I'm having trouble with proving this theorem: A metric space is separable iff it is homeomorphic to a totally bounded metric space. There is a link on Wikipedia to book by S. Willard, but it is stated there as a fact leaving it to the reader as an exercise to prove it. Any help would be appreciated.
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To Brian M. Scott: Separable => homeomorphic to a totally bounded space is causing trouble. To kahen: Isn't every totally bounded space separable? I'm not sure, but is it necessary to use the argument with completion? – Josef Ondřej Nov 29 '12 at 20:58
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@kahen: U cannot claim that "Any completion of Y is then compact" cos closure of Y may not be equal to completion of Y! – Bear and bunny Nov 27 '13 at 17:23
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First, you’re right that a totally bounded metric space is automatically separable, so that there’s no need to go through the completion: the union of finite $2^{-n}$-nets for $n\in\omega$ is a countable dense subset.
Now assume that $\langle X,d\rangle$ is a separable metric space. Without loss of generality assume that $d(x,y)\le 1$ for all $x,y\in X$, and let $D=\{x_n:n\in\omega\}$ be a dense subset of $X$. Define the map
$$f:X\to[0,1]^\omega:x\mapsto\big\langle d(x,x_n):n\in\omega\big\rangle\;.$$
Now show that $f$ is an embedding of $X$ into the compact metrizable space $[0,1]^\omega$, the Hilbert cube; being compact, the Hilbert cube is totally bounded in any compatible metric, and total boundedness is hereditary, so $f[X]$ is totally bounded in any metric inherited from $[0,1]^\omega$.
Brian M. Scott
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1Can you tell me what's the meaning of $2^-n$-nets? More precisely, what's the meaning of a net? – Bear and bunny Nov 27 '13 at 04:13
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@Frank: In this context an $\epsilon$-net in a metric space $\langle X,d\rangle$ is a set $A\subseteq X$ such that $X=\bigcup_{x\in A}B(x,\epsilon)$. Another way to say it is that for each $x\in X$ there is a $y\in A$ such that $d(x,y)<\epsilon$. – Brian M. Scott Nov 27 '13 at 17:38
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@Frank: $\omega$ is the set of natural numbers, ${0,1,2,\ldots}$; you can replace it by $\Bbb N$ if you like. $[0,1]^\omega$ is the product of countably infinitely many copies of the space $[0,1]$. – Brian M. Scott Nov 27 '13 at 17:59
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U mean the amount of elements in D is uncertain and actually determined by the separpable set that u choose in X. So u use power N over [0,1] to build the hilbert cube. Besides, if the amount of elements in D is finite and supposed by m, then the power N should be fixed to m, is that right? – Bear and bunny Nov 28 '13 at 01:04
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@Frank: No, that’s not right. $\Bbb N={0,1,2,\ldots}$ is the set of all non-negative integers. It’s an infinite set, and so is $D$. – Brian M. Scott Nov 28 '13 at 17:19
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@M.scott: oh. Actually, $[0,1]^N$ recalls me of the power set $2^N$ which represents all subset of N. So is there any difference between them? I mean I cannot understand why a set can be a power that should be a number. – Bear and bunny Nov 28 '13 at 17:41
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@Frank: $[0,1]^{\Bbb N}$ and $2^{\Bbb N}$ are completely different things. $[0,1]^{\Bbb N}$ is the Cartesian product of countably infinitely many copies of the closed unit interval; equivalently, it’s the set of all infinite sequences of real numbers in $[0,1]$, since a point of the Cartesian product is a sequence of such real numbers. $2^{\Bbb N}$ is actually the Cartesian product of countably infinitely many copies of ${0,1}$ or, equivalently, the set of all infinite sequences of $0$’s and $1$’s. There is a $1$-$1$ correspondence between such sequences and subsets of $\Bbb N$, so some ... – Brian M. Scott Nov 28 '13 at 17:48
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... people us the notation $2^{\Bbb N}$ for the power set of $\Bbb N$. This is potentially confusing until you really know what you’re doing, so I recommend that you use $\wp(\Bbb N)$ instead. – Brian M. Scott Nov 28 '13 at 17:49
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@Frank: Provided that you’re thinking of them as Cartesian products, yes. – Brian M. Scott Nov 28 '13 at 18:56
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From the point of a set of all infinite sequences of 0’s and 1’s, {0,1}^card(N) will not be equal to ${0,1}^N$, right? – Bear and bunny Nov 28 '13 at 19:01
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@Frank: They are both the set of all infinite sequences of zeroes and ones. – Brian M. Scott Nov 28 '13 at 19:02
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Sorry, I made a typo. I mean 2^card(N) is a number but 2^N is a set, right? – Bear and bunny Nov 28 '13 at 19:08
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Sorry to bother u cos Im very interested in this problem. I've noticed the comment that "Provided that you’re thinking of them as Cartesian products, yes" and so from what, the answer will be no? – Bear and bunny Nov 28 '13 at 19:15
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@Frank: In the most common formalizations of set theory everything is a set, and that includes cardinal numbers. For me $\Bbb N=\omega$, and $2^\omega$ can be a cardinal number or a set of sequences depending on context, though I prefer the notation ${}^\omega 2$ for the latter. This is really not something that can be resolved in a short conversation, especially when I don’t know how much formal set theory you know. – Brian M. Scott Nov 28 '13 at 19:16
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@Frank: Part of the problem is that I’m not sure exactly what ‘it’ is. Some of your questions are mostly about notation, and for that you just have to learn the conventions of the sources that you use. Some of it seems to be about cardinal arithmetic, and for that you want a serious introduction to set theory; this one is good, but it goes way beyond the minimum needed to get some idea of cardinal arithmetic and builds up a lot of machinery before getting to that point. – Brian M. Scott Nov 28 '13 at 19:27
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BTW, i've noticed that someone comment above with the opinion that "X homeomorphic to Y totally bounded. Any completion of Y is then compact and thus separable--by kahen". Actually, from my point of view, that "Y is then compact" is not correct because closure of Y may not be equal to completion of Y, is that right? – Bear and bunny Nov 28 '13 at 19:33
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@Frank: No, kahen is correct, because a metric space is compact if and only if it is complete and totally bounded. If $Y$ is totally bounded, so is its completion $\widetilde{Y}$, and since $\widetilde{Y}$ is also complete, it must be compact. – Brian M. Scott Nov 28 '13 at 19:38
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Y is isometric to a dense set D, the subset of completion Y' and since Y is totally bounded, D is totally bounded. And due to Y' = closure of D, Y' is totally bounded, is that right? – Bear and bunny Nov 28 '13 at 19:47
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I see ur claim of $[0,1]^N$ as hilbert cube. Actually, i've read the definition on wiki that is "The Hilbert cube is best defined as the topological product of the intervals [0, 1/n] for n = 1, 2, 3, 4, ... That is, it is a cuboid of countably infinite dimension", which is a little different from $[0,1]^N$. I haven't learned about hilbert cube before. So any form of Cartesian product of countably infinitely many closed intervals can be called as a hilbert cube, is that right? – Bear and bunny Nov 28 '13 at 21:43
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Im still a little confusing on ur mapping, the f: X→$[0,1]^N$, that is how did u figure out this idea? – Bear and bunny Nov 28 '13 at 22:35
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@Frank: This embedding of a separable metric space in the Hilbert cube is standard; I don’t know when I first encountered it, but it was many years ago. – Brian M. Scott Nov 28 '13 at 22:42
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wait, how to prove the f is one to one and onto, also continuous in order to show homeomorphism between X and the hilbert cube? – Bear and bunny Nov 29 '13 at 01:05
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@Frank: Show that if $x\ne y$, there is an $n\in\omega$ such that $d(x,x_n)\ne d(y,x_n)$; it follows immediately that $f$ is $1$-$1$. In all likelihood $f$ is not onto; this doesn’t matter. Continuity is just a matter of checking that if $U$ is a basic open set in the product, then $f^{-1}[U]$ is open in $X$, which is pretty straightforward. – Brian M. Scott Nov 29 '13 at 01:10
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Ohhh, this f is really interesting: for example, X=[0,1] with usual metric and D is the set of rational numbers in [0,1] and then D is dense in X. However, if xn=0.1, x=0.05 and y = 0.15, d(x,xn)=|x-xn|=0.05=d(y,xn)=|y-xn|. However, there does exist many xn' not equal to 0.1 such that d(x,xn') ≠ d(y,xn'). What happened here? – Bear and bunny Nov 29 '13 at 01:30
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@Frank: Nothing to worry about: as long as there is at least one $n$ such that $d(x,x_n)\ne d(y,x_n)$, you’ll have $f(x)\ne f(y)$. And since there is always such an $n$ whenever $x\ne y$, $f$ is $1$-$1$. – Brian M. Scott Nov 29 '13 at 01:32
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Why does it not matter when f is not onto? Cos onto is necessary for homeomorphism? – Bear and bunny Nov 29 '13 at 01:37
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