- #1

Bachelier

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

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- Thread starter Bachelier
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- #1

Bachelier

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

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- #2

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

- #3

Bachelier

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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

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]

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- #5

Bachelier

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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

if you take the numbers 1 and 2, then there exists no such number between them in [tex]\mathbb{N}[/tex] ?!

- #6

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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

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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

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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

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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

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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

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Thank you teacher.

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