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numerable

numerable, a. (ˈnjuːmərəb(ə)l) [ad. L. numerābilis, f. numerāre to number. So It. numerabile, Sp. -able, Pg. -avel.] 1. Capable of being numbered.1570 J. Dee Math. Pref. *j, The Glas of Creation, the Forme of Formes, the Exemplar Number of all thinges Numerable. 1629 J. Cole Of Death 107 That hee mu... Oxford English Dictionary

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Numerous vs Numerable - What's the difference? | WikiDiff

Adjective (en adjective) Indefinitely large numerically, many. *{{quote-magazine, year=2012, month=March-April , author=Colin Allen , title=Do I See What You See? , volume=100, issue=2, page=168 , magazine=(American Scientist) citation, passage=Numerous experimental tests and other observations have been offered in favor of animal mind reading, and although many scientists are skeptical ...

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Numerous vs Numerable: Which Should You Use In Writing?

Numerous is the proper word to use when you want to describe something that is very many or existing in large quantities. On the other hand, numerable is used to describe something that can be counted or enumerated. For example, you can say that there are numerous stars in the sky, meaning that there are a lot of stars.

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$X$ a numerable set, $f$ a positive function, then $\mu(A) = \sum_{x\in A} f(x)$ $\sigma$-finite? Let $(X$, ${\mathscr P}(X), \mu)$ a measurable space, where $X$ is a numerable set, $f: X \rightarrow [0, \infty]$ a po...

Actually, $\mu$ will be $\sigma$-finite if and only if $f(x)<+\infty$ for all $x\in X$. Indeed, if $f(x)<+\infty$ for all $x\in X$, then $\\{\\{x\\},x\in X\\}$ is a countable partition of $X$ and each element as a finite measure. If $\mu$ is $\sigma$-finite, there exists a sequence of subsets of $X$...

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Exercise over numerable sets Let $A$ be a infinite set and let $D\subseteq A$ be a numerable set so that $A-D$ in infinite. Prove that $(A-D)\sim A$.

I'll also assume you mean countable by "numerable".

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Existence of a numerable family given a element of a $\sigma-$algebra I'm trying to prove the next stament: Let $E\subset\mathcal{P}(X)$ fixed. Then for all $A\in\sigma(E),$ there exists a numerable subfamily $E_{0}\...

The punchline is that ${S}$ contains ${E}$ because for any $A \in {E}$, $A \in \sigma(A)$ hence $A \in S$ by definition of $S$.

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Numerability on subset of $\mathbb{R}$ with induced topology A friend of mine has this statement, that I cannot establish wether it is true or false. Let us consider $\mathbb{R}$ with the euclidean topology. Let us c...

For each $p\in A$ there is a $U_p$ such that $A \cap U_p = \\{p\\}$; the collection $\\{U_p\\}$ is uncountable. In each $U_p$ we can find an interval $V_p$ with rational endpoints containing $p$. Since also $A \cap V_p = \\{p\\}$, the $V_p$ are distinct and the collection $\\{V_p\\}$ is uncountable....

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Very specyfic linear order not isomorphic with $\langle\mathbb N,\le\rangle$ Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the s...

Consider $\Bbb N$ followed by $\Bbb Z$, i.e., $\\{0\\}\times \Bbb N\cup \\{1\\}\times \Bbb Z$ with lexicographic order. Or if you prefer it embedded into the real line: $\\{\,\arctan n\mid n\in\Bbb N\,\\}\cup\\{4+\arctan k\mid k\in\Bbb Z\,\\}$.

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Connected set cannot be written as the union of numerable proper connected separated sets. Is there a proof or a counterexample to such an affirmation? For example, we know that no finite union of proper connected s...

Let $X$ be any countably infinite set, with the cofinite topology. Then $X$ is connected, and also, $X$ is the countable union of its singleton subsets, which are proper connected sets. Also, any two distinct singletons sets are separated.

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Can a real value function, defined for every real number, have finite (or numerable) points of continuity? Can a real value function, defined for every real number, have finite (or countable) points of continuity ? A...

Here's a concrete example: let $g(x)$ be the characteristic function of the rationals, and let $f(x)=xg(x)$. Then basically the graph of $f$ looks like a big dotted "V" (the part corresponding to the rationals) with a dotted line running underneath it (the part corresponding to the irrationals), and...

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Sets of null (Lebesgue)-measure and sigma compacts Let $E\subset{\mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure. ¿It is possible to find $F\subset{\mathbb R}$ such that $E\subset F$, $m(...

The answer is no. There exists counterexample. The detailed answer for the related question is here. A short answer from Andreas Blass: > A closed set of Lebesgue measure zero has empty interior. So a countable union of such sets, a $_$ of measure zero, is meager (also called "first Baire category),...

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A $\sigma$ algebra at most of numerable cardinality is finite. Let $\mathcal{A}$ a $\sigma$ algebra at most of numerable cardinality (i.e $|\mathcal{A}|\leq \aleph_0$). Show that $\mathcal{A}$ is finite. I try to bui...

Suppose that $\mathcal A$ is a $\sigma$-algebra of subsets of the set $X$. Suppose further that $\mathcal B$ is a countably infinite subcollection of $\mathcal A$. For each $x\in X$, consider the sets $$E_x=\bigcap\\{E\in\mathcal B:x\in E\\}.$$ To show that $\\{E_x:x\in X\\}$ is a infinite disjoint ...

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Question about a corollary of Bessel inequiality I'm trying to prove: > Let $(X,<,>)$ be a vector space with inner product and $E \subseteq X$ be a orthonormal subset of $X$. Prove that > > **$\mathcal A$ ={$x \...

Since $A_n$ is finite, $A=\bigcup_{n\in \mathbb{N}}A_n$ is numerable (being a numerable union of finite sets)

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Infinite Family of Open Intervals disjoint, then The family must be numerable i'm new on this. If we have a infinite family $\\{A_{\lambda}\\}_{\lambda\in\Lambda}$ of open intervals such $$A_{\lambda_1}\cap A_{\lambda...

Hint: each interval contains a rational number.

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Counter Example of $\cup$ A$_i$ is not compact. True or false: The union of an arbitrary family of compact subsets is a compact subset. Well firstly I think for a while if it is true or not, definitely I take the sec...

In $\Bbb R$ in the discrete metric/topology, a subset $A$ is compact iff $A$ is finite. So e.g. take $A_n = \\{n\\}$, all compact, and then $\Bbb N = \bigcup_n A_n$ is not compact. So it already fails with countable unions of finite sets.

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numerable

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numerable

Numerous vs Numerable - What's the difference? | WikiDiff

Numerous vs Numerable: Which Should You Use In Writing?

$X$ a numerable set, $f$ a positive function, then $\mu(A) = \sum_{x\in A} f(x)$ $\sigma$-finite? Let $(X$, ${\mathscr P}(X), \mu)$ a measurable space, where $X$ is a numerable set, $f: X \rightarrow [0, \infty]$ a po...

Exercise over numerable sets Let $A$ be a infinite set and let $D\subseteq A$ be a numerable set so that $A-D$ in infinite. Prove that $(A-D)\sim A$.

Existence of a numerable family given a element of a $\sigma-$algebra I'm trying to prove the next stament: Let $E\subset\mathcal{P}(X)$ fixed. Then for all $A\in\sigma(E),$ there exists a numerable subfamily $E_{0}\...

Numerability on subset of $\mathbb{R}$ with induced topology A friend of mine has this statement, that I cannot establish wether it is true or false. Let us consider $\mathbb{R}$ with the euclidean topology. Let us c...

Very specyfic linear order not isomorphic with $\langle\mathbb N,\le\rangle$ Give an example of a numerable linear order such that each element has a successor, there is the smallest element, each element except the s...

Connected set cannot be written as the union of numerable proper connected separated sets. Is there a proof or a counterexample to such an affirmation? For example, we know that no finite union of proper connected s...

Can a real value function, defined for every real number, have finite (or numerable) points of continuity? Can a real value function, defined for every real number, have finite (or countable) points of continuity ? A...

Sets of null (Lebesgue)-measure and sigma compacts Let $E\subset{\mathbb R}$ be a set such that $m(E)=0$, that is, with zero Lebesgue measure. ¿It is possible to find $F\subset{\mathbb R}$ such that $E\subset F$, $m(...

A $\sigma$ algebra at most of numerable cardinality is finite. Let $\mathcal{A}$ a $\sigma$ algebra at most of numerable cardinality (i.e $|\mathcal{A}|\leq \aleph_0$). Show that $\mathcal{A}$ is finite. I try to bui...

Question about a corollary of Bessel inequiality I'm trying to prove: > Let $(X,<,>)$ be a vector space with inner product and $E \subseteq X$ be a orthonormal subset of $X$. Prove that > > **$\mathcal A$ ={$x \...

Infinite Family of Open Intervals disjoint, then The family must be numerable i'm new on this. If we have a infinite family $\\{A_{\lambda}\\}_{\lambda\in\Lambda}$ of open intervals such $$A_{\lambda_1}\cap A_{\lambda...

Counter Example of $\cup$ A$_i$ is not compact. True or false: The union of an arbitrary family of compact subsets is a compact subset. Well firstly I think for a while if it is true or not, definitely I take the sec...