Nowadays, I am just studying the book wrote by Joram Lindenstrauss and Yoav Benyamini,i.e. Geometric Nonlinear Functional Analysis. The putfroward "maximal ε-separated set".I really can not understand this generalization.But I can not find this word in any other books. If you know this definition...
I should just print screen my attempt. I know I screwed up somewhere. http://prntscr.com/j4kfso http://prntscr.com/j4kfvi I know it doesn't sound right at allllllllllllllllllllllllllll I even tried a few examples very carefully
@Semiclassical I still don't know a unifying combinatorial model for many of the times they pop up though. The most compelling one is the number of Dyck words (binary words such that at each truncation, the number of 1's is more than the number of 0's) of length n, but I don't know how that connects with the fact that the n-th Catalan number also counts the number of triangulations of a convex (n+2)-gon with numbered vertices.
And in doing so we did some stuff like affine spaces, which related to the card game SET. Also projective planes (both as incidence structures and the specific case of Galois planes) have been a thing
Graphs, a bit of groups came up as well in the context of automorphism groups of graphs, and today he very quickly reviewed linear algebra and discrete probability
I imagine soon we're gonna be talking about spectral graph theory and some of the ways in which you use probability to talk about combinatorics
$R = \Bbb Q^\Bbb N$ is a ring. The Cauchy sequences form a subring $R'$. Those that converge to zero form an ideal $I$ of $R'$. Then, $\Bbb R:=R'/I$ is a subquotient of $R$
@TedShifrin Had a look. So you simply proved that ambient isomorphism doesn't preserve the ring structure associated to the algebraic variety in the differentiable category, which is quite expected.
Without regular map there's no hope of having that.
I don't know what restriction a polynomial embedding poses on the differential geometry of the embedded submanifold, but Nash's theorem says it poses no restriction on the topology
(Unrelated) I wonder if it is possible to have some kind of notion of "remainder group" G' or "remainder ring" R' such that $G = G/N (some operator) N \cup G' and R=R/I (some operator) I \cup R'$ analogous to polynomial division. But then one obvious issue is that each equivalence class in the quotient objects may not necessary have the same number of members thus to recover the details before they are being quotiented, something more than I and R has to be "multiplied"
ok let me try again. Recall in the ring of polynomials, we have this notion of division where given polynomial p and q where deg(p) > deg(q) then we can express p=qd+r where d and r are also polynomials
What I am thinking about is whether there exists some kind of generalisation such that given some abstract algebraic structure $A$, and something that is a sub algebra $B \subset A$, we can write $A=C * B + D$ where $C$ is obtained by taking some equivalence class on $A$ with the equivalence relation determined by B and then there is some "leftovers" $D$ (and * is some operator between two algebraic structures such that it "undoes the equivalence class").
Thus for the case of quotient groups, rings etc., C will become a trivial object so that A=C * B
With quotients, the $C\cdot B$ stuff is the stuff that goes to $0$ in the quotient, and the $D$ stuff is what gives you the representatives of the elements of the quotient.
I guess that question arises because I am wondering why in quotient groups Q=G/N, n must be a normal subgroup and then in quotient rings R/I, I must be an ideal. I knew that normal subgroups have the property that the left and right cosets are equal, but I don't really know what will happen if we have some Q'=G/H where H is a subgroup but not normal.
Using the analogy of division, I will expect there is some "leftovers" because quotient objects A/B sounds really like A can be "divided" by B in some general sense since every element is being binned into equivalence class and we only take one representative from it
so for e.g. Q'=G/H it seems some elements of G get binned into more than one equivalence class but I don't know whether we can get some notion of "leftovers" from this somehow
@LeakyNun Yes I know, but what I am more interested and trying to understand is if we try to run the definition of quotient groups over something that is not normal. How does it gone wrong
Hello, guys. Please, take a look at this question if you can. It's pretty "basic" topology and doesn't require much prerequisites (just the definition of topology and nets convergence) math.stackexchange.com/questions/2734644/…
Ah I see, so unlike in the case of polynomials, integers and more generally euclidian domains, trying to do A/B = {x,y in A, ~ in B | (x,y) in ~} when B is e.g. not a normal subgroup, not an ideal, not "divisible" will in general change the nature of the object thus there is no way to define something such that the formal expression "C = A/B * B + D" makes sense
The question is, briefly speaking: I have a "good" convergence notion: a rule $\scr C$ on a set $X$ that says which nets converge to which points. This gives me a topology $T(\scr C)$. Then I can consider the convergence notion $C(T(\scr C))$ induced by this topology. The claim is: $C(T(\scr C))=\scr C$.
We could think in the question about topologies: i.e. $T(C(\tau))=\tau$, for any topology $\tau.$
@TedShifrin The four conditions are needed in order to $T(\scr C)$ be indeed a topology. In fact, this is motivated because every convergence notion from a topology have these four as properties
(a) If $x_i=x$ for each $i\in I$, then the net $(x_i)$ converges to $x$. (b) If $(x_i)$ converges to $x$, then every subnet of $(x_i)$ converges to $x$. (c) If every subnet of $(x_i)$ has a subnet converging to $x$, then $(x_i)$ converges to $x$. (d) (Diagonal principle) If $(x_i)$ converges to $x$ and, for each $i\in I$, a net $(x^i_j)_{j\in J_i}$ converges to $x_i$, then there is a diagonal net converging to $x$; i.e., the net $(x^i_j)_{i\in I,\,j\in J_i}$, ordered lexicographically by $I$, then by $J_i$, has a subnet which converges to $x$.
(a) is basically constant nets analogous to constant sequences
@Secret What I did up to now: suppose $(x_i)$ does not $\scr C$-converge to $x$ and let's try to show that $(x_i)$ does not $C(T(\scr C))$-converge to $x$. Note that for this we just need to find a sub-net of $(x_i)$ that does not $C(T(\scr C))$-converge to $x$.
By $\mathscr{C}$ convergent, do you mean the convergence as defined by (a),(b),(c),(d) and by $C$ convergent, you mean the convergence as defined by that closure $\bar{E}$?
Then, this implies that there is some $i_0\in I$ s.t. $i_0\prec i\implies x_i\neq x$ (otherwise, we could construct a subnet constant to $x$, which by axiom (a) is $\scr C$-convergent to $x$. But $x_i$ does not have any $\scr C$-convergent subnet to $x$!)
@Zee If you can point me out to where the fact that $\lim_{x \to \infty} \cos(x)$ doesn't exist poses a counterexample to what I said (which is entirely possible), I'd appreciate if you point it out. If you're trolling, shove off.
@Secret $\scr C$ convergent is the given original convergence notion. $T(\scr C)$ the induced topology by $\scr C$ and then $C(T(\scr C))$ the usual convergence notion induced by a topology, in the case, $T(\scr C)$.
So, in the end, if we suppose $(x_i)$ does not $\scr C$-converge to $x$, we can also suppose, possibly passing to a subnet, that $(x_i)$ does not admit any $\scr C$-convergent subnet to $x$ and that $x_i\neq x$, for all $i\in I$. But then...?
I think if someone points out I have made a flawed argument I have a responsibility to be vigilant about it than ignoring the person who caught the error.
Now it's entirely possible that the opponent is trolling, which gets apparent quite soon.
Well, in this case $\{x_i\,:\, i \in I\}$ is not closed in the topology $T(\scr C)$ but I simply don't see how to get a subnet of it that does not $C(T(\scr C))$-converge to $x$... :'(
@BalarkaSen "I think if someone points out I have made a flawed argument I have a responsibility to"... right, except when the name of that "someone" starts with "Z" and ends with "ee"
I think it's easier for me to think of a $(x_i)$ such that it diverges yet it contains at least one subnet $(y_i)$ that converges. For example, take the graph of the function $x \sin x$ as $(x_i)$ and pick the values which form the harmonic sequence in descending order to make $(y_i)$. Then (c) is still violeted and there are some neighbourhoods in the induced topology that it fails to converge
@Secret Yes, given any topology $\tau$, you can define the usual convergence notion $C(\tau)$: $\forall U\in \tau, x\in U, \exists i_0\in I\,:\, i_0\prec i\implies x_i\in U$.
hmm so that means (a),(b),(c),(d) should give us all the neighbourhoods and open sets of $T(\mathscr{C})$, which should since we are dealing with nets here not just sequences. Let me think...
@Secret Yes, they do. We don't work only with sequences because this restricts the possible topologies. In fact, it works if the topology obtained is metrizable.
Ah right, I mixed up open sets with neighbourhoods, because I like to work with open sets in the past
(b) is like the generalisation of Cauchy sequences, thus it makes sense that for these nets, all their subnets converge to the same point(s)
(c) is kinda like the reverse of (b), in fact, it feels like the subnets form closed sets thus their union will also be closed for these nets with the same limit points
So that leaves (d), which kinda reads like each element in the net has some nets $(x_j^i)_{j \in J_i}$ that converge to them and induced by a family $\{J\}_i\in I$, and thus we can just take one representative from each of these and form a net which is then guarantee to have a subset that converges to the same point
But why is this necessary, what kind of neighbourhood it defines, and isn't such diagonal nets already included in (b)?
Since what you try to prove is $C(T(\mathscr{C})) = \mathscr{C}$, it means if some $(x_i)$ is not $\mathscr{C}$ convergent, then $(x_i)$ will not be $C$ convergent as well (likewise, if $(x_i)$ I $\mathscr{C}$ convergent, then $(x_i)$ will be $C$ convergent). Thus your counterexample should be set up as some $(y_i)$ such that it is $C$ convergent but not $\mathscr{C}$ convergent (and vise versa)
For example, suppose I say: a net $(x_i)\subset \Bbb R$ converges to $x\in \Bbb R$ if $x_i$ is even, for all $i \in I$. This is not what we would like to call a convergence notion, since the constant net $x_i=1$ does not converge to $1$.
So (a),...,(d) are reasonable conditions for a definition of convergence notion.
so dropping (d) will probably mean there are constant sequence (i.e. points) that the only neighbourhood is the whole set
which is indeed very strange but still sound, but I digress
But anyway, I think your counterexample need to begin with a sequence such that (x) is C convergent but not $\mathscr{C}$ (and the reverse), instead of starting with (x) that is not $\mathscr{C}$ or $C$ convergent
Well, I have to go now. Thank you for the discussion, @Secret. If you have some idea, please, answer in the question link! (am thinking about starting a bounty, maybe)
because the proof means the convergence properties are preserved by the two topologies, so picking some (x) that is not convergent and then trying to prove whether it converge in another topology won't work since it can converge (if the counterexample exists) or not converge
Hmm... the definition of $\bar{E}$ pick out the nets in $E$ such that these nets converge in the sense that for all neighbourhood of x, $(x_i)$ will $C$ converges. Then (a) satisfies trivially, (b), (c) also satisfies by the definition of $C$ convergence in a net (thus BSc forms an iff condition). finally since every $x \in \bar{E}$ has some net $(x_{ij})$ $C$ converge to them, then we can by axiom of choice, pick one representative from each of these $(x_{ij})$ in descending preorder to form the net $(x_j^i)_{j \in J^I}$. Then it must have a subset that converges to x. Thus (d) is satisfie…
There are many similar questions to mine in the site, but I'm still not sure.
Let $X$ be a vector space with $T_1$ and $T_2$ two topologies that make $X$ a TVS (Hausdorff). If I want to show that $T_1=T_2$, does it suffice to show that every converegent net $(x_{\lambda})_{\lambda}$ in $X$ to som...
The way the question is posed, it seems implied that we can use the first part to prove the second. But that is impossible as far I can tell: the induced composition $R \cong R^4$ is not $R$-linear.
However, the first isomorphism can be made $R$-linear by using the following structure:
$\alpha...
you need some assumptions on A
okay maybe the linked question is not the best explanation
I don't see how A^m -> A^n can be injective for m>n if A is finite, but if A is infinite then by cardinality alone I can see how. However, if you want it to be an injective homomorphism, then I am not sure yet
In mathematics, more specifically in the field of ring theory, a ring has the invariant basis number (IBN) property if all finitely generated free left modules over R have a well-defined rank. In the case of fields, the IBN property becomes the statement that finite-dimensional vector spaces have a unique dimension.
== Definition ==
A ring R has invariant basis number (IBN) if for all positive integers m and n, Rm isomorphic to Rn (as left R-modules) implies that m = n.
Equivalently, this means there do not exist distinct positive integers m and n such that Rm is isomorphic to Rn.
Rephrasing the...
In mathematics, more specifically abstract algebra and ring theory, a noncommutative ring is a ring whose multiplication is not commutative; that is, there exists a and b in R with a·b ≠ b·a. Many authors use the term noncommutative ring to refer to rings which are not necessarily commutative, and hence include commutative rings in their definition. Noncommutative algebra is the study of results applying to rings that are not required to be commutative. Many important results in the field of noncommutative algebra area apply to commutative rings as special cases.
Although some authors do not assume...
Reading the definition of partition of unity:
Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:
For each $x \in A$ we have $0 ...
Well, partitions of unity are defined here for subsets of $\Bbb R^n$, and any subspace of a second-countable space is second-countable.
Unless what you mean is, if an arbitrary space (or arbitrary Hausdorff space, or something) has partitions of unity, is it second-countable? That I'm not sure of.
Right, but partition of unity allow us to extend local properties to global ones, this help us to contrust topological spaces, but no every topological spaces need to be second countable.
@Fargle Yeah, I meant exactly that, sorry if I did not put it on a fully context.
It's no problem. I'll have to do some serious thinking on this one.
Ah--it seems the implication doesn't always hold.
The real line under the lower limit topology (Sorgenfrey line) is paracompact Hausdorff, and therefore has partitions of unity, but it fails to be second countable.
@Fargle I see. Then, if we suppose that a Hausdorff space has paritions of unity, how should I proceed in order to build second countable for that space?
@Fargle I know, but I do not want to assume anything, actually I am trying to build piece by piece a hausdorff, second-countable mtrizability manifold.
@user17629 I think what that comment means is that you can still do partitions of unity without demanding that your space be second-countable, as long as you assume Hausdorff, paracompact, and locally Euclidean. However, you don't get metrizability unless you have second-countability, as the Sorgenfrey line is an example of a Hausdorff, paracompact, locally Euclidean space which fails second-countability.
And isn't metrizable as a result.
Or, no, wait, I may have some details wrong here. Ignore me for a sec.
When I study topological manifold, I think some property of manifolds are so important that they can "almost characterize" manifolds. But I know a topological manifold is not easily to be characterized because of its locally Euclidean property.
I want to find a "weird" example of a locally compa...
The way the question is posed, it seems implied that we can use the first part to prove the second. But that is impossible as far I can tell: the induced composition $R \cong R^4$ is not $R$-linear.
However, the first isomorphism can be made $R$-linear by using the following structure:
$\alpha...
Reading the definition of partition of unity:
Let $A\subset \Bbb R^n$ and let $\mathcal{O}$ be an open cover of $A$. Then there is a collection $\Phi$ of $C^\infty$ functions $\varphi$ defined in an open set containing $A$, with the following properties:
For each $x \in A$ we have $0 ...
