This is a long question, so I thank anyone willing to help me sort it out. That said, the material is elementary (to anyone who's taken real analysis) and I'm just long winded. The catalyst of the question is the following theorem and proof, which I repeat here verbatim:

Compact subsets of metric spaces are closed.


Let $K$ be a compact subset of a metric space $X$. We shall prove that the complement of $K$ is an open subset of $X$.

Suppose $p \in X$, $p \not\in K$. If $q \in K$, let $V_q$ and $W_q$ be the neighborhoods of $p$ and $q$, respectively, of radius less than $\frac{1}{2} d(p,q)$. Since $K$ is compact, there are finitely many points $q_1, \ldots, q_n$ in $K$ such that $$ K \subset W_{q_1} \cup\cdots\cup W_{q_n} = W. $$ If $V= V_{q_1} \cap\cdots\cap V_{q_n}$, then $V$ is a neighborhood of $p$ which does not intersect $W$. Hence $V \subset K^c$, so that $p$ is an interior point of $K^c$. The theorem follows. \end{proof}

Yes, $W_q$ of a neighborhood with center $q$ while $V_q$ is a neighborhood of center $p$, not great notation IMHO. Here, I have included the entire proof for the sake of completeness, but what I do not understand is simply the line: ''Since $K$ is compact, there are finitely many points $q_1, \ldots, q_n$ in $K$ such that $ K \subset W_{q_1} \cup\cdots\cup W_{q_n} = W. $''

This confuses me for two unrelated (I believe) reasons:

  1. As far as I can understand, the definition of a compact subset $K$ of a metric space $X$ is not that there exists a finite cover $W_1 \ldots , W_n$ of $K$ in $X$ (which is what the author seems to say) but, rather, that if there exists an open cover of $K$ in $X$ then there exists a finite sub-cover of $K$ in $X$. So, the question is: Before considering the finite open cover $W_{q_1}, \ldots, W_{q_n}$ of $K$ in $X$, is it not necessary to, first, justify the existence of any open cover?

  2. It seems strange to me that the author asserts that the set $\{q_1, \ldots, q_n\}$ which indexes the open cover $W_{q_1}, \ldots, W_{q_n}$ of the compact set $K$ is, itself, a subset of $K$. This is important because the author defines the neighborhoods in terms of these $\{q_1, \ldots, q_n\}$: $\forall q \in K, \hspace{1mm} W_q := \left\{x \in X : d(x,q)< r \right\}, \hspace{1mm} r <\frac{1}{2} d(p,q)$. Here, I have written out the definition of a neighborhood with center $p$ and radius $r <\frac{1}{2} d(p,q)$ for clarity; $d$ is the (arbitrary) metric in the definition of the (arbitrary) metric space $X$. So, the question is: Taking for given that there exists a finite open cover $W_{q_1}, \ldots, W_{q_n}$ of $K$ in $X$, how does one justify that $q_1, \ldots, q_n \in K$?

For reference, the book I'm using is Principles of Mathematical Analysis 3e by Walter Rudin: one of the most popular texts at the undergraduate level, if I understand correctly. The above is Theorem 2.34.

For reference, I will also give this author's definition of a compact subset of a metric space:

  • ''A subset $K$ of a metric space $X$ is said to be compact if every open cover of $K$ contains a finite subcover. More explicitly, the requirement is that if $\{G_\alpha\}$ is an open cover of $K$, then there are finitely many indices $\alpha_1, \ldots, \alpha_n$ such that $K \subset G_{\alpha_1} \cup\cdots\cup G_{\alpha_n}$.''

Notice that he does not say 'indeces in $K$'. Also, this requires an open cover, as well as the notation $\{G_\alpha\}$ to be defined.

  • ''By an open cover of a set $E$ in a metric space $X$ we mean a collection $\{G_\alpha\}$ of open subsets of $X$ such that $E \subset\bigcup_\alpha G_\alpha.$''

  • ''Let $A$ and $\Omega$ be sets, and suppose that with each element $\alpha$ of $A$ there is associated a subset of $\Omega$ which we denote by $E_\alpha$. The set whose elements are the sets $E_\alpha$ will be denoted by $\{E_\alpha\}$.''

Notice that the set $\Omega$ is, in general, not the same as the set $A$ which contains the indices $\alpha$. Hence, my second question.

Thanks, already, if you made it this far! I tried to make that to be not excessively dense reading material.

Edit: I mistook a $p$ for a $q$ and fixed it (I think)

  • 3
    $\begingroup$ existence of open covers is not in question, there are many. A space is compact, if every open cover has a finite subcover. In particular an author may define an open cover that fits the purpose of proving something, and use the definition of compactness to extract a finite subcover. $\endgroup$ – Mirko Jun 11 at 2:49
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    $\begingroup$ Anent your first question: the author expects you to see for yourself that $\{W_q:q\in K\}$ is an open cover of $K$. (Do you see why that is?) Then, since $K$ is compact, that open cover has a finite subcover. By the way, I think the notation $V_q,W_q$ is fine. $\endgroup$ – bof Jun 11 at 2:49
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    $\begingroup$ As for your second question, the reason the finite subcover is indexed by elements of $K$ is that the open cover the author has defined, from which the subcover is extracted by compactness, was indexed by elements of $K$. $\endgroup$ – bof Jun 11 at 2:53
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    $\begingroup$ The open cover $\{W_q:q\in K\}$ was not given a name; it doesn't need one, since it won't be referred to again as such, after the finite subcover is pulled out. It is certainly not the same as $K$; $K$ is a set of points in the metric space $X$, the open cover is a collection of open sets, completely different things. $\endgroup$ – bof Jun 11 at 3:07
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    $\begingroup$ Yep, 5 minute limit for editing comments. You can of course get around that by deleting your comment and reposting it correctly. $\endgroup$ – bof Jun 11 at 3:09

The $p \notin K$ is fixed throughout the proof. So $V_q$ and $W_q$ only "depend on" $q \in K$ and all we really need to know about them is that $p \in V_q$, $V_q,W_q$ are open, $q \in W_q$ and $V_q \cap W_q = \emptyset$ for every single $q$. So the only "variable" is that we do this for all $q \in K$ at the same time (and the explicit form of open balls of that radius is to avoid invoking AC, I suppose). So as to 1. we consider the cover $\{W_q: q \in K\}$ as the open cover of $K$. We can use any set we like as an index set, so here we use $K$. So finitely mnay indices (i.e. points of $K$) exist such that $K \subseteq W_{q_1} \cup \ldots W_{q_n}$.

Now you see the advantage of indexing by the $q$, because we can take the corresponding finitely many (!) $V_{q_1},\ldots, V_{q_n}$ neighbourhoods of $p$ and intersect them and the pairwise disjointness of all $W_q,V_q$ then ensures it is disjoint from the finite union and thus from $K$.


Follow up: Using the insight provided by @bof, I've attempted to flesh out the details in Walter Rudin's proof. I also use some of my own notation (I don't like superfluously naming objects which already have names, if it can be avoided). I also use logical notation like $\exists$, $\implies,$ etc., although I know it is considered inappropriate in mathematical writing (sorry).

My TeX macros aren't compatible with this website, as far as I know (being a new user). So, I don't know of a better way to share this than to share a link to the PDF file in my google drive account:


The same link as above: My Follow Up Answer

One of those two links must work. If anybody wishes to confirm that this answer is correct, correct this answer, or otherwise offer suggestions on writing proofs (besides to not use logic notation), it is welcome and appreciated.

Thanks, again, to all those who comment and/or answer.

  • $\begingroup$ Nothing wrong with using notation like $\exists$ and $\implies$ in mathematical writing. These are $mathematical$ symbols. One common defect of students' writing is to present a sequence of (apparently ) disconnected sentences, because they omit the words or symbols for $\implies$ or $\iff$ that should connect them $\endgroup$ – DanielWainfleet Jun 15 at 1:50

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