Rudin partition definition: Why non-strict inequalities?

In Definition 6.1 in Principles of Mathematical Analysis, Rudin writes:

Let $$[a, b]$$ be a given interval. By a partition $$P$$ of $$[a, b]$$ we mean a finite set of points $$x_0, x_1 , \dotsc, x_n$$, where $$a = x_0 \le x_1 \le \dotsb \le x_{n-1} \le x_n = b.$$

Why does he use non-strict inequalities? I don't see how we could have $$x_{k-1} = x_k$$ if $$P$$ is a set, in contrast to a multiset. Also, the definition on Wikipedia uses strict inequalities. Is this just Rudin being a little bit sloppy (though I guess, technically, his definition reduces to the other one due to $$P$$ being a set), or am I actually missing something?

• As you mention yourself, it is technically correct because it's a set. Unless he later makes a mistake, e.g., by saying that the cardinality of $P$ is always $n$, it should be fine.
– ajr
Mar 25 at 15:08
• @ajr Maybe that's all there is to it. Feel free to write your comment as an answer if you want the points. I might still wait a bit to close the question, to see if anyone offers a different opinion.
– ummg
Mar 25 at 15:43
• @ajr Here is an idea; maybe he uses non-strict inequalitites to allow for the possibility of referring to the same element of $P$ by different indices? That seems to make sense, and I can see how it might be useful in some contexts.
– ummg
Mar 25 at 16:11

The usage of $$\le$$ in the partition \begin{align*} a=x_0\leq x_1\leq\cdots\leq x_n=b \end{align*} is admissible and just a matter of convenience. If $$x_{k-1} = x_{k}$$ we have \begin{align*} \{x_{k-1},x_k\}=\{x_k\}=\{x_{k-1}\} \end{align*} in accordance with the definition of a set. If you continue reading you can see the derivation of upper and lower Riemann integral of $$f$$ based upon infimum and supremum of partitions $$P$$ of $$[a,b]$$ is fine.