What is Compactness and why is it useful?

Question

I would like to gain a better understanding of the notion of compactness (topology). The wiki definiton defines a compactness of an interval as closed and bounded.

In mathematics, specifically general topology, compactness is a property that generalizes the notion of a subset of Euclidean space being closed (containing all its limit points) and bounded (having all its points lie within some fixed distance of each other).

Question 1: I cannot imagine a subset of the real line that would be closed and unbounded, therefore in this case I dont see why they add the boundedness requirement, why not just say compact=closed, and therefore I don't see the use oh the definition of "compact" at least if we stay in 1D (on the real line) ?

Qquestion 2: is this notion of compactness only usefull in dimensions strictly greater than 1? i.e. dim 2 and above, and in any case could someone provide a simple understandable example of compact and not compacti possibly in at least 2 different dimensions to be able to compare them?

Answer 1

First, I'll answer your questions. Then, I will provide an overview of why the definition is useful.

1. An example of a closed but unbounded subset of $\mathbb{R}$ is $\mathbb{R}$ itself. Other examples include $[0, \infty)$, $\{1\} \cup (\infty, 0]$, and many more.

2. Compactness is useful in all dimensions. Examples of compact sets in higher dimensions include a sphere in 3D space (eg $\{(x, y, z) \mid x^2 + y^2 + z^2 = 1\}$), a disc in 2D space ($\{(x, y) \mid x^2 + y^2 \leq 1\}$, and many others.

Now, for why the definition is useful.

First, the definition "A compact set is a set which is closed and bounded" is actually not the optimal definition. The best definition is:

Let $X$ be any topological space. $X$ is compact if and only if for any family of open sets $\{U_i\}_{i \in I}$, if $X = \bigcup\limits_{i \in I} U_i$, then there are $i_1, i_2, ..., i_n$ such that $X = U_{i_1} \cup U_{i_2} \cup ... \cup U_{i_n}$. This is often expressed as "Every open cover of $X$ has a finite subcover."

Here $\{U_i\}_{i \in I}$ being a "family of open sets" means that for all $i \in I$, $U_i$ is an open subset of $X$.

A second definition:

If $X$ is a topological space and $S \subseteq X$, then $S$ is said to be compact if and only if $S$, equipped with the subspace topology, is also compact.

It turns out that

Theorem (Heine-Borel): A subset $S \subseteq \mathbb{R}^n$ is compact if and only if $S$ is both closed and bounded.

However, in other metric spaces, you can have closed and bounded sets which are not compact. The simplest example here is $\mathbb{N}$ equipped with the so-called "discrete metric" - that is, $d(x, y) = 0$ if $x = y$, $1$ otherwise. Under this metric, $\mathbb{N}$ is closed and bounded. However, $\{\{i\}\}_{i \in \mathbb{N}}$ is an open cover of $\mathbb{N}$ with no finite subcover.

Here's the main theorem about compactness:

Theorem: Suppose $X$ and $Y$ are topological spaces and $f : X \to Y$ is continuous. Suppose $X$ is compact. Then $f(X) = \{f(x) \mid x \in X\}$ is a compact subset of $Y$.

This allows us to prove all kinds of nice theorems.

Extreme Value Theorem: Suppose $X$ is compact and nonempty and $f : X \to \mathbb{R}$ is continuous. Then $f$ achieves a minimum value, and $f$ also achieves a maximum value.

Proof: we see that $f(X)$ is a compact, nonempty subset of $\mathbb{R}$. Therefore, $f(X)$ is closed, bounded, and nonempty. Therefore, $f(X)$ has a minimum element and a maximum element. Thus, $f$ achieves both its minimum and its maximum value.

Going a different direction with compactness, we also have

Uniform Continuity Theorem: Suppose $X$ is a compact metric space, $Y$ is a metric space, and $f : X \to Y$ is continuous. Then $f$ is uniformly continuous.

This is the crucial theorem that allows Riemann integration to function.

Answer 2

To answer the first question, if you take the whole real line, it is both closed and open by definition. However it is not bounded and thus not compact. Regarding the 2nd question, the notion of compactness is useful as a property in general topology that goes beyond $\mathbb{R}^n$. However if we talk only about $\mathbb{R}^n$, then subset of $\mathbb{R}^n$ is compact $\iff$ it is closed and bounded (note that this doesn't hold in general). I suggest you read the introductory topology book "Topology without tears" for deeper understanding. You can easily find the pdf for free online. Regarding the example, you can take the circle in $\mathbb{R}^2$ with the bound (namely the set of points $\{(x,y)\in\mathbb{R}^2 | x^2 + y^2 <=1\}$) and this set is compact. However if you exclude the bound (namely the set of points $\{(x,y)\in\mathbb{R}^2 | x^2 + y^2 <1\}$) then this set is not closed and thus not compact. Hope this helps