Can an empty set be involved in a 1-1 correspondence?

**Chizum** · February 2nd 2010, 12:43 PM

Say A is empty.

Could I do A --> B (1-1) or B --> A (1-1)?

Thanks.

**Drexel28** · February 2nd 2010, 12:52 PM

Originally Posted by Chizum

Say A is empty.

Could I do A --> B (1-1) or B --> A (1-1)?

Thanks.

Every mapping $f:\varnothing\mapsto X$ is injective. To see this merely note that for the condition that there exists $x_1,x_2\in\text{Dom}(f)$ such that $f(x_1)=f(x_2)$ we would have to say $x_1,x_2\in\varnothing$ . See a problem?

For $f:X\mapsto\varnothing$ it actually reverts to the same thing. For suppose that $X\ne\varnothing$ then for any $x\in X$ we would have that $f(x)$ does not exist? Why? Thus, if $f:X\mapsto\varnothing$ and $f$ is well-defined then by necessity we must have that $X=\varnothing$ .

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