for each ordinal $\alpha$ you well-order $\alpha \times \alpha$ by first comparing their max, and then the first coordinate, then the second @MatheinBoulomenos
then you define $\Gamma: \alpha \times \alpha \to \mathsf{Ord}$ by sending $(\beta, \gamma)$ to the order type of the initial segment of $(\beta, \gamma)$ in $\alpha \times \alpha$
then you can glue them all together and it turns out that $\Gamma(\alpha,\alpha) = \alpha$ (you assume otherwise and choose the minimum counterexample)
In ZF, the following are equivalent:
(a) For every nonempty set there is a binary operation making it a group
(b) Axiom of choice
Non trivial direction [(a) $\to$ (b)]:
The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from...
@Ultradark would you like to publish a paper with me, ring-theoretic smallest grammar? Lots of basic facts can be worked out with basic AA understanding
Usually you order by max first so that you when you think about it as a well order of $\mathsf{Ord}\times\mathsf{Ord}$ the initial segments are set sized
It's kind of a natural question after noting that the following are theorems of ZFC: "there exist an uncountable family $F$ of smooth functions $\Bbb R\to\Bbb R$ such that $\{f(x)\mid f\in F\}$ has at most two elements for every $x\in\Bbb R$", "If a family $F$ of entire functions has the property that $\{f(z)\mid f\in F\}$ is finite for every $z\in\Bbb C$, then $F$ is finite"
@MatheinBoulomenos That's a good point
Also "If $F$ is an uncountable family of polynomials, then $\{f(z)\mid f\in F\}$ is uncountable for all $z\in \Bbb C$ except at most finitely many"
he was getting impatient with some guy because he was going like "So for example $\Bbb R$ is open then?" and Banagl is like "no openness is relative" and he's like "yeah but the definition says $X$ is open so if $\Bbb R = X$ then $\Bbb R$ is open"
In ZF, the following are equivalent:
(a) For every nonempty set there is a binary operation making it a group
(b) Axiom of choice
Non trivial direction [(a) $\to$ (b)]:
The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from...
Alternative proof: Prove it by induction for finite sets. For infinite sets, consider the theory of pointed sets, given by the signature of a 0-ary function $*$ and no relations. We have an infinite model $(\Bbb N,1)$. By Skolem-Löwenheim, there are models of every infinite cardinality.
@MatheinBoulomenos consider X and {0}. by trichotomy, we have an injection f: X -> {0} or an injection g: {0} -> X. in the former case we can regard X as a subset of {0} and it is then trivial; in the latter case just pick g(0)
I have an issue about distributions: if I have a function, f(x), and I have a gaussian probability distribution for x
then what if I multiply f(x) with this gaussian? does this gaussian*f(x) curve has a special name?
I would want to have a probability distribution curve for f(x) but I don't think that this gaussian*f(x) is that, because its integral does not evaluate to 1 in the range 0 infinity
Hi I was recently in an interview and the definiteness of the matrix $X^TX$ came up. The interviewer stated that it is positive semi definite as this is a property of covariance matrices and $X^TX$ is the special covariance matrix, $cov(X,X)$
Given any set of $n$ regular convex polygons, can you algorithmically add irregular convex polygons to get a set of polygons that will tessellate the plane?
@Rithaniel Can't you simply do the following. Pick an equilateral triangle that is large enough to contain each and every one of those regular convex polygons. Then include whatever polygons you need to fill up the complement of any polygon within that equilateral triangle. And then you can tesselate the plane with those equilateral triangles. Done.
The standard proof (or I think that´s a standard one) that we can have as much as we want consecutive composite numbers is to observe the numbers $n!+2,...,n!+n$, which are $n-1$ consecutive composite numbers which also means that there are arbitrarily large gaps between two consecutive primes.
...
My question is essentially I know that the diagonalization theorem is: The diagonalization theorem states that an $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors, i.e., if the matrix rank of the matrix formed by the eigenvectors is $n$
Is this equivalent to saying that the eigenvectors form a basis for the eigenspace (aka span the eigenspace)?
In ZF, the following are equivalent:
(a) For every nonempty set there is a binary operation making it a group
(b) Axiom of choice
Non trivial direction [(a) $\to$ (b)]:
The trick is Hartogs' construction which gives for every set $X$ an ordinal $\aleph(X)$ such that there is no injection from...
