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I am reading Naive Set Theory by Paul Halmos. I saw this exact question posted earlier, here, but there is no help on it outside of the definition which I already know, and a theory on how the notation came about which could be used to prove this but it doesn't help in my understanding. The problem (from the end of chapter 8 on functions) is,

Let $Y$ be a set and $X$ be a non-empty set. Then prove

  1. $Y^\emptyset = \{ \emptyset \}$ and
  2. $\emptyset^X = \emptyset$.

So far I have: The set of $Y^\emptyset \subset P( \emptyset \times Y)$, where $P(A)$ is the power set of $A$ and $A \times B$ is the cartesian product of $A$ and $B$. $P( \emptyset \times Y) = P( \emptyset) = \{ \emptyset \}$, and the same goes for $X$.

My question is why is $\emptyset$ considered a function from $\emptyset \rightarrow Y$ but not $X \rightarrow \emptyset$. If a function in Naive set theory is seen as a set of ordered pairs such that for a function $F$ if $(x,y) \in F$ and $(x,z) \in F$, then $y = z$, and the empty set has no ordered pairs, should it not be a function for both?

Asaf Karagila
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Robert Murray
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    There is no function from $X$ to the empty set, since there is nowhere for the elements of $X$ to go. But there is a function from the empty set to $Y$, the empty function, which doesn't take anything anywhere. – Gerry Myerson Dec 21 '23 at 04:02
  • @GerryMyerson Sure but, we aren't sending any elements of $X$ anywhere with the $\emptyset$ as a function, since it doesn't take anything anywhere. – Robert Murray Dec 21 '23 at 04:05
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    A function with domain $X$ must take the elements of $X$ somewhere, that's what a function with domain $X$ is. If there's nowhere to take those elements, then there's no function. The set of all such functions is the empty set, since there are no such functions. – Gerry Myerson Dec 21 '23 at 04:10
  • So, if I understand correctly, $\emptyset: \emptyset \rightarrow Y$ is a function because it does nothing, but it has somewhere to go, but $\emptyset: X \rightarrow \emptyset$ is not a function since it can never have one to begin with? So it's just a definitional concept not an argument? – Robert Murray Dec 21 '23 at 04:15
  • As you correctly wrote $Y^{\emptyset}$ contains one element, empty function, while $\emptyset^X$ contains no elements (if $X$ is not empty), even, empty function, so there is not functions in this case and this is argument. – zkutch Dec 21 '23 at 04:23
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    Everything in mathematics is, at bottom, a definitional concept. $\theta:\theta\to Y$ is a function because it takes every element of the empty set (of which there are none) to an element of $Y$. It's a bit of abuse of notation, using the same symbol for the empty set and the empty function, but it's OK since functions are defined as sets anyway. The formulas you ask about are right because they can be deduced by an argument from the relevant definitions. – Gerry Myerson Dec 21 '23 at 04:30
  • @GerryMyerson One final question: $Y$ is specifically written to be empty or non empty, meaning $\emptyset$ works as a valid function from $\emptyset$ to $\emptyset$? Is this because both the domain and range of the function are empty? How do you prove this from the definitions because it doesn’t follow to me? – Robert Murray Dec 21 '23 at 05:33
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    The requirement of a function is to assign each element in the domain a unique element in the codomain. For $\emptyset\to\emptyset$ this can be done. – AlvinL Dec 21 '23 at 14:13
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    I don't know what Paul Halmos did in his book, but the usual definition of a function $F : X \to Y$ as a subset of $X \times Y$ has two conditions. One is the condition you stated in the OP: if $(x,y) \in F$ and $(x,z) \in F$, then $y = z$. The other is that if $x \in X$, then there is some $y \in Y$ with $(x,y) \in F$. That condition is why there is no function from $X \to \varnothing$ unless $X = \varnothing$. – Paul Sinclair Dec 22 '23 at 17:43

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From the comments (so I can close the question):

Let $Y$ be a set and $X$ be a non-empty set. Then, let $A^B$ denote the set of all functions from $B$ to $A$. The two conditions for a valid function $F:A \rightarrow B$ are: $( (x, y) \in F, (x,z) \in F) \Rightarrow y=z$ and $\forall x \in X, \exists y \in Y$ such that $(x,y) \in F$.

  1. We know $Y^\emptyset \subset \mathcal{P}(\emptyset \times X) = \{\emptyset\}$. Hence, the only possible function is $\emptyset$. This is a valid function as the empty set has no ordered pairs and the first condition is vacuously true, and $\forall x \in \emptyset, \exists y \in Y$ such that $(x, y) \in F$ as there are no $x \in \emptyset$ (also vacuously true). Hence, $Y^\emptyset = \{ \emptyset \}$ regardless of Y's elements.
  2. We know $\emptyset ^X \subset \mathcal{P}(X \times \emptyset) = \{ \emptyset \}$. However, $\emptyset$ is not a valid function from $X$ to $\emptyset$. Assume it is a valid function, then, our second condition must hold. However, by definition $(x, y) \not \in \emptyset$, this is a contradiction with our second condition and thus $\emptyset$ is not a valid function. Thus, there are no possible functions from $X$ to $\emptyset$, and $\emptyset^X = \emptyset$. Q.E.D.
Robert Murray
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