Intervals on the real line have VC dimension two

Fix $x_1<x_2$ and put $r=x_2-x_1>0$. The patterns $(0,0),(1,0),(0,1),(1,1)$ are realized respectively by $[x_1-2r,x_1-r]$ (strictly left of both points), $[x_1,x_1]$, $[x_2,x_2]$, and $[x_1,x_2]$. Thus the pair is shattered and the VC dimension is at least two.

For any three distinct ordered points, the labeling $(1,0,1)$ is impossible for an interval because containing $x_1$ and $x_3$ forces it to contain every point between them, including $x_2$. A repeated coordinate also prevents a tuple of length at least three from being shattered, since repeated instances receive the same label under every classifier. Hence no three-point indexed tuple is shattered and the VC dimension is exactly two.

On $m$ distinct ordered points, a nonempty interval has as its trace one contiguous block, and every such block is realized by the interval with those two sample endpoints. There are ${\sum }_{r=1}^m(m-r+1)=m(m+1)\mathrm{/}2$ such blocks; an interval outside the finite sample realizes the empty trace. For a tuple with $r\le m$ distinct points there are at most $1+r(r+1)\mathrm{/}2\le 1+m(m+1)\mathrm{/}2$ traces, while a distinct tuple attains that number. At $m=0$ the formula gives the single empty trace. Thus ${\mathrm{\Pi }}_{H_{\mathrm{i}\mathrm{n}\mathrm{t}}}(m)=1+m(m+1)\mathrm{/}2$.