In this link, Division of cardinals, someone asks a question about cardinal division and references a Wikipedia page about it. The Wikipedia page does not give a reference to their statement, but I'd really like to know one. Does anyone know specifically (preferably book and page number) where I can find this and a proof?
Reference Request: Cardinal Division
elementary-set-theoryreference-request
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To take this off the unanswered list, I would like to elaborate Paul Sinclair's comment. Outside of higher set theory or logic, there is almost never a need to deal with actual cardinals. This is because every statement of the form "$S$ has cardinality $k$" is equivalent to "$S$ bijects with $k$". And in ordinary mathematics, we almost always have no need to know what kind of set this $k$ is. In particular, the statement you quoted is equivalent to:
Every consistent generalized theory $T$ with infinite countable symbol set $S$ has a model that bijects with $S$.
This can be proven without any notion of cardinality (or similar) at all. The main reason to invoke the notion of cardinals is if we want to have a canonical representative for each collection of sets that bijects with one another, just like we have each $kââ$ as a canonical representative for the collection of all sets of size $k$. Even then, for the purpose here (i.e. countable sets) we could very well use $â$ as our natural representative...
This is a result of the spectral theorem for self adjoint, bounded operators on a Hilbert space. The theorem which can be found in Reed and Simons for example provides a way of constructing an operator out of an integral over the spectrum of the operator with respect to a special kind of measure called a spectral measure.
In the case of self adjoint $x$ we know $\sigma(x)\subset \mathbb{R}$ and moreover, the spectral radius is equal to $\|x\|$.
The $de(\lambda)$ is the spectral measure defined on $\sigma(x)$ and it allows you to integrate continuous functions or even measurable functions with respect to this measure to define the analogue of the given function on the operator $x$. For example, if $f(\lambda)=e^\lambda$ defined on the spectrum of $x$, then the spectral theorem says that $$f(x)=\int_{\sigma(x)}f(\lambda)de(\lambda)$$ So you can give meaning to the exponential of an operator (in the case of bounded operators this machinery is not necessary since you can represent it as a power series in the operator norm). In the case where $f(\lambda)=\lambda$ i.e the $id_{\sigma(x)}$ then the theorem yields $$x=\int_{\sigma(x)}\lambda de(\lambda)$$ But remember if $x$ is self adjoint then the spectral radius is equal to $\|x\|$ and $\sigma(x)\subset \mathbb{R}$ we have $\sigma(x)\subset [-\|x\|,\|x\|]$ (if $\sigma(x)\subsetneq [-\|x\|,\|x\|]$ you can extend the integration to the endpoints by declaring that the spectral measures are exactly zero in the compliment) and hence $$x=\int_{-\|x\|}^{\|x\|}\lambda de(\lambda)$$
Best Answer
This follows from the fact that cardinal multiplication is not very interesting for infinite cardinals. Namely, if $\kappa$ and $\mu$ are infinite, $\kappa \cdot \mu = \max(\kappa,\mu)$. Thus, if we're given $\lambda$ and $\kappa$ we may always solve the equation in variable $\mu$ $$ \kappa \cdot \mu = \lambda $$ if and only if $\kappa \leq \lambda$. Indeed, if $\kappa \leq \lambda$, then $$\kappa \cdot \lambda = \max(\kappa,\lambda) = \lambda.$$ On the other hand, if $\kappa > \lambda$ $$\kappa \cdot \mu = \max(\kappa,\mu) \geq \kappa > \lambda.$$