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Let Equivalence($\equiv$) be defined as having the same cardinality.

Assume, $A \equiv \emptyset $. Then, there exists a function $f: A \to \emptyset $ such that f is a bijection. This proof will now proceed by contradiction. Suppose, $ A \ne \emptyset$. Let $ x \in A$. Thus, for some $y \in \emptyset$, $(x,y) \in f$. However, the statement $y \in \emptyset$ is a contradiction. Therefore, $ A = \emptyset$

Is this proof correct? Thank you!

Pedro
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1 Answers1

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Your proof is correct. Note that it actually shows that if $f:A\to\emptyset$ is any function, then $A=\emptyset$. One point in the proof you might make more explicit is that if $A\not=\emptyset$, that means (by definition of $\emptyset$) that $A$ has an element, so you can choose one and call it $x$.

Eric Wofsey
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