# Motivation for the Definition of Compact Sets

I'm currently taking my first course in real analysis, and was recently introduced to the following definition of compact sets:

A set $S \subseteq \mathbb{R}$ is compact if and only if every open cover of $S$ contains a finite subcover.

Shortly after learning the above definition of compact sets, I learned the Heine-Borel Theorem, which states that a subset $S$ of $\mathbb{R}$ is compact if and only if $S$ is closed and bounded. Therefore, the Heine-Borel theorem presents us with an equivalent way to think of compactness. To me, it seems much simpler to define compactness in terms of closed and bounded sets instead of open covers and finite subcovers. My question is, why are compact sets defined using open covers and finite subcovers? Why were compact sets not originally defined to be subsets of $\mathbb{R}$ that are closed and bounded? So far in my study of analysis, introducing open covers and finite subcovers seems to be needlessly complicating things.

I've searched around the internet for answers, but most sources that I've come across discuss subjects from topology which are beyond what I've studied in analysis. Any answers that can be presented in the context of a third-year introductory course to real analysis will be much appreciated. Thank you!

• The concepts diverge in more general spaces. The topological definition of compactness using open sets becomes the more important one because sequences don't suffice to describe convergence in an arbitrary setting. – DisintegratingByParts Nov 2 '14 at 20:45

The Heine-Borel Theorem is valid only some "special kinds" of spaces, like a metric space as $\Bbb{R}$. If a metric space is not complete, then the theorem is not valid, for example take the space of the rational numbers (this is in fact a metric space, try prove is not hard) does not have the "Heine-Borel property".

In a general topological space you only have open sets to work with. Distance, and thus boundedness, does not necessarily make sense in that setting.

Compact sets retain many of the intuitive topological and metric properties that finite sets have. The essence of it is contained in the finiteness of the cover.

• Hmm how can you have openness witouth distance? Doesn't openness require interior points which require balls which require distance? – Ovi Dec 6 '17 at 16:01
• @Ovi When defining the topology of a metric space, sure. However, in general topology, you simply take your space, and define which sets you choose to call open (subject to certain restraints), and such a choice is called a topology. Choosing "All subsets" gives the so-called discrete topology, while choosing "Only the empty set and the whole set" gives the trivial topology. And in-between you have a lot of other options. – Arthur Dec 6 '17 at 16:26
• Oh thank you very much for the clarification. So for example, in $\mathbb{R}$ with the usual metric the set $[0, 1]$ is undisputably compact; but in topology, if we choose to define some unusual sets as open which are not usually considered open in $\mathbb{R}$ (such as $\mathbb{N} \subset \mathbb{R}$), we may find that $[0, 1]$ is not compact under those rules? – Ovi Dec 6 '17 at 16:34
• @Ovi Exactly. For instance, in the discrete topology, where every subset is open (there is a corresponding metric in this case: all points have distance $1$ to all other points), you can have an open covering of $[0,1]$ consisting of only single point subsets. There is no way to find a finite subcover in that case. But in the standard topology, $[0,1]$ is compact. – Arthur Dec 6 '17 at 16:42
• Thank you very much for your help; I am taking Analysis now and I had all these confusions about off the cuff claims about topology I was hearing; you have cleared that up now. – Ovi Dec 6 '17 at 16:44

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