We know by definition that if a bijection between two sets A and B exists,then A and B are equivalent.The book I was reading took the function f:Z→N f(x)= -2x,for x<0 2x+1,for x=>0 which is a bijection(it can be easily proven),but I'm a little confused. How can Z and N be equivalent sets if N is a subset of Z ?
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$\begingroup$"Equivalent" doesn't mean "the same", it means that they have the same cardinality, or "are equipotent", or "have the same size". One of the surprises of infinite cardinalities is that a proper subset of $X$ can have the same cardinality as $X$. This doesn't happen for finite sets, but it does for infinite ones – in fact it's one of the cleanest ways of defining what "infinite" means.
Building a mental model of how the natural numbers are equivalent to the even numbers, by mapping $n$ to $2n$, is a useful image to hold in your head. Yes, the even numbers are a subset of the natural numbers, but there's just as many of them.
$\endgroup$ $\begingroup$One immediate (though rather dangerous! .. see paragraph below) way to wrap your mind around this is to consider the fact that both sets are of infinite size. Yes, the one is a strict subset of the other, but they are both infinite ... so maybe it's a little easier to digest that way.
Then again, the really surprising thing about cardinality is that not any two infinite sets have the same cardinality .. that there different 'kinds' of 'degrees' of infinity, if you want. As it so happens, $\mathbb{N}$ and $\mathbb{Z}$ are of the 'same' kind though; in both cases, you can put all elements in a list ... something which is not possible for the real numbers. However, it's exactly this fact that makes the notion of cardinality and equipotence an interesting and useful notion!
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