Something I find really cool is that the definition for coprime ideals are like directly motivated by elementary number theory. Like you say two ideals $\mathfrak{a}$ and $\mathfrak{b}$ are coprime if $\mathfrak{a} + \mathfrak{b} = (1)$ and then you get a characterization of them in the sense that $\mathfrak{a}$ and $\mathfrak{b}$ are coprime iff there exists $x \in \mathfrak{a}$ and $y \in \mathfrak{b}$ such that $x + y =1$
which is basically a ring theoretic version of Bezouts idenitity from elementary number theory
Remember that equivalence relations are objects on sets, so if you have a monoid structure you would not expect something defined as a set construct to be compatible with the monoid structure
If $A$ is a 4x3 Matrix and {v1, v2, v3} is a linearly dependent set of vectors in R^3, then {Av1, Av2, Av3} is also a linearly dependent set in R^4.
This statement holds because of the following, or not?
If $v_1,v_2,v_3$ are linearly dependent you can find constants $a_1, a_2, a_3$ not all $0$ such that $a_1 v_1 + a_2 v_2 + a_3 v_3 = 0$. Now multiplying this by $A$ we get: $A ( a_1 v_1 + a_2 v_2 + a_3 v_3 ) = A 0 = 0$. Then we get $a_1 ( A v_1 ) + a_2 ( A v_2 ) + a_3 ( A v_3 ) = 0$ where one of $a_1, a_2, a_3$ is not equal to $0$.
Let $\{\Omega_i\}_{i\in I}$ a famille of connected sets such that $$\exists i_0\in I, \Omega_{i_0}\cap \Omega_j\neq \emptyset,\forall j\in I$$ I want to prove that $\bigcup_{i\in I} \Omega_i$ is connected.
If I suppose that $\bigcup \Omega_i$ is not connected then, there exists two non empty open sets $A,B$ from $\bigcup \Omega_i$ such that $$ \begin{cases} \bigcup \Omega_i= A\cup B\\ A\cap B=\emptyset \end{cases} $$ we have $\forall i\in I, \Omega_i\subset \bigcup_{i\in I}\Omega_i=A\cup B$ then $\Omega_{i_0}\subset A\cup B$ as it is connected we have $$\Omega_{i_0}\subset A~\text{or} ~ …
Hi all, I'm trying to learn about direct sums of subspaces from some videos on the Internet prying to digging into a textbook... Is there a relationship between direct sums and orthogonal complements?
@Vrouvrou . If you want to see tex in chat, store the first bookmark suggested there. THen you can run it when you get into the room. I just use the second bookmark manually when I need it.
@MikeMiller I try a counter example: the monoid $\{e,A,a,b\}$ with three equivalence classes $\{e\}$,$\{A,a\}$,$\{b\}$ and $ab=e$ and $Ab=a$, correct? (You're right!)
@JakeS Using the (external) direct sum, you can make a bigger vector space out of separate vector spaces, and using the internal version you can break up a vector space into smaller pieces. Really they're the same thing, from the right viewpoint.
I notice that the notes from my course only talk about sums and direct sums of vector subspaces, whereas the videos I'm watching to try and ground myself before digging into the text speak only of orthogonal complements
I understand they're different but related concepts, but is this normal?
By the way, as a complement to my first course in Linear Algebra I'm following Gilbert Strang's course and textbook, do either of you have any further suggestions?
not completely though, not much is missing as well
Its a monoid as long as all exist and $ex=xe=x$ for all $x\in$ the monoid.
Well I forgot associativity, I hope it can be
@MikeMiller @TobiasKildetoft OK lets use then the Klein four group as our monoid: $\{e,a,b,c\}$ and the eq.cl.es $\{a\}$, $\{b\}$,$\{cd\}$. Then $ab=c$ and $ac=b$, contradiction.
@JakeS: Strang's course is very non-rigorous. If you want to see some more proof-oriented lectures, also emphasizing the geometry, you can try my videos on YouTube (see my profile). It sounds like your course is a bit more advanced than Strang's. The direct sum construction has nothing to do with dot product (whereas, of course, orthogonal complement does). A special case of the direct sum construction is to point out that $\Bbb R^n = V\oplus V^\perp$, but you can do lots more things ...
I'll check out your videos @TedShifrin! Thanks! Unfortunately, I'm now in my second semester of Linear Algebra with only a very rudimentary grasp of some concepts from my first course, namely the associated matrix of a linear transformation and projections.
If you look at the right lectures of mine, they might help with that stuff plus more basic things like dimension, basis, etc. Plus proofs (whereas Strang does no proofs).
Do you cover those ideas in your course? (which I am now perusing! Looks fantastic, though it covers material that I'll be covering over several undergraduate courses)
@geocalc33 Strictly speaking their not really non sensical, they just have the empty set as solution (for $s=0$). In a way an equation in variables is the very same thing as the space of solutions. So in a strict sense your set consists of all equations assigning the empty set to $x$ and $y$. So in a sense its "not even wrong".
It looks like a great resource. I'll be keeping it bookmarked for calculus as well. I see quite a few that I'll be watching. Do you cover associated matrix of linear transformations anywhere?
Apologies to the rest of the chat for cluttering it up with my questions!
Sure, @Jake, quite early ... in the first dozen lectures, where I'm doing only the standard matrix. Later in the course there are projection matrices, etc., and then the discussion of a matrix for a map $V\to V$ with respect to an arbitrary basis and change of basis formula (that's within the last ten of the entire year), leading in to eigenvalues, eigenvectors, and diagonalization questions.
Given your book is somewhat more general than the course I'm taking, and Strang's doesn't do much as far as proofs are concerned, do you have any suggestions for resources to go with the ideas you cover on linear algebra?
Unfortunately the notes for my course are both in a second language and, well, generally awful where clarity is concerned.
I wrote an intermediate-level beginning linear algebra book (separate from the course on the videos). If you're more interested in proofs and an abstract discussion, I'm fond of Friedberg-Insel-Spence.
@MikeMiller @TobiasKildetoft OK again the Klein four group $\{e,a,b,c\}$ and equivlance classes $\{e\}$,$\{a,b\}$ ,$\{c\}$. $aa=e$ and $ab=c$. $e\not\sim c$. Contradiction.
@MikeMiller as you said I can independently of the "monoid" structure (which here is actually a group, freely choose what the classes are and it does not interfere with the structure of the monoid). Thanks!!
Any equivalence relation on a group that satisfies h ~ g implies ah ~ ag is necessarily of the form "h and g are the same left cosets of a given subgroup"
So in particular equivalence classes are all the same cardinality
@MikeMiller btw. the theorem I just try to digest is: In case of such an equivalence relation $R$ on a monoid $G$, the equivalence classes form also a monoid called $G/R$.
@TedShifrin Bonjour, pouvez vous voir si cette preuve est juste s'il vous plait
Let $\{\Omega_i\}_{i\in I}$ a famille of connected sets such that $$\exists i_0\in I, \Omega_{i_0}\cap \Omega_j\neq \emptyset,\forall j\in I$$ I want to prove that $\bigcup_{i\in I} \Omega_i$ is connected.
If I suppose that $\bigcup \Omega_i$ is not connected then, there exists two non empty open sets $A,B$ from $\bigcup \Omega_i$ such that $$ \begin{cases} \bigcup \Omega_i= A\cup B\\ A\cap B=\emptyset \end{cases} $$ we have $\forall i\in I, \Omega_i\subset \bigcup_{i\in I}\Omega_i=A\cup B$ then $\Omega_{i_0}\subset A\cup B$ as it is connected we have $$\Omega_{i_0}\subset A~\text{or} ~ \O…
Okay it seems correct to me @Vrouvrou, but I'd write it up a bit differently like this: Suppose we had a family of connected sets $\{C_i\}_{i \in I}$ with nonempty intersection $\bigcap_{i \in I} C_i \neq \emptyset$. Let $p \in \bigcap_{i \in I}C_i$. Suppose $\bigcup_{i \in I}C_i$ is disconnected.
Then there exist open sets $A$ and $B$ such that $A, B$ are nonempty, have empty intersection and their union is all of $\bigcup_{i\in I}C_i$. Consider a connected set $C_j$ in this family. It contains the point $p$. Note that we can't have both $C_j \cap A \neq \emptyset$ and $C_j \cap B \neq \emptyset$ otherwise $C_j$ would be disconnected.
So suppose WLOG that $C_j \subseteq A$. It follows that $p \in A$ and thus any other connected set $C_k$ in this family must be a subset of $A$ and have empty intersection with $B$, otherwise we get a contradiction by obtaining (in the same way as before) two open sets $C_k \cap A$ and $C_k \cap B$ which disconnect $C_k$. Thus we must have $\bigcup_{i \in I}C_i \subseteq A \implies \bigcup_{i \in I}C_i = A$ and so we get $B = \emptyset$ since $A$ and $B$ have empty intersection, a contradiction.
@TedShifrin It means that for all $\epsilon>0$ there exists an $N\in\mathbb{N}$ such that $f_n(x)-f(x)<\epsilon$ for all $n\geq N$ and all $x\in\mathbb{S}$. I don't know what subset of $\mathbb{R}$ is though.
1. a lattice is a partially ordered set 2. it is complete when arbitrary supremum and infimum exist 3. the topology is generated by sets of the form {x | a < x < b}
This is better, now you have proper nbhds of the least and greates element
the lattice of $\Bbb N$ ordered by divisibility, i.e. $a<b\iff a\mid b$ has some infinite discrete subsets though, like $\{p\}$ should be open for all primes $p$ since it is the interval $(1,p^2)$, this might be a problem for compactness
@OskarTegby If $|M_n|<\epsilon$ for all $n\ge N$, then doesn't it hold for all $x\in\Bbb R$? Suppose, instead, that $M_n = 1$ and $x_n = 1/n$. What happens then? Well, do we need more information?
I'll return home to the city tomorrow, there are no boars and good internet there but the temperature will be unbearable so I'm not convinced that's a good deal
I didn't even know 43°C was possible, we reach 40°C in the worst summers in Brescia, but it usually tops at 38°C. The real problem in Brescia is the humidity, it makes the heat suffocating
there was some discussion of the poincare-perelman proof in here awhile back esp wrt the new yorker Nasar-Gruber article. fyi reddit recently revisited/ rehashed some of the same angles https://www.reddit.com/r/math/comments/98i9j7/the_clash_over_the_poincar%C3%A9_conjecture_2006/
