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I've been wondering whether or not such functions can exist:
(1) $$f: \emptyset \rightarrow \emptyset$$ I can think of only one relation that satisfies this criterion - the empty relation defined on the empty set, namely: $R = \{\}$ but, my question is - is this even a function? The definition of a function says that $f$ is a function iff if two elements have the same first entry, they must have the same second entry - but I think this is not a problem here.

(2) $X$ is non-empty: $$g: \emptyset \rightarrow X$$ As for this function, there are no possible first entries, and so it is impossible to map anything into set $X$

(3) $X$ is non-empty $$h: X \rightarrow \emptyset$$ I cannot think of anything that would map something into the empty set. Any suggestions?

Andrés E. Caicedo
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Aemilius
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  • Correct; the empty relation on $\varnothing \times \varnothing$ does indeed function defines a function $\varnothing \to \varnothing$.

  • Incorrect; the empty relation on $\varnothing \times X$ does indeed function defines a function $\varnothing \to X$. Your observation merely shows that the image of this function is empty.

  • Correct; if $X$ is nonempty, then there are no relations on $X \times \varnothing$ that define a function $X \to \varnothing$.

  • Thanks! Can you think of any other function that satisfy (2) or this exhausts all the possibilities? – Aemilius Nov 15 '17 at 16:39
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    @Aemilius: In case (2), the empty relation is the only binary relation! –  Nov 15 '17 at 16:40