Color every point of $\mathbb{R}^2$ either red (r) or blue (b). Show some rectangle has its vertices all the same color.
I know that if you take say 3 points in a row on the x-axis that those three points can have one of nine different outcomes.
bbb bbr brb brr rrr rrb rbr rbb
So if you keep repeating this combination by stacking a new set of three on top of the three that we already had on the x-axis at some point you will end up with all the vertices of a rectangle (this don't say what size it is limited to) will be the same color.
Is that a good enough proof? Or am I far off base? It's another question that he says can be solved using the pigeon hole principle, but I'm having trouble finding out what is the pigeon hole and what are the pigeons. Thanks guys/gals for your help!
I think you have the right idea, but you need to formalize the proof.
Now let $K = \{R,B\}$ (the colours), and let $c:\mathbb{R}^2 \to K$ be the colouring of $\mathbb{R}^2$.
Choose three distinct points on the $x$-axis, $x_1,x_2,x_3 \in \mathbb{R}$, and let $\phi:\mathbb{R} \to K^3$ be the function $\phi(y) = (c((x_1,y)), c((x_2,y)), c((x_3,y)))$. This represents the colours of the points whose $x$ coordinate is one of $x_1,x_2,x_3$ and whose $y$ coordinate is $y$.
Since $K^3$ is finite and $\mathbb{R}$ is infinite, we must have some $y_1 \neq y_2$ such that $\phi(y_1) = \phi(y_2)$. Since there are only two colours, either $\phi(y_1)$ contains at least two reds or two blues (pigeon hole principle). So we have $[\phi(y_1)]_i = [\phi(y_1)]_j$ for some $i \neq j$. Consequently the rectangle whose $y$ coordinates are $y_1,y_2$ and whose $x$ coordinates are $x_i,x_j$ has all vertices the same colour (that is, $c((x_i,y_1))$).
In fact, since $\mathbb{R}$ is uncountable, the above argument shows that there are an uncountable number of rectangles whose vertices are all the same colour (since $\phi^{-1} \{z\}$ must be uncountable for some $z \in K^3$).