes through all edges of F? Show that this problem is NP-Complete"
@Vrouvrou That link is the 2D version. If you go to the "generalizations" section you find the version for all dimensions.
> Let $X$ be a topological sphere in the $(n+1)$-dimensional Euclidean space $\Bbb R^{n+1}$ $(n > 0)$, i.e. the image of an injective continuous mapping of the $n$-sphere $S^n$ into $\Bbb R^{n+1}$. Then the complement $Y$ of $X$ in $R^{n+1}$ consists of exactly two connected components. One of these components is bounded (the interior) and the other is unbounded (the exterior). The set $X$ is their common boundary.
I think we're going around in circles. The interesting question to ask here is what is the maximum number of disjoint curves you can cut out from the surface without disconnecting it.
What I have in mind when I say that is that the problem emerges from the eigenvalue problem for a certain Hermitean operator. That's bread-and-butter in quantum mechanics.
So I should think of this particular question in terms of the original formulation in terms of linear algebra, which is directly physical, rather than the algebraic-geometry reformulation I made of it.
How do I prove $\overline{\sigma_p} \subseteq \sigma_p \cup \sigma_c$ where $\sigma$ is the continuous spectrum of a bounded operator in a hilbert space.
Err, where $\sigma_c$ is the continuous spectrum, and $\sigma_p$ is the point spectrum
@SteamyRoot i have this set $N=\{u\in X\setminus\{0\}, I'(u)u=0\}$ it is the Nehari manifold , in the book he says that $N$ separate $X$ into two compenents, the compenent containing the origin contain a small ball around the origin
