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?

  • $\begingroup$ Can you explain what you mean by "$k$ finite unions of one-sided intervals"? $\endgroup$ Aug 27, 2018 at 1:18
  • $\begingroup$ @YuvalFilmus I mean union of $(-\infty, a1] U (-\infty, a2] U (-\infty, a3] $ $\endgroup$ Aug 27, 2018 at 2:21
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    $\begingroup$ This union is equal to $(-\infty, \max(a_1,a_2,a_3))$, that is, to a single interval. $\endgroup$ Aug 27, 2018 at 5:54
  • $\begingroup$ 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 $2k$. $\endgroup$
    – John L.
    Aug 28, 2018 at 8:44

1 Answer 1


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.

  • $\begingroup$ It seems more likely that you misunderstood the definition of one-sided interval. $\endgroup$ Sep 1, 2018 at 6:41

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