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Why is $|Y^{\emptyset}|=1$ but $|\emptyset^Y|=0$ where $Y\neq \emptyset$

Let $Y^X$ denote the set of all functions (i.e. relations, or ordered pairs) having signature $X \rightarrow Y$, where $X$ and $Y$ are arbitrary sets. Additionally, we should note that $Y^X$ is always a subset of $\wp({X \times Y})$.

It can be shown that $Y^{\varnothing}$ contains only one element, namely $\varnothing$ (the empty function). I would expect that similarly , $\varnothing^X$ would yield the same [1]. According to Halmos' book on Naive Set Theory this is not the case, and in fact, $\varnothing^X$ is empty—i.e. there exist no functions having signature $X \rightarrow \varnothing$.

Question: Given that $Y^\varnothing$ contains the empty function, why do we not see the existence of a concept dual to the empty function in $\varnothing^X$? Or am I just being pedantic in my reading [2] and missing the point entirely?

Essentially, I'm curious why we do not see a function having signature

$$X \rightarrow \varnothing$$

[1] Both $\wp(\varnothing \times Y)$ and $\wp(X \times \varnothing)$ result in $\{\varnothing\}$, leaving us with both sets of functions in question being a subset of $\{\varnothing\}$.

[2] A direct quote of what we're required to show:

Exercise: (i) $Y^\varnothing$ has exactly one element, namely $\varnothing$, whether or not $Y$ is empty or not, and (ii) if $X$ is not empty, then $\varnothing^X$ is empty.


marked as duplicate by Qiaochu Yuan Jul 19 '11 at 23:28

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.


$\emptyset^{X} = \{\emptyset\}$ if $X=\emptyset$, and $\emptyset^X = \emptyset$ if $X\neq\emptyset$.

The reason that if $X\neq\emptyset$ then $\emptyset^X=\emptyset$ just comes down to the definition of "function". For $R\subseteq X\times\emptyset =\emptyset$ to be a function, for each $x\in X$ there must exist $y\in \emptyset$ such that $(x,y)\in R$ (and in addition, if $(x,y),(x,y')\in R$, then $y=y'$); this is impossible for any given $x\in X$, so no subset is a function. That is, there are no functions from a nonempty set to the empty set.

However, if the domain is empty, then the requirements for $R$ to be a function are vacuously satisfied, so the empty set is indeed a function $\emptyset\to\emptyset$.

  • $\begingroup$ Thanks for the clarification. In hindsight this was quite a silly question; I essentially ignored the contradiction that arose from the $\exists$ part of $\forall x \in X, \exists y \in Y$ s.t. $(x,y) \in f$ $\endgroup$ – Raeez Jul 17 '11 at 20:38

For every set $X$ there is a unique function $\emptyset \to X$ defined as follows: .

If $X$ is non-empty, there is no corresponding function $X \to \emptyset$ because there are no elements of the range for the function to map an element of the domain to. This is not a problem in the other direction because $\emptyset$, by definition, has no elements.

If the lack of duality bothers you, it's because the correct dual notion is that for every set $X$ there is a unique function $X \to 1$ (where $1$ is a one-element set) sending every element of $X$ to the unique element of $1$. In category-theoretic terminology we say that $\emptyset$ is an initial object in the category of sets, whereas $1$ is a terminal object. In general there's no reason to expect these two things to be the same. When they are (more precisely, when the unique map $\emptyset \to 1$ is an isomorphism) we say that the category has a zero object.


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