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I understand that there cannot be a bijection between $S$, a finite set, and $S'$, a proper subset of $S$, because $S'$ will contain at least one fewer item than $S$.

What I don't understand is the definition of an infinite set, a set for which there is a bijection between itself and a proper subset of itself.

Take the set of natural numbers $\mathbb{N}$, which is infinite. Then isn't the set $A$ $=$ { 1, 2, 3 } a proper subset of $\mathbb{N}$ yet not a bijection of $\mathbb{N}$? What am I missing?

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2 Answers 2

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The condition is that there is some proper subset of $S$ in bijection with $S$. Of course, your example $\{1, 2, 3\}$ is finite and so it admits no bijection with $\mathbb{N}$. But $\mathbb{N}$ has many subsets to which it admits a bijection. One natural choice is the set $\{2, 4, 6, \ldots\}$ of positive even numbers, for which a natural bijection is given by doubling each number, i.e., $n \mapsto 2 n$.

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Note that the proper subset can also be infinite.

Example:

$$2\mathbb{Z} \subsetneq \mathbb{Z}$$ One bijection is $$ \phi : \mathbb{Z} \to 2\mathbb{Z} $$ given by $$ \phi(x) = 2x. $$

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