Quiz-Like Problem 6

I decided to take a look at problem 6, and thought I would post my efforts to get everyone's thoughts:

Part (a) asks us to prove is $A$ has the same cardinality as $B$, then $B$ has the same cardinality as $A$.

To say that $A$ has the same cardinality as $B$ is to say that there is a bijection $f:A \rightarrow B$. A bijective map is one-to-one and onto, so its inverse $f^{-1}:B\rightarrow A$ (defined by mapping $f(a)$ to $a$) is also one-to-one and onto$^\text{a}$. So, $f^{-1}$ is a bijection from $B$ to $A$, which means precisely that $B$ has the same cardinality as $A$.

$^\text{a}$ I have assumed that this fact is something we are able to assume here without proof; do you guys think this is correct?

Would anyone like to give part (b) a go?

Nice. I think the existence of an inverse is pretty well baked into the definition of a bijection, so I would say it's fine.

Proof of 6b:

Let $A$ ~ $B$ and $B$ ~ $C$. Thus we can define two bijections $f : A \mapsto B$ and $g : B \mapsto C$.

Consider $g \circ f : A \mapsto C$.

Since $f$ and $g$ are both bijections, their composition is also a bijection. Thus, $A$ ~ $C$.$\blacksquare$

As for an actual quiz/test question, I'm not sure if we could make the last statement without proof.

I think the composition mapping $A \rightarrow C$ would actually be $g \circ f$. Since we want to map$ A$ to$ C$, we would want $g(f(x))$.

But he told us in class Wed. to worry less about this problem as a proof and more as a definition type problem.

@violincounter I liked this but I do agree with @cseagrav.

I appreciate you pointing that out. If I don't think through it I usually mess up the ordering of compositions. It's fixed now.

@mark, I like it too.