As part of my Topology course, I saw a proof the following proposition (as a consequence of Baire's category theorem):
Proposition: Define $\triangledown_{t,h}$ as the interior of an equilateral triangle in $\mathbb{R}^2$, with the following three properties:
$(1)$ It's height is $h$.
$(2)$ One of it's edges is above, and parallel to, the $x$ axis.
$(3)$ One of it's vertices is located on the $x$ axis, at $(t,0)$.
Define $T:=\bigcup_{r\in\mathbb{R}\setminus\mathbb{Q}}\triangledown_{r,h_r}$, where $h_r$ are arbitrary positive numbers. Then there is exists a rectangle with an edge on the $x$ axis, whose interior is in $T$.
Intuitively, it does not look like this proposition holds for unions over $\mathbb{Q}$. However, I could not construct a counterexample. I tried the following: Let $q_n$ be an enumeration of the rationals, and set $T:=\bigcup_{q_n\in\mathbb{Q}}\triangledown_{q_n,\frac{1}{n}}$. By defining it this way, I ensure that there are only finitely many triangles above any given height. But I could not see how to continue from here.
Any suggestion will be greatly appreciated.
Edit: The original question was answered using the fact that $\mathbb{Q}$ is a countable set. However, the above proposition is only proved for sets of the 2nd category. This might imply that there are uncountable subsets of $\mathbb{R}$ of the 1st category (such as the Cantor set) for which the proposition does not hold. Can anyone give an example of such a set (or alternatively show that it holds for all uncountable subsets of $\mathbb{R}$)?
I'm not sure if your example works or not, but here's an alternate method:
Fix an enumeration of the rationals $\{r_i\}_{i=1}^\infty$.
Show that there is a collection of triangles $\{T_n\}_{n=1}^\infty$ satisfying your conditions, with the bottom vertex of $T_n$ at $( r_n,0)$, such that $T_n\cap T_m=\emptyset$ if $n\ne m$.
The union of the $T_n$ cannot contain the interior of a rectangle having an edge coinciding with the $x$-axis (this requires a small argument).
You can construct the required collection inductively. In the inductive step, assuming $T_1,\ldots, T_n$ have been constructed, note $F_n=T_1\cup\cdots\cup T_n$ is closed; so, there is an open disk centered at $(r_{n+1},0)$ disjoint from $F_n$ (of course, $(r_{n+1},0)\notin F_n$). Within this disc, you can construct $T_{n+1}$.