# Intersection of connected open sets with union $[0,1]^2$ is connected

I'm trying to show that if the union of two open connected sets equals $[0,1]^2$, then their intersection is also connected.

My attempt:

Let $U,V \subset [0,1]^2$ be open and connected, and suppose $U \cap V$ is disconnected. Then there exist disjoint open sets $A,B$ such that $U \cap V = A \cup B$. But then $U = U \backslash V \cup (A \cup B)$ and $U \backslash V \cap (A \cup B) = \emptyset$. So this means $U$ is disconnected if $U \backslash V$ is open since $A \cup B$ is open. But I don't think $U \backslash V$ is open since $U \backslash V = U \cap V^c$ so it's the intersection of an open set with a closed set.

Should I be going about this in a different way?

• Is $I^2$ a universal notation for something, or is this a question just for whoever is taking the course/reading the book with you? – Asaf Karagila May 3 '13 at 0:09
• I think $I$ is a common notation for the closed interval $[0,1]$. – Dan Rust May 3 '13 at 0:16
• Yes, $I=[0,1]$. I edited that above now. – Polly Nomial May 3 '13 at 0:33
• The intersection of an open set with a closed set need not be open, nor closed: $(0,1)\cap [1/2,2]=[1/2,1)$. – Julien May 3 '13 at 0:37
• Do you mean that $U$ and $V$ are open in $[0,1]^2$? – Matemáticos Chibchas Feb 4 '16 at 20:39

## 1 Answer

Firstly, notice that this doesn't hold for an annulus or donut shape. You take two pieces that go around different sides of the hole and you're done. So somehow the structure of $I^2$ must be relevant (apparently its genus).

So without using more properties of the space, we're going to be stuck. Try drawing some pictures.

Here's one intuitively based approach: let $W$ be $(U\cap V)^c$, the closed subset of $I^2$ not in the intersection. Then $W=U'\cup V'$ where primes indicate restriction to $W$, and $U'\cap V'$ is empty. Hence either one of the primed sets is empty (and we're done as the sets were nested) or $W$ is disconnected.

In this case, use the simply connected property of the original space to note that, picking two random connected components $A,B$ of $U\cap V$, and points $a,b$ within them, any path joining $a,b$ can be continuously deformed into any other. The restriction of these paths to $W$ can also be continuously deformed into each other. But since $W$ is a disconnected union of the primed sets, you can show that each connected component is entirely surrounded by one of the primed sets, contradicting the connectedness of the other.

Edit: Just had a chat with a smarter friend than me, and we concluded that the 'natural' way to do this is homology theory. The logic runs roughly as follows: $$0 \longrightarrow H_0(U\cap V) \longrightarrow H_0(U) \oplus H_0(V) \longrightarrow H_0(U\cup V) \longrightarrow 0$$ is a short exact sequence; since $U,V,U\cup V$ are connected, all their $H_0$s are $\mathbb Z$. But then since the sequence is exact (and in particular we have an injection for the second arrow) and since the kernel of the addition map $\mathbb Z \oplus \mathbb Z \to \mathbb Z$ is isomorphic to $\mathbb Z$, we conclude that $H_0(U\cap V) \cong \mathbb Z$ and therefore $U\cap V$ is also connected.

Simplifying this argument by stripping out all the general homological algebra would probably result in basically approximating $U,V$ by unions of simplices (with a deeply tedious argument) and then doing some (Euler-characteristic-style) counting to show that the intersection must be connected.

Note: One thing that's worth pointing out is that both the above arguments move to path-connectedness which is in general stronger than connectedness. However open subsets of $\mathbb R^n$ have the property that they are path connected if and only if they're connected. (Proof: Consider path components; show they are both closed and open. This contradicts connectedness.)

• Maybe you're thinking of $H_0$.. – Edoardo Lanari Jun 13 '13 at 21:55
• I believe I am! Thanks. – Sharkos Jun 14 '13 at 1:15
• +1 For Mayer - Vietoris sledgehammer! @OP $H_0(X)$ is a free abelian group on the number of components of $X$, if $H_0(X)$ is free on rank $1$ this means that $X$ is connected. – user38268 Jun 14 '13 at 1:22
• @Lano - I also dimly recall this fact, and I've added a quick note to explain why it doesn't matter that we weren't careful. – Sharkos Jun 15 '13 at 1:10
• Yeah sure I'm aware of that fact, my comment was a general precisation! – Edoardo Lanari Jun 15 '13 at 9:37