Here's what I'm trying to prove:
A closed rectangle in $\mathbb{R}^n$ is a closed set.
This was a statement in Michael Spivak's Calculus on Manifolds and I'm trying to write a proof for it.
Let $([a_i,b_i])_{i=1}^{n}$ be a finite sequence of closed intervals and let $C = \prod_{i=1}^{n}[a_i,b_i]$ be a closed rectangle in $\mathbb{R}^n$. To prove that this is closed, we just have to show that $\mathbb{R}^n \setminus C$ is open. This amounts to showing that for every $p \in \mathbb{R}^n \setminus C$, there is an open rectangle $A$ such that $p \in A \subseteq \mathbb{R}^n \setminus C$.
Let $p \in \mathbb{R}^n \setminus C$. Then, $p = (p_i)_{i=1}^{n}$. Since $p \notin C$, there is an $i \in \{1,2,\ldots,n\}$ such that $p_i \notin [a_i,b_i]$. So, $p_i \in (-\infty,a_i) \cup (b_i,+\infty)$.
Construct open intervals $(c_i,d_i)$ as follows:
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If $p_i \in [a_i,b_i]$, let $c_i = p_i-\epsilon$ and $d_i = p_i+\epsilon$, with $\epsilon > 0$ large enough so that $c_i < a_i$ and $d_i > b_i$.
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If $p_i \notin [a_i,b_i]$, let $c_i = p_i-\epsilon$ and $d_i = p_i+\epsilon$, with $\epsilon > 0$ small enough so that $(c_i,d_i)$ is contained in either $(-\infty,a_i)$ or in $(b_i,+\infty)$, depending on whether $p_i < a_i$ or $p_i > b_i$.
By definition, $p \in \prod_{i=1}^{n} (c_i,d_i)$ and clearly, this open rectangle is a subset of $\mathbb{R}^n \setminus C$. Since $p$ was arbitrary, $\mathbb{R}^n \setminus C$ is open so $C$ is closed. $\Box$
Does the argument above work? If it doesn't, then why? How can I fix it?
Best Answer
Your proof is correct. However, you can simplify your argument. You have $p_i \in U_i = (-\infty,a_i) \cup (b_i,\infty)$ which is open in $\mathbb R$. Hence $\pi_i^{-1}(U_i)$ is open in $\mathbb R^n$, where $\pi_i : \mathbb R^n \to \mathbb R$ denotes the projection onto the $i$-th coordinate. Clearly $p \in \pi_i^{-1}(U_i) \subset \mathbb R^n \setminus C$.