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In Munkres book Analysis on manifolds he gives the folowing definition of compactness for a subspace $X $ of $R^n $

"Let $X $ be a subspace of $R^n $. A covering of $X $ is a collection of subsets of $R^n $ whos union contains $X $ ;...This space $X $ is said to be compact if every open covering of $X $ contains a finite subcollection that also forms an open covering of of $X $."

Next he gives a theorem that this can be formulated in terms of only sets of the space $X $.

"A subspace $X $ of $R^n $ is compact iff for every collection of sets oopen in $X $ whose union is $X $, there is a finite subcollection whose union equals $X $.

Now my question: if $\{A_{\alpha }\} $ is a collection whose union equals $X $, isn't that a covering of $X $?

Instead Munkres states that for each $\alpha $ we chose an open set $U _{\alpha } $ of $R^n $ such that $A_{\alpha }=U _{\alpha } \cap X$. Then this collection $\{U _{\alpha }\} $ covers $X $....

This is clear, but why is this necessary?

I have tried to find a definition of "A contains X" but there is none in this book. But usually this is simply defined as that every set in $X $ is also a set in $A $. Thus I cannot see why a union that equal a set also contains the set. What have I missed here?

Thanks in advance!

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1 Answer 1

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The original def refers to a subset of $\mathbb R^n$. As such, the covering open sets are subsets of $\mathbb R^n$.

The theorem says that if instead we look at $X$ as a topological space in its own right (with the subspace topology ingerited from $\mathbb R^n$), then it's the same thing for $X$ to be compact in itself or as a subset of $\mathbb R^n$.

When we look at $X$ itself, the covering open sets are now open subsets of $X$. They are not necessarily open in $\mathbb R^n$, so this may not be a cover included in the original definition. To get a cover consisting of sets that are open in $\mathbb R^n$, we go to the $U_\alpha$'s. Conversely, if we have a cover in $\mathbb R^n$, can find a corresponding cover of sets that are open in $X$ by taking the intersections with $X$.

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  • $\begingroup$ Thanks, great explanation. $\endgroup$
    – Alexander
    Commented May 26, 2014 at 12:47

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