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It is obvious that $\mathbb{N}$ and $\mathbb{N}-\{a\}$ are equivalent for $a\in\mathbb{N}$. Moreover, if $A$ is a countable set and $B$ is a finite subset of $A$, then it is easy to prove that $A$ and $A-B$ are equivalent. But, if $A$ is an uncountable set and $B$ is a finite subset of $A$, then how to prove that $A$ and $A-B$ are equivalent? Question could be extended further as, if $A$ is an uncountable set, $B$ is a countable subset of $A$, then how to show that $A$ is equivalent to $A-B$?

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closed as off-topic by José Carlos Santos, Michael Rozenberg, Daniele Tampieri, cmk, ronno Jul 22 at 16:55

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  • $\begingroup$ What have you tried? Is this a homework exercise? Please share context. $\endgroup$ – Viktor Glombik Jul 22 at 11:03
  • $\begingroup$ If $A$ is countable then it is easy. My question is about when $A$ is uncountable. $\endgroup$ – Nanasaheb Phatangare Jul 22 at 11:53
  • $\begingroup$ @ViktorGlombik You have the right idea, but for clarity you should go on to define an injection (or bijection) from $A\setminus\{a\}$ to $A$. $\endgroup$ – bof Jul 22 at 12:33
  • $\begingroup$ @NanasahebPhatangare Can you prove that, if $B$ is a finite subset of an uncountable set $A$, then there is a countably infinite set $C$ such that $B\subseteq C\subseteq A$? And then a bijection from $C-B$ to $C$ can be extended to a bijection from $A-B$ to $A$? $\endgroup$ – bof Jul 22 at 12:36
  • $\begingroup$ Yes, I think this would work! $\endgroup$ – Nanasaheb Phatangare Jul 23 at 4:23
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Hint: Find an injection $A \to A-\{a\}$ and use the Schröder–Bernstein theorem.

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  • $\begingroup$ See also en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel $\endgroup$ – lhf Jul 22 at 12:40
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    $\begingroup$ I don't believe the Schröder–Bernstein theorem is useful here, because it's just as easy to define a bijection $A\to A-\{a\}$ as to define an injection. $\endgroup$ – bof Jul 22 at 12:40
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For the statement in the title: Since $A$ is infinite, there exists a injection $f: \mathbb{N} \to A$ with $f(0) = a$. Then, $\tilde{f}: \mathbb{N} \to A \setminus \{a\}$ defined by $\tilde{f}(n) = f(n + 1)$ is an injection as well, yielding the statement.

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