# VC dimension of finite unions of one-sided intervals

What is the VC dimension of $k$ finite unions of one-sided intervals:

If we take 3 one-sided intervals like $(-\infty, a_1]$, $(-\infty, a_2]$ and $(-\infty, a_3]$, I think union of these intervals can shatter $4$ points as below, assuming that $a_1>a_2>a_3>a_4$:

• Point $p_1$ at interval $(-\infty, a_1]$

• Point $p_2$ at interval $[a_1, a_2]$

• Point $p_3$ at interval $[a_2,a_3]$

• Point $p_4$ at interval $(-\infty, a_3]$

For $k$ finite unions I think answer is $k+1$, am I right?

• Can you explain what you mean by "$k$ finite unions of one-sided intervals"? – Yuval Filmus Aug 27 '18 at 1:18
• @YuvalFilmus I mean union of $(-\infty, a1] U (-\infty, a2] U (-\infty, a3]$ – Joshna Gunturu Aug 27 '18 at 2:21
• This union is equal to $(-\infty, \max(a_1,a_2,a_3))$, that is, to a single interval. – Yuval Filmus Aug 27 '18 at 5:54
• Is $[a1,+\infty)$ also a one-sided interval? Can you copy and past the full original problem statement or provide an accessible link? As implied by Yuval Filmus, it is not easy to comprehend your paraphrase of the problem. By the way, the VC dimension of the subsets of the real line formed by the union of $k$ intervals is $2k$. – John L. Aug 28 '18 at 8:44

## 1 Answer

Finite unions of one-sided intervals can shatter only 2 points, because as said by @YuvalFilmus in comments the union of Finite unions is a single one-sided interval, and a single one-sided interval can shatter only 2 points.

• It seems more likely that you misunderstood the definition of one-sided interval. – Yuval Filmus Sep 1 '18 at 6:41