• Support PF! Buy your school textbooks, materials and every day products Here!

Closed Sets

  • Thread starter jgens
  • Start date
  • #1
jgens
Gold Member
1,580
49

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
 

Answers and Replies

  • #2
256
0
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
jgens
Gold Member
1,580
49
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
256
0
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 Threads on Closed Sets

Replies
9
Views
2K
Replies
19
Views
8K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
617
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
24
Views
6K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
1
Views
2K
Top