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I know that my direction for this proof is correct, but it feels like my reasoning for few things are off because I am still trying to grasp the Schroder-Bernstein so I would appreciate the guidance! Please note that I am novice on a good day..

Let A be an infinite set and B a countable set. Prove $|A|=|A\cup B|$

Suppose $B\subset A$, meaning union of A and B are disjoint, this results in $|A\backslash B|=|A|=|A\cup B|$ according to our claim. This also means that $A\backslash B$ is also infinite, since A is infinite. Let C be a denumerable subset of $A\backslash B$, and since B is countable and C is denumerable their union must also be denumerable. This means that $|B\cup C|=|C|$ such that there is a bijection $f:B\cup C\rightarrow C$. Now suppose we split A into two pieces, i.e. $A\backslash (B\cup C)$ which is the same as $(A\backslash B)\backslash C$. We already have a bijection $f:B\cup C\rightarrow C$, and we can form another bijection $g:(A\backslash B)\backslash C\rightarrow (A\backslash B)\backslash C$ meaning to itself. Let h be the function that is $f\circ g$, which by definition of composite functions, another bijection such that $h:A\rightarrow A\backslash B.$

cemsicles
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  • I literally closed this question a couple hours ago (by the same user). – Asaf Karagila Mar 13 '18 at 21:52
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    @AsafKaragila Right, and I am trying to get critique to see if my proof is valid. Hence I am here asking rather than hijacking another user's question. Past question was deleted. So, please, easy on the trigger, not all of us are experts at subject and cant see similarities as clearly. – cemsicles Mar 13 '18 at 22:05

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I think you might be overcomplicating things, but your idea of using the Cantor–Schröder–Bernstein theorem is a good one. To apply the Cantor–Schröder–Bernstein theorem, you need to find an injection $A \to A \cup B$ and an injection $A \cup B \to A$. Well:

  • The inclusion function $i : A \to A \cup B$, given by $i(a)=a$ for all $a \in A$, is an injection.

  • Fix an enumeration of $B$, say $B = \{ b_0, b_1, b_2, \dots \}$. Since $A$ is infinite, it has a countably infinite subset $A' = \{ a_0, a_1, a_2, \dots \}$. Define $f : A \cup B \to A$ by letting $$f(a)= \begin{cases} a & \text{if } a \in A \text{ but not } A' \text{ or } B \\ a_{2n} & \text{if } a=a_n \in A' \text{ but not } B \\ a_{2n+1} & \text{if } a=b_n \in B \end{cases}$$ It is easy to prove that this function $f$ is injective.

Clive Newstead
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  • Could I ask why you defined your last function so? I think I get the even and odd part, but not the a. – cemsicles Mar 13 '18 at 21:58
  • @YoloSatoshi: The idea is that the elements of $A'$ get 'spread out' in order to fit in the elements of $B$ between them in an injective way. The reason why we define $f(a)=a$ for $a \in A \setminus (A' \cup B)$ is that you have to define $f$ on all the values of $A \cup B$, and you might as well just send them to themselves. – Clive Newstead Mar 13 '18 at 22:05
  • Thank you, thats so much easier. – cemsicles Mar 13 '18 at 22:09