# Union of a collection of sets equal to$X \subset R^n$ different from covering of $X?$

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?

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 $X$.