Thread: question on finite sets

I'm reading through a book on Real Analysis (Intro.) and got a question on finite sets. The definition I'm using for a finite set is: a set A is finite iff there is a one-to-one function f on Nk onto A (Nk is read N sub k where k is a subscript and Nk is a subset of the natural numbers from 1 to k) for some k.

Question
Let A = {1, 2, 3, 2}, Nk = {1, 2, 3, 4}, and assume that f is bijective from Nk onto A. Observe that f(2) = f(4) implies 2 ≠ 4. It appears that f is not bijective. Is A finite or not? If A is finite, could you describe a bijective function f from Nk onto A.
Thanks

Originally Posted by klmsurf

I'm reading through a book on Real Analysis (Intro.) and got a question on finite sets. The definition I'm using for a finite set is: a set A is finite iff there is a one-to-one function f on Nk onto A (Nk is read N sub k where k is a subscript and Nk is a subset of the natural numbers from 1 to k) for some k.

Question
Let A = {1, 2, 3, 2}, Nk = {1, 2, 3, 4}, and assume that f is bijective from Nk onto A. Observe that f(2) = f(4) implies 2 ≠ 4. It appears that f is not bijective. Is A finite or not? If A is finite, could you describe a bijective function f from Nk onto A.
Thanks

Hi klmsurf.

Yes, is finite, but you want to take not Then f is a bijection with, e.g.,

If a bijection exists between two sets, we say they have the same cardinality. A = {1, 2, 3, 2} has exactly four elements, regardless of the distinctness between the elements. Can you explain why the cardinality of A = {1, 2, 3, 2} is 3? Maybe I'm missing a definition or proposition on sets.

Originally Posted by klmsurf

If a bijection exists between two sets, we say they have the same cardinality. A = {1, 2, 3, 2} has exactly four elements, regardless of the distinctness between the elements. Can you explain why the cardinality of A = {1, 2, 3, 2} is 3?

it just has an element listed twice.
has only two elements.

May I suggest that you study a foundations of mathematics text before you try Intro to Analysis.