Show that if $A=\varnothing$, $B\neq\varnothing$, then, according to the set-theoretic definition, there is precisely one function from A to B. Show that if $A\neq\varnothing$, $B=\varnothing$, there are none. How many functions are there from $\varnothing$ to $\varnothing$?
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In the future, rather than using $\phi$ which is the Greek letter phi, you should use $\emptyset$ or $\varnothing$. Then again, it was still understandable here, so don't worry, it's not at all dramatic. – Bruno B Nov 27 '22 at 12:19
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$\varnothing$ was not familiar to me in LaTex. Thanks! – Nikhil Nov 27 '22 at 12:46
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Let's recall the formal definition of the function: A subset, S , of $A \times B$ is called a function from $A$ to $B$ if given $a \in A$ there exists exactly one $b \in B$ such that $(a,b) \in S$. Thus, if both $A$ and $B$ are empty sets, we can take $S$ as an empty set, that corresponds to the empty function. If $A$ is not empty and $B$ is empty, again, from the definition it is clear that there is no such functions.
ALNS
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But how to prove that $f=\varnothing$ is a function here if $A=\varnothing$ and $B=\varnothing$. – Nikhil Nov 27 '22 at 12:50
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@RajKumar https://math.stackexchange.com/questions/45625/why-is-an-empty-function-considered-a-function – Asaf Karagila Nov 27 '22 at 13:15