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qchu.wordpress.com | ||
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dominiczypen.wordpress.com
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| | | | | Let $latex \omega$ denote the first infinite cardinal - that is, the set of non-negative integers. Let $latex p_0 = 2$ be the smallest prime number, and let $latex (p_n)_{n\in\omega}$ enumerate all prime numbers in ascending order. Let $latex \mathcal{U}$ be a free ultrafilter on $latex \omega$. We consider the field $latex F = \big(\prod_{n\in\omega}\mathbb{Z}/p_n\mathbb{Z}\big)/{\mathcal... | |
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cp4space.hatsya.com
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| | | | | In the early 1930s, Pascual Jordan attempted to formalise the algebraic properties of Hermitian matrices. In particular: Hermitian matrices form a real vector space: we can add and subtract Hermitian matrices, and multiply them by real scalars. That is to say, if $latex \lambda, \mu \in \mathbb{R}$ and $latex A, B$ are Hermitian matrices, then... | |
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almostsuremath.com
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| | | | | Given a sequence $latex {X_1,X_2,\ldots}&fg=000000$ of real-valued random variables defined on a probability space $latex {(\Omega,\mathcal F,{\mathbb P})}&fg=000000$, it is a standard result that the supremum $latex \displaystyle \setlength\arraycolsep{2pt} \begin{array}{rl} &\displaystyle X\colon\Omega\rightarrow{\mathbb R}\cup\{\infty\},\smallskip\\ &\displaystyle X(\omega)=\sup_nX_n(\omega). \end{array} &fg=000000$ is measurable. To ensure that this is well-defined, we need to allow X to have values in $latex... | |
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linsdoodles.wordpress.com
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| | | For XingfuMama's Pull up a seat Photo Challenge | ||
