Partition of an interval of $\mathbb{R}$

Question

A partition of an inteval $[a,b]$ of $\mathbb{R}$ is generally defined as a finite sequence of the form: $$a = x_0 < x_1 < x_2 < \dots < x_n = b$$

Then, $[a,b]$ is seen as the following union of intervals: $[x_0,x_1] \cup [x_1,x_2]\cup\dots\cup[x_{n-1},x_n]$.

I wonder if it makes sense to extend this definition to sequences of the form: $$a = x_0 \leq x_1 \leq x_2 \leq \dots \leq x_n = b$$

Then, degenerate intervals (i.e. intervals with equal bounds) are allowed.

For instance, if $a=x_0=x_1=x_2=...=x_{n-1}<x_n=b$ then $$[a,b] = [x_0,x_1] \cup [x_1,x_2]\cup\dots\cup[x_{n-1},x_n] = [x_0,x_0] \cup [x_0,x_1]\cup...\cup [x_0,x_n]$$

It this still a partition?

Answer 1

It is not a partition in the proper sense, because a partition is a family of subsets that are mutually disjoint. What you get in your example gives sets $[x_0,x_0]$ and $[x_0,b]$ that have the point $x_0$ in common. But this happens also in your original definition of partition that some authors call instead subdivisions or pointed partitions. Your question is about nomenclature, and I would use the term subdivisions or pointed partitions as some authors do when introducing Riemann sums.

Answer 2

The two definitions make sense, and yours is more general. However, I would like to elaborate on a classical confusion, also discussed here.

This confusion is due to the fact that one often thinks of partitions as disjoint subsets that cover a set: a partition $P$ of a set $S$ is a family $\{P_1,P_2,\cdots,P_k\}$ of subsets of $S$, called parts, such that: $\bigcup_i P_i = S$ and $P_i\cap P_j \neq \emptyset \Rightarrow i=j$.

This is the classical definition of partitions of finite sets, and it may very well be used for intervals of $\mathbb{R}$ too, or any other set. However, it then raises questions.

Indeed, if we consider a closed interval $[a,b]$ of $\mathbb{R}$, then such a partition is a family of intervals that have to be open on one side, except for the last one. For instance $[a,x)$, $[x,y)$, $[y,b]$, with $x$ and $y$ in $[a,b]$, is a partition of $[a,b]$. This may however be seen as unsatisfactory as we end up a mix of closed and half-open intervals.

To avoid this, one may consider partitions of half-open intervals only: $[a,x)$, $[x,y)$, $[y,b)$ is a partition of $[a,b)$. Then, we only deal with half-open intervals.

Another approach is to allow non-empty intersections between parts, but require that they have null measure: $\bigcup_i P_i = S$ and $|P_i\cap P_j| \neq 0 \Rightarrow i=j$. Then, the intervals may overlap on their end-points, and $[a,x]$, $[x,y]$, $[y,b]$ is a partition of $[a,b]$. However, this does says in which of the three parts are $x$ and $y$, which may be frustrating.

Another approach, closer to what the question suggest, is to allow any kind of intervals, including degenerate ones. Then, $[a,x)$, $[x,x]$, $(x,y]$, $(y,b)$, $[b,b]$, for instance, is a partition of $[a,b]$. This is the most general form, I think, and I wonder if it may be represented as a sequence of values in the interval and relations between them.