Proving Closed Rectangle A is a Closed Set

[jgens]

Homework Statement

Prove that a closed rectangle $A \subset \mathbb{R}^n$ is a closed set.

Homework Equations

N/A

The Attempt at a Solution

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

My questions:
1) Is this "proof" correct?
2) If so, is there an easier way to do this?

Thanks

 

[rochfor1]

No. $$\varepsilon_i$$ may be zero for some i. Consider the unit square [0,1] x [0,1] in $$\mathbb{R}^2$$. The point (1/2, 2) is not in the unit square, but your $$\varepsilon_1=0$$.

An easier proof: recall that a set is closed if and only if for a sequence $$x^{(k)}$$ in the set that converges, the limit is in the set. Now let $$x^{(k)}$$ be a sequence in A that converges to the point x. For i = 1, 2,..., n we have $$x_i^{(k)} \to x_i$$ in R. Now, $$[a_i,b_i]$$ is closed in R (this is very easy to prove), so $$x_i \in [a_i,b_i]$$ which impies that $$x \in A$$.

 

[jgens]

Ah! That's a really good point. :grumbles quitely:

I haven't learned that formulation of a closed set so I can't really follow your proof (sorry!). So, suppose I revised my proof as follows: Let $x = (x_1, \dots, x_n) \in \mathbb{R}^n - A$, then for some $x_i$ we have that $x_i \notin [a_i,b_i]$. I can then choose $\varepsilon = \inf_{c_i \in [a_i,b_i]}\frac{|x_i - c_i|}{2} > 0$ and since $(x_i - \varepsilon, x_i + \varepsilon) \cap [a_i,b_i] = \emptyset$ I can define the open rectangle $B = (x_1 - \varepsilon, x_1 + \varepsilon) \times \dots \times (x_n - \varepsilon, x_n + \varepsilon) \subset \mathbb{R}^n - A$. Hence, for any $x \in \martbb{R}^n - A$ there exists an open rectangle $B$ such that $x \in B \subset \mathbb{R}^n - A$, completing the proof.

Does this work any better? I just started learning about open and closed sets today so I appreciate all the help that I can get. Thanks again!

 

[rochfor1]

This almost works like a charm. But the fact that $$\varepsilon > 0$$ requires you to prove that $$[a_i,b_i]$$ is closed in R. There's an easier way to pick $$\varepsilon = \frac{ 1 }{ 2 } \min \{ |x_i-a_i|, |x_i-b_i| \}$$.