Is there a cover of the disk in the plane by three open connected sets $U_1, U_2, U_3$ such that any two of these sets have nonempty intersection, but the triple intersection $U_1 ∩ U_2 ∩ U_3$ is empty?

Clearly this is possible for the disk's boundary, i.e. a circle, but it does not seem possible for the disk itself. I can tell that such a cover will not be a good cover, since this would imply the disk is homotopically equivalent to the covering's nerve, in this case a triangle.

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    $\begingroup$ Do you mean three open sets in the plane (in which case @Janik's answer works) or three relatively-open sets in the disk (which seems more likely)? $\endgroup$ Commented Aug 25, 2016 at 21:29
  • $\begingroup$ I suppose I should merely have said "the disk" with no mention to the plane. I only care about the topology of the disk; its coincidental embedding in the plane is unimportant here. $\endgroup$
    – Feryll
    Commented Aug 26, 2016 at 1:52
  • $\begingroup$ Ah,..in that case @Janik's answer is no longer useful. But it was clever! My guess is that the answer is "no, there's no such thing", in part because you've required that the $U_i$ be connected. If you said "path connected," I think that it might be not too hard to make a fundamental-group argument why it's impossible, but I can't do so off the top of my head. $\endgroup$ Commented Aug 26, 2016 at 1:59
  • $\begingroup$ Path components in a (weakly) locally connected space are open anyway, so connected implies path connected in this case. $\endgroup$
    – Feryll
    Commented Aug 26, 2016 at 3:23

2 Answers 2


No it is not true. Mayer-Vietoris sequence for homology will give a contradiction.

Suppose $D^2 = U\cup V\cup W$ where $U,V,W$ are connected open set. Let $Z=V\cup W$ is connected open set . Then the fact that these three open sets have no common point implies $Z\cap U$ has at least two open component ( since $U\cap V$ and $U\cap W$ are non-empty). Now Mayer-Vietories long exact sequence for the pair $(U,Z)$

$\cdots \to H_1(D^2)\to H_0(U\cap Z)\to H_o(U)\oplus H_0(Z)\to H_0(D^2)\to 0$


$\cdots \to 0\to H_0(U\cap Z) \to \mathbb{Z\oplus Z} \to \mathbb{Z}\to 0$

Since the above s.e.s splits, this implies $H_o(U\cap Z)= \mathbb Z$ which is a contradiction.

EDIT: Actually the above solution tells us that not only for $D^2$ but for any simply connected space, we cannot cover it by using $3$ open connected sets which satisfy the given property.

  • $\begingroup$ Forgive my ignorance but why does the s.e.s. at the end split? $\endgroup$ Commented Aug 26, 2016 at 14:19
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    $\begingroup$ @PeteCaradonna Do you know projective module en.wikipedia.org/wiki/Projective_module ? You read the following link, and if you have any doubts, you can ask, I;ll try to answer you. $\endgroup$ Commented Aug 26, 2016 at 15:35
  • $\begingroup$ I haven't encountered them before, no. Sorry for what I suspect is a trivial question but how do they get relate to this? $\endgroup$ Commented Aug 26, 2016 at 15:39
  • $\begingroup$ @PeteCaradonna Any free module is a projective module, and for any projective module, the s.e.s splits. $\endgroup$ Commented Aug 26, 2016 at 15:41
  • $\begingroup$ Oh, ok, I was overthinking it. Thanks! $\endgroup$ Commented Aug 26, 2016 at 15:42

So I'm no expert on this, but here's a thought that might work.

Let $U_{ij} := U_i\cap U_j$.

Let $V_i$ denote the trivial line bundle on $U_i$. We may glue $V_1$ to $V_2$ via some isomorphism $\alpha_{12} : V_1|_{U_{12}}\rightarrow V_2|_{U_{12}}$, and similarly we may glue $V_2$ to $V_3$ via an isomorphism $\alpha_{23} : V_2|_{U_{23}}\rightarrow V_3|_{U_{23}}$.

Let $s_1$ be a section of $V_1$.

Since the $V_i$'s are trivial, $\alpha_{12}$ is determined by where it sends the section $s_1|_{U_{12}}\in V_1|_{U_{12}}$. Let $s_2|_{U_{12}} := \alpha_{12}(s_1|_{U_{12}})$, and let $s_2$ be the extension of $s_2|_{U_{12}}$ to all of $V_2$ (makes sense since $V_2$ is trivial). Similarly, let $s_3 := $"$\alpha_{23}(s_2)$". Finally, glue $V_1$ to $V_3$ via an isomorphism $\alpha_{13} : V_1|_{U_{13}}\rightarrow V_3|_{U_{13}}$ which does not send $s_1\mapsto s_3$.

If $U_1\cap U_2\cap U_3 = \emptyset$, then the cocycle condition is automatically satisfied, and these gluings (ie, "transition functions") should combine to yield a line bundle on the disc which is NOT trivial, since the local section $s_1$ can't extend to a global section (since $\alpha_{13}(s_1)\ne s_3$). On the other hand, there are no nontrivial line bundles on the disc, contradicting the assumption that $U_1\cap U_2\cap U_3 = \emptyset$.


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