How to prove that every infinite set is "as big as" the set of non-negative integers.
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1"at least as big as". That being said, it depends on your definition of inifinte set and whether or not you are assuming AC. – Oct 29 '17 at 16:07
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What do you mean by "assuming AC?" – Nate123 Oct 29 '17 at 16:41
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AC is axiom of choice. – Oct 29 '17 at 17:10
1 Answers
It depends very much on your definition of "infinite set". My favourite definition is "a Hilbert's hotel-like set", id est a set $X$ such that does exists a $f:X\longrightarrow X$ such that $f$ is injective but not surjective (hence $X$ is equipotent to a proper subsets of itself). Another definition tells that $X$ is infinite if it contains a copy of $\mathbb{N}$, i.e. if exists a $h:\mathbb{N}\longrightarrow X$ injective, assuming that the countable cardinality is the minimum infinite one. Let's use the first one and construct the injection $h:\mathbb{N}\longrightarrow X$ to prove what you need (which in turn could be taken as definition of infinite set). We assume to have $f:X\longrightarrow X$ injective but not surjective. Let's define our (injective) $h$ by recursion. Choose $h(0)\in X\setminus f(X)\neq\emptyset$, now suppose $h$ already defined for each $m<n$ and choose $h(n)\in f^{n-1}(X)\setminus f^n(X)$ (prove that this set is non-empty). By induction we defined an injection from $\mathbb{N}$ onto our set $X$ (the second definition we gave of infinite set) which is more convenient for the last step. We want to define $s:X\longrightarrow\mathbb{N}$ surjective, the most simple thing to do is do define $s(x)=\begin{cases} 0\quad\text{if }x\notin h(\mathbb{N})\\n\quad\text{if }x=h(n)\end{cases}$. This $s$ is a surjection.
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I have to prove it based on this proof: Let A be a set and b is not in set A. Then A is infinite iff A bij A U {b} Proof. Since A is not the same size as A U {b} when A is finite, we only have to show that A U {b} is the same size as A when A is infinite. That is, we have to find a bijection between A U {b} and A when A is infinite. Here’s how: since A is infinite, it certainly has at least one element; call it a0. But since A is infinite, it has at least two elements, and one of them must not equal to a0; call this new element a1... (same logic continues for an infinite amount of elements). – Nate123 Oct 29 '17 at 16:57
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Continuing in this way, we conclude that there is an infinite sequence a0, a1, a2... of different elements of A. Now it’s easy to define a bijection: e(b) ::= a0 e(an) ::= an + 1 (there is an element, n, in the non-negative set of #s) e(a) ::= a (there's an element, a, in the set A) – Nate123 Oct 29 '17 at 17:03
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You precisations are not so clear to me, actually. It seems that you use a definition of infinite set and than that you try to demonstrate it using an intuitive idea of infinite set. I don't understand well. Please write your accepted definition of infinite set and your (presumed) equivalent condition. Thanks – Tancredi Oct 29 '17 at 17:59