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Although the title is the same as one in Suppose $X$ is infinite and $A$ is a finite subset of $X$. Then $X$ and $X \setminus A$ are equinumerous, but the proof is different. Hence it's not a duplicate.


Suppose $X$ is infinite and $A$ is a finite subset of $X$. Then $X$ and $X\setminus A$ are equinumerous


My attempt:

Lemma 1: If $A$ is finite and $B$ is countably infinite, then $A\cup B$ is countably infinite.

Lemma 2: If $X$ is infinite and $A$ is finite, then $X\setminus A$ is infinite.

Lemma 3: If $Y$ is infinite, then there exists $B\subsetneq Y$ such that $B$ is countably infinite. (Here we assume Axiom of Countable Choice)


Since $X$ is infinite and $A$ is finite, then $X\setminus A$ is infinite by Lemma 2.

Since $X\setminus A$ is infinite, there exists $B\subsetneq X\setminus A$ such that $B \sim \Bbb N$ by Lemma 3.

Since $A$ is finite and $B$ is countably infinite, then $A\cup B \sim \Bbb N$ by Lemma 1.

Since $B \sim \Bbb N$ and $A\cup B \sim \Bbb N$, $B \sim A\cup B$ and thus there exists an bijection $f_1:B \to A\cup B$.

Let $f_2:(X\setminus A)\setminus B \to (X\setminus A)\setminus B$ be the identity map on $(X\setminus A)\setminus B$. Then $f_2$ is a bijection.

We define $f:X\setminus A \to X$ by $f(x)=f_2(x)$ for all $x \in (X\setminus A)\setminus B$ and $f(x)=f_1(x)$ for all $x \in B$. Thus $f$ is a bijection.

Hence $X\setminus A \sim X$.


Does this proof look fine or contain gaps? Do you have suggestions? Many thanks for your dedicated help!

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  • $\begingroup$ What does X/A/B mean? $\endgroup$ – William Elliot Sep 13 '18 at 11:30
  • $\begingroup$ @WilliamElliot $X\setminus A\setminus B=(X\setminus A) \setminus B$. $\endgroup$ – LE Anh Dung Sep 13 '18 at 13:03
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    $\begingroup$ Perfect $\;\!$, just one note, the identity map of $Y$ is usually denoted by $\mbox{id}_Y$, in this case you have $f_2\equiv \mbox{id}_{(X\setminus A)\setminus B}$, but the name doesn't affect the proof $\endgroup$ – ℋolo Sep 13 '18 at 14:20
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    $\begingroup$ Thank you so $\infty$ @Holo! You have confirmed many of my recent attempts. Your help is really really helpful to me, removing my uncertainty. Since I'm self-studying without any teacher besides MSE, your help is extremely meaningful to me. By that noble conduct and your generosity, you have distinguished yourself from other users. I'm deeply grateful. $\endgroup$ – LE Anh Dung Sep 13 '18 at 14:36
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    $\begingroup$ @LeAnhDung I'm happy I was useful$\ddot\smile$ $\endgroup$ – ℋolo Sep 13 '18 at 14:57

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