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Q: Is the union of finite countable sets finite?

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here is the proof: http://planetmath.org/encyclopedia/ProductOfAFiniteNumberOfCountableSetsIsCountable.html

CHECK THIS OUT http://www.mathstat.dal.ca/~hill/2112/assn7sol.pdf

No, it is uncountable. The set of real numbers is uncountable and the set of rational numbers is countable, since the set of real numbers is simply the union of both, it follows that the set of irrational numbers must also be uncountable. (The union of two countable sets is countable.)

The cardinality of finite sets are the number of elements included in them however, union of infinite sets can be different as it includes the matching of two different sets one by one and finding a solution by matching the same amount of elements in those sets.

The union of two sets.The union of two sets.The union of two sets.The union of two sets.

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here is the proof: http://planetmath.org/encyclopedia/ProductOfAFiniteNumberOfCountableSetsIsCountable.html

Yes, finite numbers are always countable.

all finite set is countable.but,countable can be finite or infinite

CHECK THIS OUT http://www.mathstat.dal.ca/~hill/2112/assn7sol.pdf

No, it is uncountable. The set of real numbers is uncountable and the set of rational numbers is countable, since the set of real numbers is simply the union of both, it follows that the set of irrational numbers must also be uncountable. (The union of two countable sets is countable.)

It is a measure, but it isn't always sigma-finite. Take your space X = [0,1], and u = counting measure if u(E) < infinity, then E is a finite set, but there is no way to cover the uncountable set [0,1] by a countable collection of finite sets.

They are sets with a finite number of elements. For example the days of the week, or the 12 months of the year. Modular arithmetic is based on finite sets.

The cardinality of finite sets are the number of elements included in them however, union of infinite sets can be different as it includes the matching of two different sets one by one and finding a solution by matching the same amount of elements in those sets.

Closed sets and open sets, or finite and infinite sets.

Proof By Contradiction:Claim: R\Q = Set of irrationals is countable.Then R = Q union (R\Q)Since Q is countable, and R\Q is countable (by claim), R is countable because the union of countable sets is countable.But this is a contradiction since R is uncountable (Cantor's Diagonal Argument).Thus, R\Q is uncountable.

the number of steps of an algorithm will be countable and finite.

sets