Cohomology of punctured euclidean spaces

I'm running into a problem with a proof in From Calculus to Cohomology but I suspect I'm seeing not seeing something extremely simple. In Proposition 6.11, they prove that $H^{p+1}(\mathbb{R}^{n+1}-A) \cong H^p(\mathbb{R}^n-A)$ and they begin by defining open subsets of $\mathbb{R}^{n+1}$: $$U_1=\mathbb{R}^n \times (0,\infty) \cup (\mathbb{R}^n-A) \times (-1,\infty)$$ $$U_2=\mathbb{R}^n \times (-\infty,0) \cup (\mathbb{R}^n-A) \times (-\infty,1)$$ They then claim that $U_1 \cup U_2 = \mathbb{R}^{n+1} -A$, and $U_1 \cap U_2 = (\mathbb{R}^n-A) \times (-1,1)$.

I'm having trouble seeing why $U_1 \cup U_2$ is $\mathbb{R}^{n+1}-A$. For example why isn't $U_1$ simply equal to $\mathbb{R}^n \times (-1,\infty)$, and similarly why isn't $U_1 \cup U_2$ simply $\mathbb{R}^{n+1}$? An important part of the proof relies on the fact that $U_1$ and $U_2$ are contractible; is this "obvious" in the sense that I should be able to see it once I understand what $U_1$ and $U_2$ are?

Sorry if this is a very simple question. It's frustrating to me that I can't understand the simplest part of the proof! (The following parts, involving homotopy invariance and the Mayer-Vietoris sequence make sense to me if I take the above as given.)

EDIT: Sorry, I forgot to mention. $A$ is an arbitrary closed subset of $\mathbb{R}$ (not $\mathbb{R}^n$, it seems) and $A \ne \mathbb{R}^n$.

• What is $A$ ${}{}$? – Andres Mejia Aug 8 '17 at 18:52
• @AndresMejia I'm sure that $A$ is a closed subset of $\Bbb R^n=\Bbb R^n\times\{0\}$. – Lord Shark the Unknown Aug 8 '17 at 18:55

$U_1$ isn't $\Bbb R^n\times(-1,\infty)$; it does not contain, say $(-1/2,a)$ for $a\in A$.
$U_1$ deformation retracts to $\Bbb R^n\times\{1\}$ along "vertical lines", and $\Bbb R^n\times\{1\}$ surely is clearly contractible.