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From Section 8 of Halmos' Naive Set Theory.

(i) Show that $Y^\emptyset$ has exactly one element, namely $\emptyset$. (ii) Show that if $X\neq\emptyset$, then $\emptyset^X=\emptyset$.

I'm not completely sure I understand what this means nor how to prove it.

For (i), we want to show that the set of functions $f:\emptyset\to X$ is exactly $\{\emptyset\}$. To me, this means that there is one "map", the "empty map", which associates nothing to something. I'm not really sure how to show this. We know that $Y^\emptyset\subseteq\mathcal P(\emptyset \times Y)=\{\emptyset\}$. Then we can show that $\{\emptyset\}\subseteq Y^\emptyset$ to obtain the equality. So this is vacuously true?

For (ii), since $X\neq\emptyset$, to me this exercise means that we're showing that there are no maps which associate something to nothing. But how to show that there is no function $X\to\emptyset$? Is this just immediate from the definition of function?

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marked as duplicate by Lord_Farin, MathOverview, Cameron Buie, Zhen Lin, Dan Rust Nov 4 '13 at 18:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The second part is the easier part, perhaps. Note that a function $f\colon A\to B$ must have an ordered pair $(a,f(a))$ for every $a\in A$. If $A$ is non-empty and $f\subseteq A\times\varnothing$, can $f$ be a function whose domain is $A$?

For the first part, yes. You have to show that $\{\varnothing\}=Y^\varnothing$. By showing that $\varnothing$ itself is a function from $\varnothing$ into $Y$ you will obtain this equality. But this is not vacuously true, you have to show that the definition of a function holds for $\varnothing$, when the domain is taken to be $\varnothing$.

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(I use $0$ for the empty set.)

(Note although $A$ can be a proper class in expression $A^B$, $B$ has to be a pure set.)

Since $0:0\rightarrow Y$, so $\{0\}\subseteq Y^0$. Now suppose $f\in Y^0$, then $\mathrm{dom}(f)=0$, which implies $f=0$.

For the second part, there is a stronger (if and only if) result I found: $A^z=0\leftrightarrow A=0\wedge z\neq 0$.

$Proof:$

$(\leftarrow)$: Suppose $\exists f(f\in A^z)$. $z\neq 0$ implies $f\neq 0$ and $A=0$ implies $f=0$, contradiction.

$(\rightarrow)$: Suppose $z=0\vee A\neq 0$. If $z=0$, then $0:0\rightarrow A$, so $A^0\neq 0$. If $A\neq 0$, then $\exists y(y\in A)$. We could let $f=\{\langle x,y\rangle\mid x\in z\}$, then $f$ is a function. We claim $\mathrm{dom}(f)=z$. Indeed, $x\in\mathrm{dom}(f)\leftrightarrow y=f(x)\leftrightarrow x\in z$. Hence $f:z\rightarrow A$ and thus $A^0\neq 0$.

$\mathrm{QED}$

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  • $\begingroup$ Use \emptyset (to get $\emptyset$) or \varnothing (to get $\varnothing$). $\endgroup$ – Pece Nov 4 '13 at 18:12
  • $\begingroup$ I know, I just don't want to type these long commands every time. :) $\endgroup$ – Kaa1el Nov 4 '13 at 18:19