I will start my question by giving Wikipedia information of Sierpinski Triangle first. https://en.wikipedia.org/wiki/Sierpi%C5%84ski_triangle
I was proving why the Sierpinski triangle is compact in metric space $\mathbb{R}^{2}$, and came up with the idea that
"since the area of $F_{n}$, which is the set of remaining triangles after $n^{th}$ process, is greater than $0$ $\forall n\in \mathbb{N}$, $F_{n}$ is non-empty. If we call the Sierpinski triangle $F=\bigcap_{n=1}^{\infty }F_{n}$, then its area is still greater than 0 because of 'the area is greater than $0$ property.' Thus, $F$ is non-empty."
My concern is this: can we really say the set $F$ is non-empty in $\mathbb{R}^{2}$ just because the set's area is not $0$? Basically, can we ever claim that there is an area of some closed and bounded subset of Metric Space $\mathbb{R}^{2}$?
I know the proof using subsequences and sets, but I just wonder if this idea works. Or, since this is about the metric space, we are only allowed to use 'elements, sequence, sets, distance, etc' terms?
Thanks a lot for your help.
If the area (or measure) of $F$ were positive, you could. This is so simply because the area of the empty set id $0$ and hence would be different from that of $F$. However, the area of $\bigcap F_n$ cannot be larger than the area of any of the $F_n$, and these get arbitrarily small. Thus the area of $F$ is in fact $0$.
To show that $\bigcap F_n$ is non-empty, it suffices to exhibit a single common element of all $F_n$, and at least the three vertices come to mind as easy candidates.