Analysis involving Cardinality of Infinite sets

cchatham
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1. If X is an infinite set and x is in X, show that X ~ X \ {x}



A~B if there exists a one-to-one function from A onto B.



Attempt at a solution
I'm pretty much completely stumped on this problem. I know that since X is infinite then it contains a sequence of distinct points. So x in X maps onto x1 of X\{x} and xn maps onto xn+1 of X\{x}. Is this enough to show that is 1-1 and onto?
 
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If by 'containing a sequence of discrete points' that you mean that X contains a subset that can be put into 1-1 correspondence with the positive integers and contains x, yes, that's exactly what you do. You should probably specify what your mapping does to points that aren't in the 'discrete sequence' as well, right? So can you SHOW that's 1-1 and onto?
 
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Yes that's what I meant by containing a sequence of distinct points. I think I got it now but I'm still not 100% convinced about the answer. Basically I want to say that there is a sequence of distinct points in X \ {x} where x->x1 and xn->xn+1 and for any other element y in X that y->y. That definitely shows onto since each element in X\{x} has a corresponding element in X that maps to it and I suppose it shows 1-1 as well.
 
What could go wrong with 1-1? This is the same as showing the map from {0,1,2...} to {1,2,3,...} defined by i->i+1 is 1-1. The points that aren't in those sets are automatically 1-1, since y->y. Why so hesitant about 1-1?
 
I'm not really hesitant about 1-1. Sorry my syntax was kind of confusing there. I think I understand the math behind it, but it just doesn't make intuitive sense to me. We're essentially saying that the cardinality of an infinite set is the same as the cardinality of that same set minus one element. Sure, it can be proved mathematically, I just don't like it.
 
You can choose not to like it. But the logic is hard to argue with, isn't it? A={0,1,2,3...} and B={1,2,3,4...}. B is just A 'moved over 1'. Just adding a number to each element of a set can't change the number of elements in the set, can it? How can their sizes really be different? Infinite sets take some getting used to, I'll admit that.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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