Prove a subset from 1000 points contains one point that is strictly larger than the other one

Question

I have been working on the following problem(from Supplementary Exercises of Chapter 1 of A Walk Through Combinatorics 4th edition) for a while.

Let $K$ denote the 1000 points in the three-dimensional space whose coordinates are all integers in the interval [1, 10]. Let $S$ be a subset of $K$ that has at least 272 points. Prove that $S$ contains two points $u$ and $v$ so that each coordinate of $v$ is strictly larger than the corresponding coordinate of $u$.

I know I should use pigeon hole principle and try to find 271 holes. I also can find a subset(planes: $x=1, y=1, z=1$) which cantians 271 points and meets requirement, but I don't know how to use this special case.
Any hints? Thanks.

Answer 1

Let's introduce an equivalence relation on $[10]^3$. $(a,b,c),(d,e,f) \in [10]^3$ are equivalent if there exist $\lambda \in \mathbb{Z}$ such that $(a,b,c) + \lambda (1,1,1) =(e,d,f)$. Tis a doddle to quickly check RST. A representative for each equivalence class could be to keep subtracting $(1,1,1)$ from a given vector until on of the coordinates is $1$.

You now need to show there are $271$ such equivalence classes (I will show this if requested) and also observe that $2$ elements in the same equivalence class will have the property that all the coordinates of one majorises the other ... and we are dumb.

Comment: When $[n]^3$ is visualised in terms of the lines parallel to $(1,1,1)$, we can now see the importance of Rezha Adrian Tanharja's comment.

Answer 2

For those curious about why the number of equivalence classes is 271 = $10^3$ - $9^3$, consider which elements of $[10]^3$ are NOT representatives of their equivalence class (i.e. their least coordinate is not a 1). It is all the points with all three coordinates of in the range of 2 to 9, of which there are $9^3$.