# Tag Info

3

This holds for arbitrary (measurable) subsets $A,B \subset \Bbb{R}^2$, because $$A^\circ \cap B^\circ \subset A \cap B$$ is an open set (finite intersection of open sets) of measure zero, thus empty, where I denoted the (topological) interior relative to $\Bbb{R}^2$ of a set $M$ by $M^\circ$.

3

Since we are describing $C$ as the union of two sets, why not simply refer to $A$ as the first set in this union of two sets. But, honestly, I'd simply refer to these sets by name. If you state that $C = A\cup B$, why not go on to assert: "Set $A$ ensures...., while $B$ guarantees..."

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Neither term exists, so define it however you like. A google search yields zero results for implications tuple and merely three for tuple of implications. One of those results is another Math.SE question regarding a chain of implications, which I now see you yourself started.

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A geodesical ball $B$ centered in $p\in M$ of radius $r>0$, in a connected Riemannian manifold $(M,g)$, is a set of the form $$B:= \{x\in M \:|\: d(x,p) < r\}\:,$$ where $d(x,y) = \inf\left\{ L(\gamma)\:|\: \gamma: [0,1] \to M\:, \gamma \in C^1([0,1])\:, \gamma(0)=x\:, \gamma(1)=y \right\}$ and $L(\gamma)$ is the arch length of $\gamma$, L(\gamma) ...

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This is a special case of the tensor product of vectors known as outer product. It has some interesting properties, for example that the trace of the matrix is the square of the (Euclidean) norm of the vector. And as you point out it is always symmetric.

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Following my answer to your previous post, we can say that a formal system is made by an alphabet (the set of symbols), a gramamr (the formation rules, defining the "correct" expressions, i.e. the set of well-formed formulas) and a proof system or deductive calculus. See Enderton, page 110 : We will introduce formal proofs but we will call them ...

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"regular" here means holomorphic. Each connected component of the space of quasi-characters is just a copy of the complex plane; if I understand your notation correctly, then $c(\alpha)$ is some (unitary) character in this component, and then all the other quasi-chars. in the component have the form $c(\alpha)|\alpha|^s$. The $\zeta$-function can thus be ...

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