TITLE: Can open sets be an open cover, for itself?
QUESTION [7 upvotes]: I have Baby Rudin's book with me and it clearly defines a cover to be open. In a followup, it defines a set $K$ to be compact if every open cover of $K$ contains a finite subcover.
And the rest I quote

More explicitly, the requirement is that if $\{ G_{\alpha}\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $$K \subset G_{\alpha_1} \cup \dots\cup G_{\alpha_n}$$

From the notation, it would seem to suggest that we can't have $K$ to be an improper subset of the covers. Now in another book (which I referenced another user whom referenced the book) by Richardson (I don't know the full name sorry), it's definition of a cover allows a set to cover itself.
So what subtlety could I have possibly overlooked? Sorry if this mysterious reference is vague, but I couldn't get enough information on the book.

REPLY [3 votes]: If $X$ is a metric space and $K$ is a subset, then I believe Rudin (in his Principles book) defines $K$ to be compact if for every collection of open sets $\{U_i:i\in I\}$ of $X$ with the property that $K\subseteq\bigcup_iU_i$, there are finitely many $i_1,\ldots,i_n\in I$ with $K\subseteq U_{i_1}\cup\cdots\cup U_{i_n}$. That is to say, the definition involves coverings by sets open in $X$. Compact subsets in metric spaces are closed, and if the metric space is connected, like $\mathbf{R}^n$, non-empty, proper closed sets are not open, so, because an arbitrary union of open sets is open, if $X$ is connected and $K$ is a non-empty, proper closed subset, you will never have $K=\bigcup_iU_i$ for $U_i$ open in $X$.
The aspect of this definition that I don't really like is that it makes it seem like compactness is a relative property, i.e., that it depends somehow on the ambient space $X$. This is not the case. By restricting the metric on $X$ to $K$, $K$ becomes a metric space in its own right, and one can check that a subset $A$ of $K$ is open in the metric topology of $K$ if and only if $A=K\cap U$ for some open subset $U$ of $X$. With this in mind, you can verify that $K$ is compact in Rudin's sense given above if and only if for every collection $\{V_j:j\in J\}$ of sets open in $K$ with the property that $K=\bigcup_jV_j$ (note the equality), there are finitely many $j_1,\ldots,j_n\in J$ with $K=V_{j_1}\cup\cdots\cup V_{j_n}$.
This carries over to arbitrary topological spaces without change (I just phrased it in terms of metric spaces since I think that is the generality of Rudin's book). The point is that compactness is an intrinsic property of topological spaces, and is in no way relative to some larger ambient space.