# If $U$ and $V$, having $U \cap V = \emptyset$, are open in $\mathbb R$ with the standard metric, then there exists $c \notin U \cup V$

If $$U$$ and $$V$$, having $$U \cap V = \emptyset$$, are open in $$\mathbb R$$ with the standard metric, then there exists $$c \notin U \cup V$$

Proof:

Let $$A = [a,b] \cap U$$ where $$a \in U$$ and $$b \in V$$. Since $$a \in A$$ and $$a \in U$$, $$A$$ is non-empty and for all $$x \in A$$, $$x \leq b$$ so $$A$$ is bounded above. Thus $$c = \text{sup} A$$ exists.

If $$c \in V$$ we have $$B(c,r) \subset V$$ for some positive $$r$$. Now $$t < c$$ for $$t \in A$$, and if $$t < x < c$$, $$x \in A$$. Thus for every positive $$r$$, some there is some $$t \in A$$ for which $$t \in B(c,r)$$. This especially means $$t \in U$$ so $$U \cap V$$ is nonempty which is a contradiction. Thus $$c \notin V$$.

If $$c \in U$$ we have $$B(c,r) \subset U$$ for some positive $$r$$. Now $$t \leq c$$ for all $$t \in A$$. If $$c \in [a,b]$$ then for all $$v \in (c,b]$$ we have $$v \in V$$, which implies $$v \in B(c,r)$$ for any positive $$r$$. Thus $$c \notin [a,b]$$ and $$b < c$$. Thus for all positive $$r$$, $$b \in B(c,r)$$ which is a contradiction. Thus $$c \notin U$$ and this implies $$c \notin U \cup V$$. $$\blacksquare$$

Are there any major errors or unnecessarily complicated arguments?

• I don't see where you defined $c$. I think you forget to mention that you are taking $c$ to be $sup(A)$. Jun 14, 2019 at 9:51
• @KaviRamaMurthy Thanks for mentioning my erratum. I edited the answer to include the definition of $c$. Jun 14, 2019 at 9:54
• What if $\;U=V=\emptyset\;$ ? Jun 14, 2019 at 10:12
• @DonAntonio This is quite trivial since any $r \in \mathbb R$ will do. Jun 14, 2019 at 10:55

If $$c\in[a,b]$$ then for all $$v\in(c,b]$$ we have $$v\in V$$, which implies $$v\in B(c,r)$$ for any positive $$r$$.
You cannot really conclude that $$v\in V$$, but if there would be a $$v\in (c,b]$$ such that $$v\notin V$$ then you would have $$v\notin U\cup V$$, giving you the result.
• Ah of course, since $v \in (c,b]$ might not necessarily be in $V$ itself (although, as you point out, in this case it is not problematic). Thanks for the correction! Jun 14, 2019 at 11:23