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If $V(x)$ is a separable Hilbert space, is $\bigcup_{x \in X}V(x)\times\{x\}$ separable when $X$ is an uncountable set? How to make it separable if it's not? What assumptions do I need?

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What is the answer if $V(x)=\{0\}$ for all $x$? –  Jonas Meyer Feb 8 '13 at 17:37
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So you want the final union to bbe just a topological space or something like the space of vector-valued funcitons? What is the relationship between the topology you want and topologies of $V(x)$? –  Norbert Feb 8 '13 at 17:38
    
@Norbert I want it to be like vector-valued functions. I am not really firm on topologies.. basically I want a function space in which I can define functions $f$ such that $f(x)$ is in $V(x)$ for $x \in X$. –  soup Feb 8 '13 at 17:40
    
@JonasMeyer I guess yes, but for non-trivial case I don't know –  soup Feb 8 '13 at 17:48
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I changed $\cup_{x\in X}V(x)\times\{x\}$ to $\bigcup_{x\in X}V(x)\times\{x\}$. I take that to be standard when you have subscripts or superscripts on $\bigcup$, and use $\cup$ only for things like $A\cup B$ and $A\cup\cdots\cup B$, etc. –  Michael Hardy Feb 8 '13 at 19:15

2 Answers 2

If you are referring to the product space together with the product topology then I am afraid it is not true. Take for instance the space $[0,1]^{[0,1]}$, which means uncounably many copies of $[0,1]$. This space is not separable. Since this space is compact and Hausdorff being separable is actually equivalent to being metrizeable. For this reason I don't think there is a reasonable way to "make" this space separable, since I don't see a reasonable way to make it a metric space without changing the topology completely.

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Suppose $X=[0,X_0]$ in my notation. What topology can I give it? –  soup Feb 8 '13 at 17:31
    
I might be missing something, because I don't see how this relates to the question exactly. There is a union of copies of Hilbert space in the question, not an infinite product of $[0,1]$. (Separable compact Hausdorff spaces need not be metrizable. E.g. $\beta \mathbb N$.) –  Jonas Meyer Feb 8 '13 at 17:36
    
Actually, Edward, your space is separable by the theorem Paul cites. Being separable and being metrizable are not equivalent in compact Hausdorff spaces. What would be equivalent in a CHS is being separable and metrizable and being second countable. –  Syd Henderson Oct 11 '13 at 20:13

There is a result from Pondiczery, Hewitt and Marczewski:

If there are not more than $\mathfrak{c}$ ( which is the continuum), separable topological spaces, then their product is still separable.

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