- #1
evinda
Gold Member
MHB
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Hey! (Mmm)
Proposition
The Cartesian product of two at most countable sets is countable.
Proof
Let $A,B$ sets both of which are at most countable.
That means that there are functions: $f : \omega \overset{\text{surjective}}{\rightarrow} A, \ g : \omega \overset{\text{surjective}}{\rightarrow} B$.
We define the function $h: \omega^2 \to A \times B \\ \langle m, n \rangle \mapsto \langle f(m), g(n) \rangle$
We can easily see that $h$ is surjective.
Thus, $A \times B$ is at most countable.
In order to justify that $h$ is surjective could we say that since $f,g$ are surjective we know that $\forall f,g, \exists m, n \in \omega$ such that $f(m)=m, g(n)=n$?
Also we have shown that $A \times B$ is at most countable, but according to the proposition it is just countable.
So can it be that it $A \times B$ is finite in this case or not?
Proposition
The Cartesian product of two at most countable sets is countable.
Proof
Let $A,B$ sets both of which are at most countable.
That means that there are functions: $f : \omega \overset{\text{surjective}}{\rightarrow} A, \ g : \omega \overset{\text{surjective}}{\rightarrow} B$.
We define the function $h: \omega^2 \to A \times B \\ \langle m, n \rangle \mapsto \langle f(m), g(n) \rangle$
We can easily see that $h$ is surjective.
Thus, $A \times B$ is at most countable.
In order to justify that $h$ is surjective could we say that since $f,g$ are surjective we know that $\forall f,g, \exists m, n \in \omega$ such that $f(m)=m, g(n)=n$?
Also we have shown that $A \times B$ is at most countable, but according to the proposition it is just countable.
So can it be that it $A \times B$ is finite in this case or not?