Set of finite partitions of the unit interval

Question

Is the set of partitions of a given length, say $n$, of the unit interval compact?

Answer 1

I subsequently realized that it is not compact.

First of all, let us think of a partition of length $n$ as a vector of length $n$ with given endpoints: $0$ and $1$. Therefore a partition is a vector of length $n-2$, of points chosen from $(0,1)$, ordered from lowest to highest. Therefore the set of all such partitions, $S \subset (0,1)^{n-2}$, is bounded.

But it is not closed because it is easy to show that there exist sequences of such partitions which collapse on to $(0,0,...,1,1)$ in the limit, which is a degenerate partition (i.e. not a partition of length $n$). For example, suppose $n$ is even and consider the sequence of partitions $\{(\frac{1}{N},\frac{2}{N},...,\frac{(n-2)}{2N},\frac{N-\frac{n-2}{2}}{N},...,\frac{(N-2)}{N},\frac{(N-1)}{N})\}^{\infty}_{N=10}$, which collapses on to $(0,0,...,1,1)$ in the limit. Similarly it is easy to show for odd $n$ as well.

However, following the above logic, the set of all finite partitions of $[0,1]$ upto length $n$ is a compact set.