Problem: Let $ R $ be the set of partitions of a real interval. Then for all elements in $ R $, every pair of elements has an upper bound.
I am having trouble structuring the proof; and intuitively understanding why every pair of elements has an upper bound.
Let the interval be $[a,b]$. A partition $X$ of $[a,b]$ is a finite sequence $X=\langle x_0,x_1,\ldots,x_n\rangle$ of points of $[a,b]$ such that $x_0=a$, $x_n=b$, and $x_0<x_1<\ldots<x_n$. A partition $Y=\langle y_0,\ldots,y_m\rangle$ refines $X$ if and only if $\{x_0,\ldots,x_n\}\subseteq\{y_0,\ldots,y_m\}$: every partition point of $X$ is a partition point of $Y$ or, equivalently, each interval of $Y$ is contained in an interval of $X$.
For reflexivity, you need only show that each partition of the interval is a refinement of itself; that’s completely trivial.
For transitivity you must show that if $Y$ refines $X$, and $Z$ refines $Y$, then $Z$ refines $X$; that also is a very straightforward application of the definition of refines.
Suppose that $X=\langle x_0,\ldots,x_n\rangle$ and $Y=\langle y_0,\ldots,y_m\rangle$ are partitions of $[a,b]$; you want a partition $Z=\langle z_0,\ldots,z_r\rangle$ that refines both $X$ and $Y$ and has the further property that if $W$ refines $X$ and $Y$, then $W$ refines $Z$. In other words, you want $Z$ to be the coarsest possible common refinement of $X$ and $Y$.
What partition points does $Z$ absolutely have to include in order to refine in order to refine both $X$ and $Y$?
Does $Z$ have to include any partitions points beyond that bare minimum?