Infinite set

Infinite set :-

A set is said to be an infinite set whose elements cannot be listed if it has an unlimited (i.e. uncountable) by the natural number 1, 2, 3, 4, ………… infinite , for any natural number n is called a infinite set. 

A set which is not finite is called an infinite set.
example of infinite sets:-
set of natural numbers , N={1,2,3,4,.........................}
set of whole numbers  ,W= {0,1,2,3,.........................}
set of real numbers    , R etc

Since we cannot ever finish counting an infinite set, we need a different approach to

thinking about “how many elements” such a set contains. In the early 1900s, Georg

Cantor presented an idea that has clarified thinking about this issue and had enormous

impact on opening new pursuits in logic.

Two (finite or) infinite sets A and B are said to have the same cardinality (equipotent) if

there is a one-to-one, onto function (i.e., a bijection) f: A Æ B.

 Since it is impossible to ascribe an ordinary number to the size of an infinite set, we

don’t try. Instead, the comparative idea above is used. If we can “line the elements up” in

exactly corresponding pairs, we say the two sets have the same cardinality (i.e., size).

So, for instance, the set of natural numbers N = {0, 1, 2, … } has the same cardinality as

the collection of all strings over the alphabet.

 ifnfinite sets that have the same cardinality as N = {0, 1, 2, … } are called countably

infinite. A set that has a larger cardinality than this is called uncountably infinite.

Assertion: there are uncountably infinite sets.

Proof by contradiction:

Let [0,1] denote the interval of all real numbers x, 0≤x≤1 — this set is uncountably

infinite. First of all, note that there is an injection f: N Æ [0,1] defined by f(n) = n/(n+1).

This maps N to a proper subset of [0,1], so [0,1] has cardinality at least as large as N.

To show that the cardinality of [0,1] is actually larger than N, we proceed by

contradiction. Suppose that [0,1] is countably infinite. Suppose that g :N Æ [0,1] is a

bijection, and therefore {g(n) | nŒN} is the set of all real number in the interval [0,1]. We

show that there must be at least one real number missing in this enumeration providing

a contradiction to the existence of g, and hence there is no bijection. Specifically, we

construct the decimal expansion of a number a that is missing from {g(n) | nŒN}. Take

the decimal expansion of a to be .d1d2d3 …, where dn = 5 if the nth digit of g(n) is 6, and

dn = 6 otherwise. Then for each n, the nth digit of a differs from the nth digit of g(n).

Therefore a differs from every number in the enumeration in at least one digit and so

aœ{g(n) | nŒN}, a contradiction since this set was assumed to be all such numbers.

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