- #1

jgens

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## Homework Statement

Prove that a closed rectangle [itex]A \subset \mathbb{R}^n[/itex] is a closed set.

## Homework Equations

N/A

## The Attempt at a Solution

Let [itex]A = [a_1,b_1] \times \dots \times [a_n,b_n] \subset \mathbb{R}^n[/itex], then [itex]A[/itex] is closed if and only if its complement, [itex]\mathbb{R}^n - A[/itex], is open. Now, let [itex]x = (x_1, \dots, x_n) \in \mathbb{R}^n - A[/itex] and choose [itex]\varepsilon_i = \inf_{c_i \in [a_i,b_i]}\frac{|x_i - c_i|}{2} > 0[/itex]. Clearly we have that [itex](x_i - \varepsilon_i,x_i + \varepsilon_i) \cap [a_i,b_i] = \emptyset[/itex]; thus, we can define an open rectangle [itex]B = (x_1 - \varepsilon_1, x_1 + \varepsilon_1) \times \dots \times (x_n - \varepsilon_n, x_n + \varepsilon_n)[/itex] [itex]\subset \mathbb{R}^n - A[/itex] with the property that for any [itex]x \in \mathbb{R}^n - A[/itex], [itex]x \in B \subset \mathbb{R}^n - A[/itex], completing the proof.

My questions:

1) Is this "proof" correct?

2) If so, is there an easier way to do this?

Thanks