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Let $\mathcal{F}$ be an open cover of a metric space $(X,d)$. I have a question concerning some algebraic aspects of $\mathcal{F}$ and $X$.

First of all, is it always possible to consider $(X,d)$ a subspace of some bigger space $(X^{\prime},d)$, where $d$ is the inherited metric from $X^{\prime}$? If this is possible, can we say that $\mathcal{F}$ is a subset of the power set of $X^{\prime}$? What are the conditions that we must state for this to be true?

Moreover, if all of that is true, can we say that $\mathcal{F}$ is an algebra of sets? Of course, for it to be true, $\mathcal{F}$ should contain both $X^{\prime}$ and $\emptyset$, but, whatever the cover is, we can always add those two sets and not change anything about the cover, right? So, finally, whatever $\mathcal{F}$ is, can we say that $\mathcal{F} \cup X^{\prime} \cup \emptyset$ is an algebra of sets over $X^{\prime}$?

Some motivation: in one of my exams I could prove something just saying "It is enough to notice that the cover is an algebra of sets", but I was unsure if that was always true.

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Your first question: You can construct $X'$ bu adding one new point to $X$ which is at distance $1$ from everything in $X$. Your 1nd question: sure, no conditions: to see why, just use the definitions to check it. –  Mariano Suárez-Alvarez Oct 4 '10 at 14:11
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You can say that $\mathcal F$ is an algebra of sets iff it is an algebra of sets. It is not true that all open covers are algebras –  Mariano Suárez-Alvarez Oct 4 '10 at 14:11
    
As for your motivation: please oh please look for an example of an open cover of your favorite metric space and see if the conditions for it to be an algebra are satisfied or not. Whatever the answer is, find next an example where the answer is the opposite. –  Mariano Suárez-Alvarez Oct 4 '10 at 14:12
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@Mariano: In your one-point construction you will violate the triangle inequality if X has two points of distance greater than 2. A Cartesian product will prove your point, though. –  whuber Oct 4 '10 at 14:21
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up vote 3 down vote accepted

Your first question is rather vague: you can always take $X'=X$, $d'=d$, and then $(X,d)$ is a subspace of $(X',d')$ with the inherited metric. Presumably you first of all want $X$ properly contained in $X'$ ("bigger" is a too vague, sorry!). But - do you also want some other property for $X'$, besides merely containing $X$? Perhaps "complete", or something along those lines?

If all you want is a metric space that properly contains $X$ and with induced metric, then the answer is yes. For example, if $X$ has only one element, then you can let $X'=X\cup\{p\}$, where $p\notin X$, and take the discrete metric. If $X$ has more than one element, you can take $X'=X\times X$, and define a metric by $d'((a,b),(x,y)) = d(a,x) + d(b,y)$ (the "taxicab norm"). To verify that this is indeed a metric, note that since $d(r,s)\geq 0$ for all $r,s\in X$, we have $d'((a,b),(x,y))\geq 0$; and $d'((a,b),(x,y))=0$ if and only if $d(a,x)=d(b,y)=0$, if and only if $a=x$ and $b=y$, if and only if $(a,b)=(x,y)$. Also, $d'((a,b),(x,y)) = d(a,x)+d(b,y) = d(x,a)+d(y,b) = d'((x,y),(a,b))$. Finally, for the triangle inequality, we have that $$\begin{array}{rcl} d'((a,b),(r,s))+d'((r,s),(x,y)) &=& d(a,r)+d(b,s) + d(r,x) + d(s,y)\\ & \geq & d(a,x) + d(b,y) = d'((a,b),(x,y)). \end{array}$$ So $(X',d')$ is a metric space.

Now fix $x_0\in X$, and embed $X$ into $X'$ by mapping $a\in X$ to $(a,x_0)$. This is an embedding, and for all $a,b\in X$, $d'((a,x_0),(b,x_0)) = d(a,b)+d(x_0,x_0)=d(a,b)$, so if we identify $X$ with its image in $X'$ you get that $d$ is the restriction of $d'$ to $X$. Since we are assuming that $X$ has more than one element, (the image of) $X$ is properly contained in $X'$.

When you say "$\mathcal{F}$ is an open cover for a metric space $(X,d)$", I interpret this to mean that each element of $\mathcal{F}$ is an open subset of $X$ and that $\cup\mathcal{F}=X$. You seem to be thinking that the sets may be strictly larger than $X$, but in the absence of a given overset, this makes no sense. Thus, we can talk about open covers of $[0,1]$ in $\mathbb{R}$ (open subset of $\mathbb{R}$ whose union contains $[0,1]$); but if we just say "an open cover of $\mathbb{R}$", we assume the sets are subsets of $\mathbb{R}$ absent any other context.

Now, if $X$ is a set, $X'$ contains $X$, and $\mathcal{F}$ is a family of subsets of $X$, then $\mathcal{F}$ is also a family of subsets of $X'$, hence a subset of the power set of $X'$; so the answer to your second question ("can we say $\mathcal{F}$ is a subset of the power set of $X'$?") is an absolute yes. The fact that we are dealing with a family of open sets or any such thing is completely immaterial; this is simply a consequence of the meaning of "subset".

As for being an algebra: an algebra of sets must be closed under unions and complements. If you have an open cover, it is not necessarily closed under unions, and it will usually not be closed under complements (the complements will be closed; if they are also part of the open cover, then each element of the open cover except for $X$ and $\emptyset$ (if they are there) would disconnect the space). So "most of the time" an open cover will not be closed under complementation and hence not be an algebra. ("Closing" the cover under unions is fairly easy; just throw in the unions of any two elements until you are closed under unions; all of them will be open, and enlarging an open cover by adding open sets will not change the fact that it is an open cover).

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Of course, by "bigger" I meant "which properly contains". As for the third question, I see it now; somehow I completely forgot about complements, which of course make everything invalid in general. Thanks for the answer. –  Andy Oct 4 '10 at 15:07
    
(I'm actually feeling a bit stupid for asking this...it's pretty basic, I should have seen it myself) –  Andy Oct 4 '10 at 15:13
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@Andy: part of my point was that the "of course" is not really implied. "Strictly bigger" and "bigger" and not generally synonyms in mathematics, so you really want to specify which one you mean to avoid misunderstandings. (In some areas, the default is strict ordering, so "bigger" will mean "strictly bigger"; in other areas, the default is the nonstrict ordering, so "bigger" will just mean "larger than or equal"). –  Arturo Magidin Oct 4 '10 at 15:56
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