quick question... IF we have eight people showing up for free concert tickets and we want to figure out how many ways can exactly 3 of them get tickets, isn't that just 8 choose 3 or $\binom{8}{3}$?
ok so if we wanted to find out how many subsets of size k are there from a set of size n isn't it just $\binom{n}{k}$ or am I missing something here since we need different subsets of size k ... maybe it's $\binom{n}{k_{1}}\binom{n-k_{1}}{k_{2}}$ and so forth?
In above argument, we can conclude that $K_1K_2$ is spanned by $\alpha_i\beta_j$ over F, because, closed set under addition and multiplication of $\sum\alpha_i\beta_j$ is a field, right?
Are we implicitly assuming that $K_1,K_2$ are contained in some larger field $L$? Because otherwise we need to check closure under inversion as well, perhaps
On the other hand, I don't think that this works if $K_1$ and $K_2$ are infinite-dimensional. Like, I don't think you can write $\dfrac1{\pi+e}$ as a finite sum $\sum a_{ij}\alpha_i\beta_j$ where $a_{ij}\in\Bbb Q$, $~\alpha_i\in\Bbb Q(\pi)$, and $\beta_j\in\Bbb Q(e)$
(assuming $\pi$ and $e$ are algebraically independent, which they probably are)
In this abstract version of $K_1K_2$, both $x+y$ and $x-y$ are nonzero, so that means $x+y$ is a zero divisor
whereas if they were embedded in a larger field, either $x+y$ or $x-y$ would have to be zero (we don't know which if we don't know what the embeddings are)
because if $Ax=Ay$ then $A(x-y)=0$ and multiplication by $1/A$ (which might not be in $\{\sum a\alpha\beta\}$ but is in $L$) gives $x-y=0$ and thus $x=y$
Unrelated thought
In the quaternions, $ij=-ji$ is a direct consequence of $(ij)^2=-1$
That is, once we have $i^2=j^2=k^2=-1$ and $ij=k$, the rest of the multiplication table is forced
In Hatcher's book on Alg. Top., he calls $e_{\alpha}^n$ a cell. I tried searching through his book for an explanation of this, but I couldn't find anything. Would someone mind explaining what $e_{\alpha}^n$ is; i.e., what is the definition of $e_{\alpha}^n$?
Let $A$ be euclidian ring and $K$ be its field of a fraction.Let $(V,B)$ be a nonzero IPS (inner product space) over $K$. A finitely generated A-submodule
$L ⊆ V $is said to be an $A$ -lattice in $V$ if $L$ contains a $K$-basis of $V$ . As we have
already observed, $L$ must be $A$-free since it ...
You're taking the disjoint union of a bunch of (closed) disks, and then quotienting the boundaries of those disks together under an equivalence relation defined in terms of the "attaching maps"
The (open) interiors of those closed disks are your cells
Do you know what it means to quotient by an equivalence relation?
@LeakyNun Remind me why that is? I think I knew that at one point but forgot it
What, like how if $p(x)=\sum a_nx^n$ and $q(x)=\sum b_nx^n$ then $\sum a_nb_nx^n$ can be found by doing $p(e^{it})*q(e^{it})$ and then substituting in $t=\frac1i\ln x$?
Is there a closed form for the number of $n\times n$ matrices with integer coefficients from $\{-m,\cdots,m\}$ whose determinant is $\pm a\in\mathbb{Z}$?
The problem statement, all variables and given/known data:
Show that if $X$ is a subset of $M$ and $(M,d)$ is separable, then $(X,d)$ is separable. [This may be a little bit trickier than it looks - $E$ may be a countable dense subset of $M$ with $X\cap E = \varnothing$.]
Definitions
Per our boo...
@AkivaWeinberger This is very false for arbitrary topological spaces, for every space $X$ you can put a topology on $X\cup\{p\}$ such that $\{p\}$ is dense, if you start with a nonseparable $X$ then you have a counterexample
In set theoretic topology there was (there is?) a lot of interest in S-spaces, which are regular and hereditarily separable but not Lindelöf spaces and L-spaces, which are regular and hereditarily Lindelöf but not separable spaces
What is the weakest $P$ such that $P$+separable$\implies$hereditarily separable? $P=$metric space works but I wonder if it can be weakened
@AkivaWeinberger Yeah apparently people expected S-spaces and L-spaces to be very symmetric but now it doesn't seem to be the case
I think that "least $\kappa$ such that every open cover of $X$ has a subcover of cardinality $\kappa$" should even have a name. There are a lot of similar cardinal functions with names
I'm actually getting ready to apply to grad school for the spring of 2020. I'm trying to collect as much data on different fields of math as I can as I think about specializations.
The extension of $S$ behaves in a way such that given any fixed polynomial $P$ that fit the n numbers, the n+1 th number is always different from P(n+1)
Model theory is about studying models of theories (no really), a lot of set theory is about studying models of $ZF(C)$, a particular first-order theory, but there's also a lot more being done in set theory
What I want to know is, would $a$ look pseudorandom? Or would it have some sort of pattern that the network would fail to pick up on
I think it will be pseudorandom, cause say the network predicts (given bits),1,0,0,1,0,1,... then a has to be (given bits),0,1,1,0,1,0,.., so there is a perfect anti correlation to the prediction dependent on the seed and settings of the network (because that controls the prediction)
Sometimes I am wondering: If our computers are analogue and does not suffer from the problems of analogue computers in history, will circuit diagrams for e.g. addition have to be that complicated
Problem: For $n \in \Bbb{N}$, let $X_n(\Bbb{Z})$ be the simplicial complex whose vertex set is $\Bbb{Z}$ and such that vertices $v_0,...,v_k$ span a $k$-simplex if and only if $|v_i-v_j| \le n$ for all $i,j$. Using induction, show that $X_n(\Bbb{Z})$ is contractible by showing that it deformation retracts onto $X_{n-1}(\Bbb{Z})$.
I've been thinking about this problem for at least a week and have made no progress. I could use some help.
$X_1(\Bbb{Z})$ is contractible because it is a connected graph.
Consider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$
Edit: A weak vector f...
Consider the vector field restricted to the rationals, $\vec F_\Bbb {Q^2}=(x,y).$ This is a vector field $\vec F:\Bbb Q^2\to \Bbb Q^2.$ Is this vector field weak with respect to the same vector field over the reals: $\vec F_\Bbb {R^2}=(x,y),$ $\vec F: \Bbb R^2\to \Bbb R^2?$
Edit: A weak vector f...