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math771
Jun7-11, 12:46 PM
A set is countable if a 1-1 correspondence can be constructed between that set and the set of all positive integers J.
Suppose we have a set S consisting of all positive integers plus a "copy" of the element 1: i.e., S={1,1,2,3,4,5,6...}. I have encountered several proofs of basic topological theorems that would say that S is countable. However, something strange happens if we attempt to construct a 1-1 correspondence between S and J. We may either associate with each 1 in S distinct elements of J, in which case 1 will be associated with two different positive integers and the correspondence will not be 1-1. Or we may associate a single element of J with both 1s in S, in which case two elements of S will be associated with a single element of J and the correspondence will not be 1-1.
On the other hand, I'm wondering whether the former method does indeed create a 1-1 correspondence. Perhaps the two 1s can be distinguished from each other so that the element of J associated with one is not necessarily associated with the other. Then again, if S is not ordered, how would one distinguish between the two 1s? (Perhaps this last question is of a more philosophical than mathematical nature and warrants an answer along the lines of "you just do!", but I'll pose it anyway out of curiosity.)
Any advice and/or corrections of my comments would be much appreciated. Thanks!

micromass
Jun7-11, 12:53 PM
Hi math771! :smile:

It's hard to see what exactly you mean here. You propose the set S={1,1,2,3,4,...}, but any set is determined by it's members. Thus, we actually have that

\{1,1,2,3,4,...\}=\{1,2,3,4,...\}

These two sets are equal because any element in the left-hand side is in the right-hand side and conversly.

The same thing happens with finite sets really. We have for example that
\{0,1\}=\{0,1,1\}=\{0,1,1,1\}

The order and multiplicity in which the elements occur in the set doesn't matter. If it did, then you could say that the above set has 2 elements, but als 3 and 4 elements.

On the other hand, if you add a new element to {1,2,3,4,...}, then you can't call this element 1, since it is already in the set. You may call it a, for example, and we get {a,1,2,3,4,...}. This is countable by the bijection

a\rightarrow 1~\text{and}~n\rightarrow n+1

Does this answer your question??

math771
Jun7-11, 01:00 PM
Yes, it does answer my question. Thanks.