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enter image description here I want to know what is the criteria for two sets (finite and infinite to be equivalent) .Does the same criteria hold for both type of sets? I read that two sets are equivalent if their no. Of elements( cardinality) is same. Its alright for finite sets. But again in case of infinite sets as in this question cardinality is infinite. Here I read again that one to one correspondence should be there. But in the question above for an infinite set if a subset is proper first of all its cardinality with its superset will never be equal .secondly for one - one correspondence how come a proper subset have one to one correspondence. It goes on till infinity. How can we be sure that there are no term which do not have one to one corresponce. If someone can please make me understand in simple language and convince me I ll be greatful.

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The set of positive integers $\mathbb N^+$ does have a one-to-one correspondence with a proper subset.

For example, the set of squares $S=\{1,4,9,\ldots\} $. Just pair $n $ with $n^2$. Since every integer has a unique square, and every square has a unique positive square root, nothing is missed or duplicated.

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    $\begingroup$ So the set of real , the set of complex no. Which are infinite are all equivalent to one of its proper subsets. Can you give an example for both sets? $\endgroup$ – shadow kh Jan 20 '17 at 10:52
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    $\begingroup$ @shadowkh, since those sets contain the positive integers, can you think of a way to use the correspondence in my post to get one for those sets? $\endgroup$ – Mark S. Jan 20 '17 at 11:56
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    $\begingroup$ The positive terms can be squared but it will have the same value for the square of the negative so it will be many one not one - one . $\endgroup$ – shadow kh Jan 20 '17 at 12:21
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    $\begingroup$ In fact if we cube it then both the positive and negative terms will be one - one . yes the proper subset of the cube of this infinite series do have a one to one correspondence with the set . correct me if I am wrong here $\endgroup$ – shadow kh Jan 20 '17 at 12:23
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    $\begingroup$ @shadowkh, You're right that squaring everything is generally two-to-one (ignoring 0). Cubing works on the integers, but not on the reals, because the set of cubes of reals isn't a proper subset: cube root is defined for all reals. Try to use the fact that the reals and complex numbers contain the integers: if your correspondence keeps all the nonintegers the same, then... $\endgroup$ – Mark S. Jan 20 '17 at 12:56

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