Counter example to a Sequentially Compact question

  • Thread starter Bachelier
  • Start date
  • #1
Bachelier
376
0
Can you provide one to show a separable complete boundd metr. space X is not always seq. compact.
 
Last edited:

Answers and Replies

  • #2
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,314
What did you try already? What metric spaces do you know?
 
  • #3
Bachelier
376
0
What did you try already? What metric spaces do you know?

I tried a subset of either [tex]\mathbb{N} \ or \ \mathbb{R}[/tex] with the discrete metric

Like consider [tex](A, d) \ such \ that \ A \subset \mathbb{R} \ something \ like \ [0,n] \with \ n =1, ..., 100. \ and \ d \ is \ the \ discrete \ metric.[/tex]
 
  • #4
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,314
Yes, that is very good! Try [tex]\mathbb{N}[/tex] with the discrete metric. Isn't that the counterexample you're looking for?
 
  • #5
Bachelier
376
0
Yes, that is very good! Try [tex]\mathbb{N}[/tex] with the discrete metric. Isn't that the counterexample you're looking for?

I wasn't sure about [tex]\mathbb{N}[/tex] being separable, but after thinking about it, it is dense in itself. Because every natural number is either in [tex]\mathbb{N}[/tex] or is a limit point.

But the definition that states that A is dense in B iff for all b1 and b2 in B, with b1<b2 there exists a in A st b1<a<b2. still confuses me.

if you take the numbers 1 and 2, then there exists no such number between them in [tex]\mathbb{N}[/tex] ?!
 
  • #6
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,314
You have several (non-equivalent) definitions for dense. The one you mention is dense for ordered sets. However, what we need here is topological dense. Then the definitoin states:

A set D is dense in X if the closure of D is X (or equivalently, that every non-empty open set in X contains a point of D).

With that definition, it can be easily checked that N is indeed dense in itself, and thus separable!
 
  • #7
Bachelier
376
0
You have several (non-equivalent) definitions for dense. The one you mention is dense for ordered sets. However, what we need here is topological dense. Then the definitoin states:

A set D is dense in X if the closure of D is X (or equivalently, that every non-empty open set in X contains a point of D).

With that definition, it can be easily checked that N is indeed dense in itself, and thus separable!

Thank you micromass. Hey I understand the following part 100%: that every non-empty open set in X contains a point of D. But I must admit that the concept of Closure of a set D still eludes me. I understand the definition that it is the intersection of closed sets containing D. I'd like to think of it in terms that if we add its boundary to D (which may be scattered around like in the case of [tex]\mathbb{Q}[/tex] then you get the closure).

Can you please elaborate on closure a little bit more.
Thanks
 
  • #8
Bachelier
376
0
Can you provide one to show a separable complete boundd metr. space X is not always seq. compact.

BTW the converse of the question is true. If X is seq. cmpact it is separ. bounded and complete. Right?
 
  • #9
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,314
Yes, your intuition is certainly correct. The closure is simply the set with it's boundary added. But it still remains tricky, because we have no formal definition of the word "boundary". Of course, we could define the boundary of a set as the points you add when you close the set (and this is what they indeed do), but it doesn't bring us closer to understanding what the closure is.

There are several different approaches that I like to use when discussing closure:
1) You already said one: the closure is the intersection of all closed sets containing the set. This is a nice property, but it doesn't really help us much in concrete examples.

2) When working in general topological spaces, then I always use the following property: A point x is in cl(A) if each open set around x intersects A. This is a very, very handy property. I always use this when checking whether something is in the closure.

3) I always use property (2), EXCEPT when working in metric spaces. In the case of metric spaces, there is a property that is much more handier: a point x is in the closure of A if and only if there exists a sequence (x_n) in A that converges to x.
For example: 1 is in the closure of ]-1,1[, since 1-1/n is a sequence in ]0,1[ that converges to 1. As you see, the property is super-easy to check and is very handy! (However: it is only valid in metric spaces).

An advice that I would like to give you to get used to closures is to calculate much closures. After a while you just get used to it. At first, I must admit that it's a bit abstract, but it's quite easy once you're used to it!
 
  • #10
micromass
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
22,178
3,314
BTW the converse of the question is true. If X is seq. cmpact it is separ. bounded and complete. Right?

In metric spaces, yes!
 
  • #11
Bachelier
376
0
Thank you teacher. :smile:
 

Suggested for: Counter example to a Sequentially Compact question

Replies
12
Views
523
Replies
0
Views
258
Replies
1
Views
250
  • Last Post
Replies
6
Views
940
Replies
12
Views
475
Replies
12
Views
670
Replies
6
Views
1K
Replies
4
Views
493
  • Last Post
Replies
5
Views
1K
Top