# If $A\sim B$(both dedekind infinite), is it then that $A\sim B\cup \{x\}$

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?:$$A \sim B \Rightarrow A \sim B\cup\{x\}$$

• Which of the definitions of Dedekind-infinite are you using? – Asaf Karagila Jan 6 at 13:44
• If A is dedekind infinite,then there is an injection (which is not surjective), from A to itself. – ArminAshrafi Jan 6 at 13:47

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.

• The axiom of choice has no business being involved here. – Asaf Karagila Jan 6 at 13:44
• It could not, but I can. And it works. – Lucas Henrique Jan 6 at 13:45
• @AsafKaragila isn't it the case that AC is somewhat involved here? math.stackexchange.com/a/1396713 – Stupid Questions Inc Jan 7 at 6:28
• @StupidQuestionsInc: No, choice is involved in making sure every infinite set is Dedekind infinite. But here it is given, thus no choice is needed. – Asaf Karagila Jan 7 at 7:52
• @AsafKaragila thank you $\mathsf{AC}$ master – Stupid Questions Inc Jan 7 at 9:21

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.

• 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 6 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 6 at 13:51
• P.S. I've changed the notation in the answer slightly to make it more readable. – Clive Newstead Jan 6 at 13:53