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cæcal
cæcal, a. Phys. (ˈsiːkəl) [f. cæc-um + -al1.] Pertaining to, or of the nature of, the cæcum; having a blind end.1826 Kirby & Sp. Entomol. IV. xl. 121 Their cæcal appendages are numerous. 1858 Lewes Sea-side Studies Index, Cæcal prolongations of the intestines are..ramifications without openings at t... Oxford English Dictionary
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cœcal
@@@LINK=cœcum Oxford English Dictionary
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C. Cal Evans
C. Cal Evans was born May 23, 1905, to Elijah Charles Evans and Sarah Elizabeth McConnaughey. Mayor C. Cal Evans lost to Arthur Selland in a field of eight candidates. wikipedia.org
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$F^{op}$ functor I'm trying to make sens of category theory. In my course it is stated that a functor is a map $$F:{\cal{C}}\rightarrow{\cal{D}}:X\mapsto F(X)$$ and which sends $f\in \text{Hom}_{\cal{C}}(X,Y)$ to $Ff\...
Let $F:\mathcal C\to \mathcal C$ be the identity functor on this category. Now your concern is valid -- there's no functor $\mathcal C^\text{op}\to \mathcal C$ that remains the identity on objects but reverses the arrow, simply
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Is every sigma algebra generated by some subclass. Let $\Omega$ a non empty set and consider and $\cal{A}$ a sigma algebra in $\Omega$. Is there always a set $\cal{C}\subset\mathscr{P}(\Omega)$, $\cal{C}$ strictly con...
If you take $\mathcal{C}=\mathcal A$, then you'll have a set that generates $\mathcal A$. If you want a smaller set, take $\mathcal{C}=\mathcal{A}\setminus\\{\emptyset\\}$.
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Construct a projection satisfying a certain property > Let $\cal G$ be a group of finite order $n$. For every prime divisor $p$ of $n$, construct a projection $P\in \cal N(G)$ such that $\operatorname{tr}_{\cal N(G)}(...
If $g$ is an element of order $p$ (this always exists by Cauchy's Theorem), let $$ Q=\frac1p\,\sum_{j=0}^{p-1} g^j. $$ Then $\text{tr}(Q)=1/p$ and $$ Q^*=\frac1p\,\left(\sum_{j=0}^{p-1} g^j\right)^*=\frac1p\,\sum_{j=0}^{p-1} g^{-j} =\frac1p\,\sum_{j=0}^{p-1} g^{p-j}=\frac1p\,\sum_{k=1}^{p} g^k=\frac...
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If $T \in \cal{L}(V)$ and every vector in $V$ is an eigenvector of $T$ then $T=cI$ for some $c \in \cal{F}$ Suppose $T$ is an operator on a vector space $V$ (i.e. a linear map from $V$ to itself) such that every vecto...
Choose a basis $B$ of $V$ and show that two members of $B$ have the same Eigenvalue: If $v_1,v_2\in B$ and $\lambda_1,\lambda_2$ are their Eigenvalues it follows: $T(v_1+v_2) = \lambda_1v_1 + \lambda_2v_2$ On the other hand $v_1+v_2$ ist an Eigenvector too. Also an Eigenvalue $\eta$ exists with $T(v...
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Questions about multiplier algebra and corona algebra When I read N.E. Wegge-Olsen's book K-theory and C-star-algebras_ A friendly approach I meet the following two problems about standard isomophisms: 1. For any $...
For part 2, we just apply part 1, and the isomorphism given: $$ \mathbb{M}_2(\cal{C}(\cal{A}\otimes\cal{K(H)}))=\mathcal C(\mathbb{M}_2(\cal{A}\otimes\ cal{K(H)})\simeq \mathcal C(\cal{A}\otimes\cal{K(H)}) $$
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Is a family of disjoints atoms in $\sigma$-finite neasurable space at most countable? Let $\mu$ be a positive measure on a $\sigma$-algebra $S$ in $X$. A subset $A\subset X$ is called an atom if $\mu(A)>0$ and for e...
Let $X = \cup_n E_n$, where $\mu(E_n) <+\infty$. Then $E_n$ being a finite measure space, can only intersect countably many of the atoms, by your own argument for finite measure spaces. So $X$ can also have at most countably many atoms in a family.
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Comparability of topologies on $\Bbb R$ > Let $\cal C=\\{a,b):a<b; a,b\in \Bbb Q\\}$ be a basis for a topology on $\Bbb R$. [This says that topology generated by $\cal C$ is not comparable to the countable complement ...
From the fact that $\mathbb R-\mathbb Q$ is evidently not an element of $\tau(\mathcal C)$ (in contrast to non-empty elements in $\tau(\mathcal C)$ it does not contain an interval) it can be concluded that $\tau(\mathcal C)$ is not finer than the cocountable topology.
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Boundary point sequence proof > Let $A\subset \mathbb{R}^p$ and $x\in \mathbb{R}^p$. Then $x$ is a boundary point of $A$ if and only if there is a sequence $(a_n)$ of elements in $A$ and a sequence $(b_n)$ of elements...
A similar argument applies to $A^c$. Now choose $U_n = B(x,\frac{1}{n})$, and pick $a_n \in U_n \cap A$, $b_n \in U_n \cap A^c$. Similarly $b_n \in A^c$ and $b_n \to x$.
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3-dim Lie algebra with two commutative elements Let $\cal g$ be a Lie algebra and let $a,b,c\in \cal g$ be such that $ab=ba$ and $[a,b]=c\not =0$. Let $\mathcal h=span\ \\{a,b,c\\}$. How to prove that $\mathcal h$ is ...
The Lie algebra with $[a,b]=c$ is the $3$-dimensional Heisenberg Lie algebra $\mathfrak{h}_1$. pmatrix} 0 & 1 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0 \\\ \end{pmatrix}, \quad b = \begin{pmatrix} 0 & 0 & 0 \\\ 0 & 0 & 1 \\\ 0 & 0 & 0 \\\ \end{pmatrix}, \quad c
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Showing a functor has no right adjoint Let ${\cal C}$ be the category of groups. Let ${\cal C}'$ be the full subcategory of ${\cal C}$ with objects the class of abelian groups. Let $F$ be the inclusion of ${\cal C}'...
You are correct that abelianization is left adjoint to the forgetful/inclusion functor. If $F$ had a right adjoint, then $F$ would itself be a left adjoint, which would imply that it preserved colimits. The easiest way to show that $F$ has no right adjoint is to give an example of a colimit that is ...
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Measure extension theorem(unique) Please give an example of two probability measures $\mu \not = \nu$ on $\cal{F} $= all subsets of {1, 2, 3, 4} that agree on a collection of sets C with $\sigma(C)=\cal{F}$ . thanks ...
Obviously, $C$ should _not_ be a $\pi$-system ( _do you know why?_ ). Let me provide some collection $C$ that works and let you find $\mu$ and $\nu$ (there are infinitely many such pairs): $$C=\\{\\{1,2\\},\\{1,3\\},\\{1,2,3,4
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Let $(X, {\cal E}, \mu)$ be a measure space and let $B,C \in {\cal E}$. Show $\mu_B + \mu_C \leq \mu$. Let $(X, {\cal E}, \mu)$ be a measure space and let $B,C \in {\cal E}$. Define $\mu_B:{\cal E} \to [0,\infty]$ by ...
{B}$ and $\mathrm{C}.$ Hence, $\mu_{\mathrm{B}} + \mu_{\mathrm{C}} = \mu_{(\mathrm{B} \setminus \mathrm{C}) \cup (\mathrm{C} \setminus \mathrm{B})} \leq 2 \mu(\mathrm{B} \cap \mathrm{C}) \leq \mu(\mathrm{B} \cap \mathrm{C}),$ hence $\mu(\mathrm{B} \cap \mathrm{C}) = 0.$ Q.E.D.
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