Counter example to a Sequentially Compact question

[Bachelier]

Can you provide one to show a separable complete boundd metr. space X is not always seq. compact.

 

[micromass]

What did you try already? What metric spaces do you know?

 

[Bachelier]

What did you try already? What metric spaces do you know?

I tried a subset of either $$\mathbb{N} \ or \ \mathbb{R}$$ with the discrete metric

Like consider $$(A, d) \ such \ that \ A \subset \mathbb{R} \ something \ like \ [0,n] \with \ n =1, ..., 100. \ and \ d \ is \ the \ discrete \ metric.$$

 

[micromass]

Yes, that is very good! Try $$\mathbb{N}$$ with the discrete metric. Isn't that the counterexample you're looking for?

 

[Bachelier]

Yes, that is very good! Try $$\mathbb{N}$$ with the discrete metric. Isn't that the counterexample you're looking for?

I wasn't sure about $$\mathbb{N}$$ being separable, but after thinking about it, it is dense in itself. Because every natural number is either in $$\mathbb{N}$$ 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 $$\mathbb{N}$$ ?!

 

[micromass]

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!

 

[Bachelier]

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 $$\mathbb{Q}$$ then you get the closure).

Can you please elaborate on closure a little bit more.
Thanks

 

[Bachelier]

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?

 

[micromass]

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!

 

[micromass]

BTW the converse of the question is true. If X is seq. cmpact it is separ. bounded and complete. Right?

In metric spaces, yes!

 

[Bachelier]

Thank you teacher.