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Johan_K
I've been sitting reading too much math online and my eyes are starting to melt I think. I blame it all on that cursed "is 0.9... = 1?"-thread laugh.gif. Wikipedia, which I think very little of as a reliable source for correct information of any kind, is a dangerous thing with all its links leading you farther and farther away like some will-o'-the-wisp. Anywho! I clicked around and found myself in a metric space and my question is as follows: (actually it's more like a question of clarification):

A metric space, (M,d), is bounded if there exists an r>0 such that d(x,y) ≤ r for all x and y in M, and totally bounded if there for every r>0 exists a finite number of 'open balls' with radius r whose union totally covers M.

So far so good and it's not hard to see why a totally bounded metric space is also bounded. However I have some trouble seeing why the reverse - why a bounded space isn't also totally bounded, is true and I suppose my question boils down to:

Is it because an infinite metric space, (R,|x-y|) say, is bounded by r=∞ but you can't cover the infinite set with a finite number of open balls (unless the distance metric is something like d(x,y)={∞ if x≠y or 0 otherwise} then it is also totally bounded? but that is hardly a useful metric I suppose and not very 'general')? Does that mean then that all metric spaces are bounded?

I'm not sure that is making sense - at all.

Thank you.
Stage left.

AlphaNumeric
Bounded means that ALL the elements in M are a finite distance away from one another (r=infinity doesn't count as bounded!). Totally bounded effectively means that the elements always come in clumps.

For instance the intervals (0,1) and (2,3) are seperately bounded and M = {(0,1),(2,3)} is TOTALLY bounded, because it's covered by a finite set of finte open coverings.

Consider (n,n+1) and (-n-1,-n). Again, they are each seperately bounded, the set (n,n+1) is bounded since all elements are within r=1 of one another. However, taking their union, M = {(-n-1,-n),(n,n+1)} is totally bounded since (-n-1,-n) is bounded with r=1 and (n,n+1) is bounded with r=1. It is also bounded for n=finite, because r=2n+2 is enough to cover ALL the elements, but in the limit n->infinity, this bound is destroyed, M is no longer bounded, but the individual sections are still totally bounded.

Bounded says ALL the elements clump together in a finite patch. Totally bounded says the elements clump together in a finite number of finite patches, but it doesn't matter how far apart those patches are. If the number of patches in a totally bounded metric space is 1 then total bounded is the same as bounded, but it's a less strict statement in general.
Alpha
QUOTE (Johan_K+Mar 12 2007, 02:58 PM)
Wikipedia, which I think very little of as a reliable source for correct information of any kind...

Nature magazine doesn't agree with you.

Maybe you want to dispute certain articles?
Johan_K
QUOTE
Nature magazine doesn't agree with you.

Maybe you want to dispute certain articles?


Nah, that vishy-washy tabloid. Dont they also produce the "National Enquirer"? wink.gif *hehe* Ofcourse not Everything in wikipedia is wrong rolleyes.gif , but actually I found Something that was a bit off, hold on, let me get it for you... the article about subsets. Under "definitions" it sais,

A < B (< is my subset sign-thingie cause it's the closest I have tongue.gif )
A = {α,β,γ,δ,ε,ζ}
B = {α,β,γ}

I thought that was strange. Maybe I misunderstood something like a line that said "this is wrong" - has happened before you know.

And by induction I proclaim that the rest on wiki is wrong too! laugh.gif

(watch out fo the irony wink.gif )
Johan_K
@AlphaNumeric
Thanks for the clarification AlphNum. It seems quite obvious that any finite space is bounded and also totally bounded. (Though it makes me kind of uneasy with the statement "if there for every r>0 exist finitely many open balls of radius r whose union covers M", seems to me like theres some kind of lim r->0 that would make the number of balls->∞? (There has been too much talk about limits tongue.gif)

What confused me about an infinite set was:
QUOTE
The converse does not hold, since any infinite set can be given the discrete metric (the first example above) under which it is bounded and yet not totally bounded.

but now that I read it more closely I found that the discrete metric was defined as "d(x,y)=1 for all x not equal to y and d(x,y)=0 otherwise" dry.gif I guess that's what happends when you stay up all night and try to learn math with a brain turned into cotton, but with that distance function it isnt so strange that set is bounded too!

I thought it sounded weird if one could just take r=∞ and make things bounded and totally bounded. would sort of render the whole point of the distinction useless!
AlphaNumeric
QUOTE (Johan_K+Mar 12 2007, 05:43 PM)
(Though it makes me kind of uneasy with the statement "if there for every r>0 exist finitely many open balls of radius r whose union covers M", seems to me like theres some kind of lim r->0 that would make the number of balls->∞? (There has been too much talk about limits tongue.gif)

I can see why you have the slight qualm, but it's a similar thing to 0.9, 0.99, 0.999 etc. Every element in the sequence is not 1, but the limit is something 'different', just as you can define irrational numbers as the limit of rational sequences (indeed, some points of view define it as such!).

Consider the sequence (r=1, r=0.1, r=0.01, ....). For any finite position element in that sequence the interval (0,1) is coverable by finitely many 'balls' of that radius. The limit is however not an open ball (it's a point, which is open and closed) and so the limit of the sequence isn't valid, hence why it's not defined by sequences but by the stating of a specific non-zero r.
QUOTE (Johan_K+Mar 12 2007, 05:43 PM)
, but with that distance function it isnt so strange that set is bounded too!
The discrete metric is not a very often used metric for physical considerations, it doesn't have the properties you'd expect from physical 'distances', but it's a common thing in mathematical analysis since it has all the same properties of a metric but often is a counter example. Same goes for the more general notion of a topology. The discrete topology is much like the discrete metric and is a counter example to many people's ideas.
QUOTE (Johan_K+Mar 12 2007, 05:43 PM)
I thought it sounded weird if one could just take r=∞ and make things bounded and totally bounded. would sort of render the whole point of the distinction useless!
Lol, true. Increasing without bound while being bounded would be a little strange wink.gif
rpenner
QUOTE (Johan_K+Mar 12 2007, 04:17 PM)
Nah, that vishy-washy tabloid. Dont they also produce the "National Enquirer"? wink.gif *hehe*

I think the Anthrax Killer (USA, September/October 2001) stopped that. tongue.gif
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