# If $A$ is an infinite set and $a \in A$, then how to show that $A$ and $A-\{a\}$ are equivalent? [closed]

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 $$A-B$$?

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

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• What have you tried? Is this a homework exercise? Please share context. – Viktor Glombik Jul 22 at 11:03
• If $A$ is countable then it is easy. My question is about when $A$ is uncountable. – Nanasaheb Phatangare Jul 22 at 11:53
• @ViktorGlombik You have the right idea, but for clarity you should go on to define an injection (or bijection) from $A\setminus\{a\}$ to $A$. – bof Jul 22 at 12:33
• @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 $A$? – bof Jul 22 at 12:36
• Yes, I think this would work! – Nanasaheb Phatangare Jul 23 at 4:23

Hint: Find an injection $$A \to A-\{a\}$$ and use the Schröder–Bernstein theorem.
• 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. – bof Jul 22 at 12:40
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.