Set notation and functions into sets

Question

Let $\mathbf 2 = \{0, 1\}$ and $X , Y$ be sets.

(i) Describe the elements of $\mathbf 2^{X\times Y}$ and $\left({\mathbf 2^X}\right)^Y$.

(ii) Find a bijective function from $\mathbf 2^{X\times Y}$ to $\left({\mathbf 2^X}\right)^Y$.

(iii) Let $\mathbb N$ and $\mathbb R$ be, respectively, the sets of natural numbers and real numbers. Describe and compare the cardinalities of: $\mathbb N$, $\mathbb N \times \mathbb N$, $\mathbf 2^{\mathbb N}$ , $\mathbb R$ and $\mathbf 2^{\mathbb R}$.

First of all, apologies for being incompetent with Latex (^) indicates "to the power of" so in the case of 2 ^ (X x Y) - this is the set of all functions from the cartesian product of (X x Y) into 2, 2 being defined as {0, 1}. Also; N and R are the naturals and reals respectively.

Can someone please explain the answers to this question? Many thanks.

In your apology, you already solved half of (i). What else did you find by yourself? — Hagen von Eitzen, Jan 12 '13 at 14:11
Please add parenthesis to clarify, $2^{X^Y}$ means $2^{(X^Y)}$ or $(2^X)^Y$? — Asaf Karagila, Jan 12 '13 at 14:59
@AsafKaragila Note that LaTeX already makes a tiny distinction between $\mathbf 2^{X^Y}$ and ${\mathbf 2^X}^Y$, so just spending an hour with a magnifying glass would have told you ;) — Hagen von Eitzen, Jan 13 '13 at 17:37
@Hagen: Yes. I am well aware of that. I still think it's best to have explicit parenthesis rather than half-point difference in the font. — Asaf Karagila, Jan 13 '13 at 18:00
@AsafKaragila After checking the histoy, I noticed that it was in fact my fault: The OP had no LaTeX dollars in it, hence made the parentheses clear and this got lost after my adding LaTeX markup ... — Hagen von Eitzen, Jan 13 '13 at 18:29
@Hagen: Well then, that's great. Now all that's left is to direct the OP to this particular thread. — Asaf Karagila, Jan 13 '13 at 18:31

score 1 · Answer 1 · answered Jan 13 '13 at 17:48

(i) As $A^B$ is the set of functions $B\to A$, we see that $$\mathbf2^{X\times Y}$$ is the set of functions from $X\times Y$ to $\mathbf 2$, whereas the elements of $$\left(\mathbf2^X\right)^Y$$ are functions from $Y$ to the set of functions from $X$ to $\mathbf 2$. One may also note that one can biject $\mathbf 2^A$ with the powerset $P(A)$ via $f\mapsto f^{-1}(\{1\})$, so via this identification the first set "is" $P(X\times Y)$ and the second "is" the set of maps $Y\to P(X)$.

(ii) The only map $X\to \mathbf 2$ that comes to mind if you are given a map $f\colon X\times Y\to \mathbf 2$ and and element $y\in Y$, is $x\mapsto f(x,y)$. Showing that this map $f\mapsto (y\mapsto(x\mapsto f(x,y)))$ is a bijection, is straightforward.

(iii) You probably have already shown statements (Cantor diagonal arguments) essentially equivalent to $|\mathbb N|=|\mathbb N\times \mathbb N|<\left |\mathbf 2^{\mathbb N}\right|=|\mathbb R|<\left |\mathbf 2^{\mathbb R}\right |$.

Set notation and functions into sets

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