Topology: An Inquiry-Based Approach

Closed and bounded intervals have important properties in calculus. Recall, for example, that every real-valued function that is continuous on a closed interval $[a,b]$ attains a maximum and minimum value on that interval. The question we want to address in this section is if there is a corresponding characterization for subsets of topological spaces that ensure that continuous real-valued functions with domains in topological spaces attain maximum and minimum values. The property that we will develop is called compactness.

The word “compact” might bring to mind a notion of smallness, but we need to be careful with the term. We might think that the interval $(0,0.5)$ is small, but $(0,0.5)$ is homeomorphic to $\mathbb{R}\text{,}$ which is not small. Similarly, we might think that the interval $[-10000,10000]$ is large, but this interval is homeomorphic to the “small” interval $[-0.00001,0.000001]\text{.}$ As a result, the concept of compactness does not correspond to size, but rather structure, in a way. We will expand on this idea in this section.

Since a topology defines open sets, topological properties are often defined in terms of open sets. Let us consider an example to see if we can tease out some of the details we will need to get a useful notion of compactness. Consider the open interval $(0,1)$ in $\mathbb{R}\text{.}$ Suppose we write $(0,1)$ as a union of open balls. For example, let $O_n=(\frac{1}{n},1-\frac{1}{n})$ for $n\in {\mathbb{Z}}^{+}$ and $n\ge 3\text{.}$ Notice that $(0,1)\subseteq \bigcup _{n\ge 3}{O}_n\text{.}$ Any collection of open sets whose union contains $(0,1)$ is called an open cover of $(0,1)\text{.}$ Working with a larger number of sets is generally more complicated than working with a smaller number, so it is reasonable to ask if we can reduce the number of sets in our open cover of $(0,1)$ and still cover $(0,1)\text{.}$ In particular, working with a finite collection of sets is preferable to working with an infinite number of sets (we can exhaustively check all of the possibilities in a finite setting if necessary). Notice that $O_n\subset O_{n+1}$ for each $n\text{,}$ so we can eliminate many of the sets in this cover. However, if we eliminate enough sets so that we are left with only finitely many, then there will be a maximum value of $n$ so that $O_n$ remains in our collection. But then $\frac{1}{2n}$ will not be in the union of our remaining collection of sets. As a result, we cannot find a finite collection of the $O_n$ whose union contains $(0,1)\text{.}$ Note that there may be some collections of open sets that cover of $(0,1)$ for which there is a finite subcollection of sets that also cover $(0,1)\text{.}$ For example, if we let $U_n=(n-\frac{3}{4},n+\frac{3}{4})\text{,}$ then $(0,1)\subseteq \bigcup _{n\in \mathbb{Z}}{U}_n\text{,}$ and $(0,1)\subseteq U_0\cup U_1\text{.}$ The main point is that there is at least one collection of open sets that covers $(0,1)$ for which there is no finite subcollection of sets that covers $(0,1)\text{.}$

Let’s apply the same idea now to the set $[0,1]\text{.}$ Suppose we extend our open cover $\{O_n\}$ to be an open cover of the closed interval $[0,1]$ by including two additional open balls in $\mathbb{R}\text{:}$ $O_0=B(0,0.5)$ and $O_1=B(1,0.5)\text{.}$ Now the sets $O_0\text{,}$ $O_1\text{,}$ and $O_4$ form a finite collection of sets that covers $[0,1]\text{.}$ So even though the interval $[0,1]$ is “larger” than $(0,1)$ in the sense that $(0,1)\subset [0,1]$ we can represent $[0,1]$ in a more efficient (that is finite) way in terms of open sets than we can the interval $(0,1)\text{.}$ This is the basic idea behind compactness.

If $(X,\tau )$ is a topological space and $X$ is a compact subset of $X\text{,}$ then we say that $X$ is a compact topological space. There is some terminology associated with Definition 17.1.

So the sets $O_0\text{,}$ $O_1\text{,}$ and $O_4$ in our previous example form a finite subcover of the open cover $\{O_n{\}}_{n\ge 3}\text{.}$

Using the terminology we have now established, we can restate the definition of compactness in the following way: a subset $A$ of a topological space $X$ is compact if every open cover of $A$ has a finite subcover of $A\text{.}$

There are two perspectives by which we can look at compactness. If $(X,{\tau }_X)$ is a topological space and $A$ is a subset of $X\text{,}$ then Definition 17.1 tells us what it means for $A$ to be compact as a subset of $X\text{.}$ From this perspective, we use open sets in $X$ to make open covers of $A\text{.}$ We can also consider $A$ as a subspace of $X$ using the subspace topology ${\tau }_A\text{.}$ From this perspective we can examine the compactness of $A$ using relatively open sets for open covers. Exercise 14 tells us that these two perspectives are equivalent, so we will use whatever perspective is appropriate for a given situation.