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If the symbol $A\sim B $signifies that there is a bijection between A and B, and We take our sets to be dedekind infinite, then is the following correct? If not, what is the counter example?:$$A \sim B \Rightarrow A \sim B\cup\{x\}$$

Andrés E. Caicedo
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ArminAshrafi
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2 Answers2

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Let $f : A \to B$ be a bijection, and fix an injection $g : \mathbb{N} \to A$, which exists since $A$ is Dedekind-infinite, and let $a_n=g(n)$ for each $n \in \mathbb{N}$.

Define $f' : A \to B \cup \{ x \}$ by letting $$f'(a) = \begin{cases} f(a) & \text{if } a \not\in \mathrm{im}(g) \\ x & \text{if } a=a_0 \\ f(a_{n-1}) & \text{if } a=a_n \text{ for some } n>0 \end{cases}$$ You need to prove that $f'$ is a bijection.

Clive Newstead
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  • Thank you for answering. This question was asked as I was trying to reason about the Schröder-Bernstein theorem. I unfortunately fail to see the insight(The idea of why you use the methods you do). – ArminAshrafi Jan 06 '19 at 13:44
  • @user13910: I came up with the solution by thinking about what I knew about Dedekind-infinite sets, and pondering whether any of the things I knew might help me answer this question. The goal was to find a way of turning a bijection $A \to B$ into a bijection $A \to B \cup { x }$. A set $A$ is Dedekind-infinite if and only if there is an injection $\mathbb{N} \to A$, and so 'bumping' all the elements of $A$ in this function opened up a 'gap' that I could fill with $x$. And then fleshing out the details to make this precise led to the function $f'$ that you see in my answer. – Clive Newstead Jan 06 '19 at 13:51
  • P.S. I've changed the notation in the answer slightly to make it more readable. – Clive Newstead Jan 06 '19 at 13:53
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Assuming the axiom of choice, cardinality is a total order; equivalently, every set is well ordered. Now think of ordinals and each element associated with exactly one ordinal.

Lucas Henrique
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