how to show that a subset of a domain is not in the range

Question

If $g$ is a function from a set $Y$ to the collection of all subsets of $Y$, how does one show that there exists a subset of $Y$ that is not in the range of $g$?

An explanation of how to attempt a proof would be great. Thanks.

Answer 1

In short. you are asking for a proof that there is no surjection $Y\to \mathcal{P}(Y)$, where $\mathcal{P}(Y)$ is the power set of $Y$ (the set of all subsets of $Y$).

This is known as Cantor's Theorem, and the proof is standard. Given $g\colon Y\to\mathcal{P}(Y)$, define $$W_g = \{ y \in Y \mid y\notin g(y)\}.$$ I call this set "$W_g$" to stand for "the witness for $g$". This is a witness to the fact that $g$ is not surjective.

To see how $W_g$ is defined, let's consider a couple of examples with finite sets and functions:

Let $Y = \{1,2,3\}$, so that $\mathcal{P}(Y)$ consists of the eight subsets: $$\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\}.$$

Let's take one function: $g\colon Y\to \mathcal{P}(Y)$ defined by, say, $g(1) = \{2,3\}$, $g(2)=\emptyset$, $g(3)=\{1,2,3\}$.

Note that $1\notin g(1)$, $2\notin g(2)$, and $3\in g(3)$. That means that $W_g=\{1,2\}$. Now notice that $\{1,2\}$ is not in the range of $g$.

Another function: now let's take $h\colon Y\to \mathcal{P}(Y)$ given by $h(1)=\emptyset$, $h(2) = \{1\}$, $h(3)=\{2\}$. Then $1\notin h(1)$, $2\notin h(2)$, and $3\notin h(3)$, so $W_h = \{1,2,3\}$... which is not in the range of $h$.

Another function: now let's take $k\colon Y\to\mathcal{P}(Y)$ given by $k(1)=\{1\}$, $k(2)=\{2\}$, and $k(3)=\{3\}$. Then $1\in k(1)$, $2\in k(2)$, and $3\in k(3)$, so $W_k = \emptyset$... which is not in the range of $k$.

Now, there are many sets that are not in the ranges of $g$, of $h$, and of $k$. But one set that is not in the range is the "witness."

So... try proving that

1. For any $g\colon Y\to\mathcal{P}(Y)$, $W_g$ is an element of $\mathcal{P}(Y)$ (easy);
2. There does not exist a $y\in Y$ such that $g(y)=W_g$ (meat of the proof);
3. Conclude that $g$ is not onto.