Do these theorems about power sets hold for the empty set?

From the definition of the power set as the set of all subsets of a given set, I realize that $\mathcal P(\varnothing) = \{ \varnothing \}$, in other words, the power set of the empty set is the set containing just the empty set. But does this work for some theorems we're proving for homework? Consider the below two theorems:

For any set $A$, there is a 1-to-1 function $f$ from $A$ into $\mathcal P(A)$.

There is no function from a set $A$ onto $\mathcal P(A)$.

I don't need help proving those theorems for non-empty sets, but I am confused about whether or not they are relevant for the empty set. I can't imagine how one could define a function from the empty set into another set, because the empty set doesn't have any elements to act as inputs for the function. Am I correct in thinking this?

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This holds for equal cardinalities of the sets, like the one you mentioned: $f:\varnothing \rightarrow P(\varnothing)$. This function is bijective. – NasuSama Jan 19 '14 at 3:09
@NasuSama There is no bijection between $\emptyset$ and $\mathcal P(\emptyset)$. The former is empty, and the latter contains one element, namely $\emptyset$. Or, written another way, $\mathcal P(\emptyset) = \{\emptyset\} = \{\{\}\}$. – Dustan Levenstein Jan 19 '14 at 3:25

The answer is yes to both. A function $A \to B$ can be defined as a subset $f \subset A \times B$ with the property that for each $a \in A$, there exists precisely one $b \in B$ such that $(a, b) \in f$, and then we notate this by $b = f(a)$. If $A$ is empty, then so is the Cartesian product, and the empty set $f := \emptyset$ vacuously defines a function $\emptyset \to B$; it is indeed true that for every $a \in \emptyset$, there exists precisely one $b \in B$ such that $(a, b) \in f$, because no such $a$ exists in the first place.

That said, your proofs of $A \hookrightarrow \mathcal P(A)$ and $A \not\twoheadrightarrow \mathcal P(A)$, if you did them right, should apply equally well, with no complication, to the corner cases such as $A$ being empty or finite.

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Thinking of a function as something that transforms an input in an output is probably not the right point of view here.

A function $A\to\mathcal P(A)$ is a subset $f$ of the Cartesian product $A\times\mathcal P(A)$ such that if $(x,y)$ and $(x,z)$ are in $f$, then $y=z$. The empty set (or "empty function", we might call it) satisfies this, and it is the only element in $\emptyset\times\{\emptyset\}$.

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I know there are three other answers already (and you've accepted one), but I want to offer an alternative point of view because I'm uncomfortable with the idea of identifying a function $f : A \to B$ with its graph, the latter being a subset of $A \times B$. These two notions are formally identified in most axiomatic set theories, but there's no need to do this to answer this question.

When kids are taught what functions are they're told that they are rules, or machines, that take things in (from their domain) and spit things out (from their codomain), in such a way that if you keep giving it the same thing from the domain it will keep spitting out the same thing in its codomain.

From this very informal notion of a function we still obtain the results in your question. Indeed, if $f : \varnothing \to \mathcal{P}(\varnothing)$ then $f$ has a very easy job: it doesn't have anything to take in, so it just sits there and does nothing. It's as good a function as any, just a very lazy one, the so-called empty function. In particular, it is vacuously injective, and it can't be surjective because there's nothing it can take in in order to spit anything out.

All of this can be made rigorous without resorting to identifying a function with its graph.

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It could be surjective if it was going into the empty set. – coffeemath Jan 19 '14 at 6:11
@coffeemath: Yes but this one's going to its power set! – Clive Newstead Jan 19 '14 at 14:54
I agree it can't be surjective to a nonempty set such as the power set. Just making the point that the empty function, considered from $\varnothing$ to itself, is not only injective as you note, but also surjective. – coffeemath Jan 19 '14 at 16:05

My favorite definition of a function is that it's a triple $(D, C, R)$, where $D$ is a set called the domain, $C$ a set called the codomain, and $R$ a subset of $D \times C$ called the "rule" or "relation" or "graph", and where $R$ has certain properties:

1. For every $x \in D$, there's an element $(x, y) \in R$ whose first element is $x$.

2. If $(x, y_1) \in R$ and $(x, y_2) \in R$, then $y_1 = y_2$.

That may seem overly formal, but in this case, the "function" you're looking for is $(\emptyset, \{{\emptyset}\}, {\emptyset})$.

You can check that it has both required properties (since both are trivially satisfied).

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