Is the Cartesian product of two uncountable sets uncountable? Is Cartesian product of two uncountable sets uncountable?
Suppose we have a set of real numbers $R$, Can't it be shown that $R$ is uncountable by Cantor's diagonalization method, so it follows that the Cartesian product of two sets of real numbers in uncountable?
 A: HINT: This is done in two steps:


*

*If $A$ and $B$ are not empty, then there is an injective function from $A$ into $A\times B$.

*If $A$ is uncountable there is an injective function from $A$ into $D$, then $D$ is uncountable.
A: Yes. Do it by contradiction. Assume there is a bijection between $\mathbb{N}$ and the cartesian product and show this would imply there is a bijection with at least one of the original sets.
A: You can use the Cantor diagonalization argument to show that $$A \text{ and } B \text{ countable } \implies A\times B \text{ countable.}$$ So: $$A \times B \text{ uncountable} \implies A \text{ or } B \text{ is uncountable}.$$
Alternatively, you can fix $b \in B$. Then $A \times \{b\}$ has the same quantity of elements that $A$, which is uncountable, and $A \times \{b\} \subset A \times B$. So $A \times B$ contains an uncountable set.
A: Here's an easy proof (if you already know every subset of $\Bbb N$ is countable) -- please only look at this after you've already attempted the problem yourself given the other hints:
Suppose $R$ and  $Y$ are uncountable sets.  Suppose by contradiction $R \times Y$ is countable.  Let $\phi :  R \times Y \to \Bbb N$ be the bijection you can find.  Then fix some $y \in Y$ and look at $ A = \{ (r, y) \mid r \in R \}$.  Note that $A \subseteq R \times Y$.  
Then the restriction map $\phi \mid_{A} : A \to \phi(A)$ is a bijection between $A$ and $\phi(A)$ (why?).  But $\phi(A)$ is a subset of $\Bbb N$, and so it is countable.  Thus, $A$ is countable.  
But there is a clear bijection between $R$ and $A$ by sending $r \mapsto (r,y)$.  You should check that this is a bijection.  Then composing this map and $\phi$ gives you a bijection between $R$ and a $\phi(A)$, and $\phi(A)$ is a subset of $\Bbb N$ and thus countable.  So $R$ is countable.  Contradiction!  Since we assumed $R$ was not countable.  So $R \times Y$ is uncountable, as desired.
A: Let $b$ be an arbitrary element in $B$. If $A$ and $B$ are uncountable and $A\times B$ is countable, then $A\times\{b\}$ is also countable, because $A\times\{b\}\subset A\times B$. But $A$ is uncountable iff $A\times\{b\}$ does. This contradicts.
