I'm taking the measure theoretic probability course this semester, I've finally seen an example of two distinct random variables with the same moments today, that stuff is weird
@TobiasKildetoft why exactly do we have to be careful with decomposable algebras? If $A=B \times C$ and $b_1, \dots, b_n \in B$ is a positive basis of $B$ that includes $1$ and $c_1, \dots, c_m$ is a positive basis of $C$ that includes $1$, then $b_1 \oplus c_1, \dots, b_n \oplus c_m$ will be a positive basis for $B \times C$, or am I missing something? At least this direction seems to work. (But $B \times C$ may have a positive basis even when $B$ and $C$ don't)
So assuming that the quiver stuff works for basic commutative algebras in general, that should give us by induction with the stuff we have done before that there is always a positive basis in the commutative case
@Lucas It says directly "If $f$ is odd, then $f(s)\cos(ns)$ is odd, so $A_n = 0.$" It only examines $A_n$, how it reduced the examination from "$A_n \cos (nt) + B_n \sin(nt)$ should be odd" to only "$A_n$ should be odd "I hope what I said is clear :)
as long as no products in one of them contain $1$ with non-zero multiplicity, then it works when we take the union of the bases but replace the first one by the sum $b_1+c_1$ (assuming it is $B$ that has this property, otherwise we replace the one corresponding to $c_1$ by it)
But without this property, it becomes a bit tricky to see how one might do a nice change to include $b_1 + c_1$
since whenever such a product exists, we need to subtract something
It might be illustrative to see what the change looks like for $\mathbb{C}^3$ when we regard it as $\mathbb{C}[\mathbb{Z}_2]\times \mathbb{C}$ and change to the basis corresponding to having it be $\mathbb{C}[\mathbb{Z}_3]$
But the union of $b_1 + c_1$ together with $b_i + 0$ and $0 + c_j$ will be a generating system, right? If I have a vector space $V$, a linearly independent subset $X$ and a generating system $Y$ such that $X \subset Y$, then there is a subset of $Y$ that includes $X$ and is a basis? Can't we apply that with $X=\{b_1 + c_1\}$?
We don't even need to work with sums of the form $b_i + c_j$ in general. We can even get a basis that is contained in the union of $1=b_1+c_1$ and the union of the bases from $B$ and $C$
guys, is the next template always okay to use? "For any ___ there exists a ___" . What if I substitute "the" instead of "a" for example? or remove "a" completely, would it be a mistake?
@MatheinBoulomenos Let's say the one coming from the standard basis on it as a group algebra (since that is the sort that causes problems)
If we picked the KL basis we would already have the nice property
Hmm, but the basis for the single copy of $\mathbb{C}$ does have that, so we need to go up to $\mathbb{C}^4$ as two copies of the group algebra with the usual bases.
That basis is $\{(1,1), (1,-1)\}$, right? So we want a basis contained in $\{(1,1,1,1),(1,1,0,0),(1,-1,0,0),(0,0,1,1),(0,0,1,-1)\}$ that includes $1$. If we choose $\{(1,1,1,1),(1,1,0,0),(1,-1,0,0),(0,0,1,-1)\}$, then $(0,0,1,-1)^2=(1,1,1,1)-(1,1,0,0)$
Let $f$ be an arbitrary complex function on $\Bbb{R}$, and define $\varphi(x,\delta) = \sup \{|f(s)-f(t)| : s,t \in (x-\delta, x+ \delta) \}$ and $\varphi (x) = \inf \{\varphi(x,\delta) : \delta > 0 \}$. Prove that $\varphi$ is upper semicontinuous, that $f$ is continuous at a point if and onl...
@TobiasKildetoft is there a natural choice for an isomorphism $\Bbb C[\Bbb Z/2\times \Bbb Z/2] \cong \Bbb C^4$? There seem to be a lot of idempotents and each complete orthogonal set gives an isomorphism
@MatheinBoulomenos Not any particularly nice choices, no (basically, any such choice will be the same as choosing a bijection bwteeen the elements of the group and its irreducible representations).
@MatheinBoulomenos So applying this successively, we get the basis $(1,1,...,1), (1,1,...,1,0), ..., (1,0,...,0)$ for $\mathbb{C}^n$ where each is an idempotent that acts as the identity on all the ones that comes later
Anyway, I need to go now, but this was quite an enlightening discussion.
For a sequence $\{f_n\}$ of measurable functions with common domain $E$, show that the following functions are measurable: $\inf \{f_n\}$, $\sup \{f_n\}$, $\lim \inf \{f_n\}$, and $\lim \sup \{f_n\}$
Here is my proof:
It suffices to show that $\sup \{f_n\}$ and $\lim \sup \{f_n\}$ are ...
The Cult of Infinity] 0. Begin by specifying an alphabet $\mathcal{A}$. This give us the symbols to be used 1. Next, define a formal grammar $\mathcal{G}$ which provide the rules in producing new strings and how they are concatenated, truncated etc. 2. Together, these form the formal language $\mathcal{L}$ where all the details of the foundation $\mathfrak{F}$ is writtened and expressed 3. Next specify a logic $L$ with its inference rules to provide the syntax of reasoning in $\mathfrak{F}$ But before we begin, we need to define a couple of primitive notions:
3d. Predicate P(S): An expression whose truth value depends on the arguements S. By itself it usually has no truth value unless it has the same truth value for all members S.
Let $f$ be an arbitrary complex function on $\Bbb{R}$, and define $\varphi(x,\delta) = \sup \{|f(s)-f(t)| : s,t \in (x-\delta, x+ \delta) \}$ and $\varphi (x) = \inf \{\varphi(x,\delta) : \delta > 0 \}$. Prove that $\varphi$ is upper semicontinuous, that $f$ is continuous at a point if and onl...
