Show that the the open unit interval and the closed unit interval have the same cardinality.
Note: Christina and Jordan have indicated one reasonable approach - I even "liked" there work! But, I do think an explicit bijection is quite feasible as well.
I seem to recall seeing this proof in the past, but we used the Schr$\ddot{o}$der-Bernstein Thm. to prove it. I know he mentioned it in class, but is that legitimate to use since we have not actually proven that theorem?
If it is fair game, then I would start by letting $(a,b)$ be the open unit interval and $[a,b]$ be the closed unit interval, noting that $a < b $.
Then constructing a function mapping $(a,b) \rightarrow [a,b]$ would be a matter of using the identity function, since $(a,b) \subset[a,b]$.
Hence, $f(x) = x$ for all $x \in (a,b)$.
Since $f(x_1) =f(x_2)$ implies that $x_1 = x_2$, then $f$ is one-to-one.
The trickier part is mapping $[a,b] \rightarrow (a,b)$. I am not sure how to do this part without some research and thought, but once you get a one-to-one mapping this direction, the theorem states that the two sets have the same cardinality.
For the other part you could use $F: [a,b] \rightarrow (a,b)$ given by the function $f(x)=\frac{x-a}{2}+\frac{b-a}{4}+a$. After you look at this equation for a little while you will notice this actually maps $[a,b]$ onto the set $[a+\frac{1}{4}(b-a),b-\frac{1}{4}(b-a)]$ which is contained within $(a,b)$. Note the first term scales $x$ and the remaining terms scale the starting point of the interval. You will also notice this function can really be written as $f(x)=\frac{x}{2}+c$ where $c$ is a constant which we can easily verify as being one to one. Therefore, from what @cseagrav wrote, we can invoke the Schroder Bernstein Thm and we're done.
As Mark was talking about in class, if we can find a bijection between our two sets, then they must have the same cardinality. The strategy we used in class was to treat $0$ and $1$ as special cases. The overall function is:
\[
f(x) =
\begin{cases}
\hfill \frac{1}{2} \hfill & \text{ if $x$ is 0} \\
\hfill \frac{1}{n+2} \hfill & \text{ if $x$ is of the form }\frac{1}{n} \\
\hfill x \hfill & \text{ if $x$ is anything else} \\
\end{cases}
\] Using this method $0$ maps to $\frac{1}{2}$ and $1$ maps to $\frac{1}{3}$. We then push everything else around so that it works. The last piece means that if something did not need to be changed we just leave it as is when we map it. Since we now have a bijection $f: (0,1)\mapsto [0,1]$ these sets must have the same cardinality.