# If $A,B \subset \mathbb{R}^n$ are disjoint non-empty and open then $\exists x \in \mathbb{R}^n$ such that $x \notin A \cup B$

I am trying to prove or disprove the statement that

If $A,B \subset \mathbb{R}^n$ are disjoint, non-empty, open sets then $\exists x \in \mathbb{R}^n$ such that $x \notin A \cup B$

My reason for asking this question is because I tried (and am struggling) to prove that $\mathbb{R}^n$ is connected. What I have is this:

Suppose $\mathbb{R}^n$ is disconnected. Then $\mathbb{R}^n = A \cup B$ where $A,B$ are disjoint non-empty open sets. But since $A,B$ are disjoint, non-empty and open $\exists x \in \mathbb{R}^n$ such that $x \notin A \cup B$. This contradicts $\mathbb{R}^n = A \cup B$.

I know that if the statement in the title is in fact true, then this completes my proof that $\mathbb{R}^n$ is connected, but I am not sure how to actually prove or disprove it.

Even if the statement is false, I think, at the very least, $A$ and $B$ are both unbounded; if one or both of $A,B$ is contained in a sphere of finite radius, we clearly cannot have $\mathbb{R}^n = A \cup B$. Perhaps $A,B$ are "half-spaces" i.e. sets of the form $\{(x_1 ... x_n) \in \mathbb{R}^n | x_i > R \}$. But since $A,B$ are disjoint I feel like there has to be least a single point (more likely an entire plane) that is not in either of them, but I honestly don't know.

• If you're not committed to this approach there's a totally general and pretty nice way to show that the product of two connected spaces is connected (e.g.here). – JonathanZ Nov 24 '17 at 18:07
• @JonathanZ I'm not sure what you mean by "committed to this approach." Can you be more specific? – cpiegore Nov 24 '17 at 18:08
• Here, $\mathbb{R}^n$ is convex, so path-connected, so connected – Max Nov 24 '17 at 18:11
• @cpiegore, most everyone who looks at your statement is going to say "Yes, it's true, because $\mathbb{R}^n$ is connected.", and they will trot out their favorite proof that it is connected. It sounds like you want a proof of your claim that such an $x$ exists (and you need proof of that -- just claiming that such an $x$ exists is not a valid argument). That's what I meant by "committed to this approach", and it sounds like yes, you are committed to it. (There's nothing wrong with that, it just means that you're not looking for proofs of connectedness that take a different approach.) – JonathanZ Nov 24 '17 at 18:20
• @cpiegore The step in your attempted proof is incomplete, First, your argument that both $A$ and $B$ must be unbounded is illogical; there's no immediately obvious reason you can't have one be bounded and the other be bounded. Next, you haven't given us any reason to say that there must be a point which is disjoint from $A$ and $B$. Of course there must be, since $\mathbb{R}^n$ is connected, but without assuming that, why would there be? – Duncan Ramage Nov 24 '17 at 18:20

## 1 Answer

The negation of the statement '$\mathbb R^n$ is connected' is :

There exists disjoint non-empty open sets $A$, $B$ such that $\mathbb R^n=A\cup B$. Period. No $x$.

To get to a contradiction, you can argue as follows: take $x\in A$, $y\in B$. Define $t:=\inf\{\lambda\in[0,1]: \ \lambda x + (1-\lambda )y\in A\}$. Such a $t$ exists, moreover $t\in (0,1)$. Consider the point $x_t := t x+ (1-t)y$. Then $x_t\not\in A$ ($A$ is open $x_t$, if $x_t$ would be in $A$, we could make $t$ smaller). But $x_t\not\in B$ as well (as there cannot exist a sequence of points in $A$ converging to $x_t\in B$). Contradiction.

• Everything else looks fine, but how exactly can we guarantee that such a $t$ exists? – cpiegore Nov 24 '17 at 19:30
• It is the infimum of a bounded nonempty set of real numbers. – daw Nov 24 '17 at 19:42
• Small problem: if we take $\lambda = 0$ then the expression $\lambda x + (1-\lambda )y\in A$ becomes $y \in A$ which seems to contradict $y \in B$ – cpiegore Nov 24 '17 at 20:33
• @cpiegore Be careful about interpretation. The expression $\left \{ \lambda\in[0,1]: \ \lambda x + (1-\lambda )y\in A\ \right \}$ means those $\lambda$ in the interval $[0,1]$ that belong to $A$. It does not necessarily mean all values in that interval should belong and in fact it isn't true obviously for the reason you just pointed out :) – John11 Nov 24 '17 at 20:47