Proving Closed Rectangle A is a Closed Set

In summary, the proof provided is not entirely correct as it does not consider the possibility of \varepsilon_i being equal to zero for some i. An easier proof would be to use the fact 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. By considering a sequence in A that converges to a point outside of A, it can be shown that A is closed. Another possible proof involves selecting an appropriate \varepsilon value to construct an open rectangle B that contains x and is contained in \mathbb{R}^n - A, thus proving that A is closed.
  • #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
 
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  • #2
No. [tex]\varepsilon_i[/tex] may be zero for some i. Consider the unit square [0,1] x [0,1] in [tex]\mathbb{R}^2[/tex]. The point (1/2, 2) is not in the unit square, but your [tex]\varepsilon_1=0[/tex].

An easier proof: recall that a set is closed if and only if for a sequence [tex]x^{(k)}[/tex] in the set that converges, the limit is in the set. Now let [tex]x^{(k)}[/tex] be a sequence in A that converges to the point x. For i = 1, 2,..., n we have [tex]x_i^{(k)} \to x_i[/tex] in R. Now, [tex][a_i,b_i][/tex] is closed in R (this is very easy to prove), so [tex]x_i \in [a_i,b_i][/tex] which impies that [tex]x \in A[/tex].
 
  • #3
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 [itex]x = (x_1, \dots, x_n) \in \mathbb{R}^n - A[/itex], then for some [itex]x_i[/itex] we have that [itex]x_i \notin [a_i,b_i][/itex]. I can then choose [itex]\varepsilon = \inf_{c_i \in [a_i,b_i]}\frac{|x_i - c_i|}{2} > 0[/itex] and since [itex](x_i - \varepsilon, x_i + \varepsilon) \cap [a_i,b_i] = \emptyset[/itex] I can define the open rectangle [itex]B = (x_1 - \varepsilon, x_1 + \varepsilon) \times \dots \times (x_n - \varepsilon, x_n + \varepsilon) \subset \mathbb{R}^n - A[/itex]. Hence, for any [itex]x \in \martbb{R}^n - A[/itex] there exists an open rectangle [itex]B[/itex] such that [itex]x \in B \subset \mathbb{R}^n - A[/itex], 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!
 
  • #4
This almost works like a charm. But the fact that [tex]\varepsilon > 0[/tex] requires you to prove that [tex][a_i,b_i][/tex] is closed in R. There's an easier way to pick [tex]\varepsilon = \frac{ 1 }{ 2 } \min \{ |x_i-a_i|, |x_i-b_i| \}[/tex].
 

Related to Proving Closed Rectangle A is a Closed Set

What is a closed set?

A closed set is a set in which all of its limit points are contained within the set itself. In other words, every sequence of points within the set converges to a point within the set.

How do you prove that a closed rectangle is a closed set?

To prove that a closed rectangle is a closed set, we must show that every sequence of points within the rectangle converges to a point within the rectangle. This can be done by showing that the limit of the sequence is contained within the rectangle and using the definition of a closed set.

Can a closed rectangle be an open set?

No, a closed rectangle cannot be an open set. By definition, a closed set contains all of its limit points, while an open set does not. A rectangle can only be one or the other, not both.

What are the properties of a closed rectangle?

A closed rectangle has the properties of being a closed set, meaning that it contains all of its limit points, and it is bounded, meaning that it has a finite area. It also has four sides, four vertices, and four right angles.

Why is it important to prove that a closed rectangle is a closed set?

Proving that a closed rectangle is a closed set is important because it allows us to use properties and theorems related to closed sets to analyze and solve problems involving the rectangle. It also helps in understanding the behavior and characteristics of the rectangle as a geometric shape.

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