By $\emptyset^{\large\emptyset}$ i mean the set of functions from $\emptyset$ to $\emptyset$.
Is it the empty set or the set containing the empty set?
I just dont know how to prove it.
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The set $\emptyset^\emptyset$ contains one and only one element, which is $\emptyset$. For more details, see this question and its answers.
José Carlos Santos
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So it actually has an element? – Garmekain Sep 22 '17 at 18:47
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1@Garmekain That's what I wrote. – José Carlos Santos Sep 22 '17 at 18:48
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5And so $0^0=1$. – Angina Seng Sep 22 '17 at 18:49
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@LordSharktheUnknown Nice remark! – José Carlos Santos Sep 22 '17 at 18:50
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2@Garmekain You've already got two great answers, but here's one more comment: remember that formally a function $f\colon A \to B$ is a subset $f \subseteq A \times B$ satisfying certain properties. So $f\colon \varnothing \to \varnothing$ is actually a subset $f \subseteq \varnothing \times \varnothing = \varnothing$, and hence can only possibly be $\varnothing$. Since $\varnothing$ satisfies the properties required to be a function, that's it. – Ivo Terek Sep 22 '17 at 19:01
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Lets approach this problem from a category theory perspective. Since $\textsf{Set}$ is a category, note that it has the following property:
For every object (set) $A$ of $\textsf{Set}$, there exists (at least) one morphism (function) $1_A \in \text{Hom}_\textsf{Set} (A,A)$, the 'identity' on $A$.
Noting that $\emptyset \in \text{Obj($\textsf{Set}$)}$, then the set $\emptyset^\emptyset = \text{Hom}_\textsf{Set} (\emptyset,\emptyset) \neq \emptyset$, that is, it contains at least one function. And yes, that element in $\emptyset^\emptyset$ is $\emptyset$.
Also, never forget the crucial thing you're taught when learning set theory! For every set $A$, $\emptyset \subseteq A$.
Andrew Tawfeek
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