8
$\begingroup$

Let $(M,d)$ be a metric space. A set $A \subset M$ is said to be compact if every open cover of $A$ has a finite subcover.

Why do we use this definition, rather than the other "definition" which holds in $\mathbb{R}^n$, that is, a set is compact if it is closed and bounded? It is a more intuitive definition, and it is hard for me to think of compact sets being separate from "merely" closed and bounded sets (probably because I can only imagine Euclidian spaces).

Is it simply because being closed and bounded (together) is not a topological property?

$\endgroup$
13
  • 7
    $\begingroup$ Boundedness is certainly not a topological property. $\endgroup$ Sep 2, 2020 at 18:35
  • 17
    $\begingroup$ The open cover definition is far more general and gives the desired properties in all topological spaces. Settling on it rather than any of its historical predecessors was the result of recognizing that it was the ‘right’ definition, the one that behaved best. $\endgroup$ Sep 2, 2020 at 18:37
  • 1
    $\begingroup$ See also math.stackexchange.com/questions/371928 and math.stackexchange.com/questions/485822. $\endgroup$
    – Qi Zhu
    Sep 2, 2020 at 18:42
  • 4
    $\begingroup$ @Student Even in general metric spaces (where boundedness is indeed well defined) the definitions are not equivalent. And we want to keep the definition which has more good properties. For example, compact sets are preserved by continuous functions. This is not true for closed and bounded sets. $\endgroup$
    – Mark
    Sep 2, 2020 at 18:42
  • 2
    $\begingroup$ Does this answer your question? math.stackexchange.com/questions/2179367/… $\endgroup$
    – Lee Mosher
    Sep 2, 2020 at 20:51

3 Answers 3

8
$\begingroup$

That definition is the more general definition - it holds in $\mathbb{R}^n$ and in things very different from $\mathbb{R}^n$. (In fact, the more concrete definition isn't even appropriate to arbitrary metric spaces!)

Specifically:

  • It makes sense in arbitrary topological spaces, even ones which are non-metrizable (that is, which cannot be thought of as coming from a metric). For example, we can say with confidence that the cofinite topology on an infinite set is compact ... even though such a topological space is never metrizable.

  • Within the context of arbitrary metric spaces, "closed and bounded" doesn't behave the way it should: consider a discrete metric space where every point is at distance $1$ from every other point. Every set in such a space is closed and bounded, but we don't have any of the phenomena associated to compactness in $\mathbb{R}^n$ which we actually want (e.g. we can have an infinite sequence with no convergent subsequence). The open cover definition, by contrast, gets things right (e.g. a subset of a discrete metric space is compact iff it is finite).

$\endgroup$
4
  • $\begingroup$ Yes I have corrected "More general" to "Other". By "More General", I meant it holds for more sets than the actual definition does. $\endgroup$
    – Student
    Sep 2, 2020 at 18:43
  • 5
    $\begingroup$ @Student Ah - well that's a drawback then, per my second bulletpoint: we're not really interested in closed-and-boundedness per se, we're interested in the overall behavior it entails in the context of $\mathbb{R}^n$. The open covers definition captures that behavior and makes it work in arbitrary topological contexts; just lifting the phrase "closed and bounded" doesn't make sense in topological spaces and doesn't result in the right behavior in all metric spaces. $\endgroup$ Sep 2, 2020 at 18:45
  • $\begingroup$ @yoyo isn't completeness of the space also required? $\endgroup$
    – Student
    Sep 4, 2020 at 7:57
  • 2
    $\begingroup$ @Student it is. Otherwise $(0,1)$ is a counterexample which is closed and totally bounded subset of itself. But it is neither complete nor compact. $\endgroup$
    – freakish
    Sep 4, 2020 at 10:56
3
$\begingroup$

The question "why we define something as something?" is a really tricky question. There is no some universal reason not to define "compact" as "closed and bounded" or as "finite" or as "empty" or as "green grass". It's just a definition, a label, nothing more.

What we actually care about is behaviour and usefulness. In the metric world it turns out that the property "every sequence has a convergent subsequence" is a very strong and desired one. We called that "(sequentially) compact". And by the Bolzano-Weierstrass theorem this definition is equivalent to being "closed and bounded" but only for $\mathbb{R}^n$. The simplest counterexample in the metric world is $\mathbb{Q}$. Indeed, not every sequence in $A=[0,1]\cap\mathbb{Q}$ has a convergent subsequence (e.g. approximation of $\sqrt{2}/2$ by rationals) even though $A$ is both closed and bounded (with respect to the Euclidean metric) in $\mathbb{Q}$.

So "every sequence has a convergent subsequence" is a great property. And in fact it is easily generalizable to non-metric spaces. In the general setup it is also known as "sequential compactness". But it turns out that for metric spaces there is another property that is equivalent to compactness, namely "every open cover has a finite subcover". And since the definition doesn't require a metric (unlike "bounded" property), it is easily generalizable to non-metric world as well. But unfortunately outside of the metric world, this definition of compactness is not equivalent to sequential compactness (in fact neither implies the other). Comparing the two definitions mathematicians came to conclusion that the "open cover" definition is actually more useful and hence it became the standard one.

It is a more intuitive definition

Well, just because something is more intuitive doesn't mean it is better. Besides, nothing is intuitive until it becomes intuitive. :) I doubt that any professional mathematician nowadays would call the open cover definition as counterintuitive. It is so common that they've got used to it.

$\endgroup$
6
  • 1
    $\begingroup$ @Student Heine-Borel states that a subset of $\mathbb{R}^n$ is closed and bounded if and only if it is compact (in the sense of open cover). While Bolzano-Weierstrass says that it is closed and bounded if and only if it is sequentially compact. Both together say that for subsets of $\mathbb{R}^n$ open cover compactness and sequential compactness are equivalent (which can be further generalized to all metric spaces). For more information see this: math.stackexchange.com/questions/275954/… $\endgroup$
    – freakish
    Sep 4, 2020 at 10:18
  • $\begingroup$ Okay, so you are assuming equivalence of sequential compactness and cover compactness (which doesn't need either HB or BW, it holds generally in all metric spaces), and hence the BW theorem is equivalent to the HB theorem $\endgroup$
    – Student
    Sep 4, 2020 at 10:20
  • 1
    $\begingroup$ @Student in this context yes. $\endgroup$
    – freakish
    Sep 4, 2020 at 10:21
  • $\begingroup$ @freakish a little question about your comment, namely, how that generalizes to all metric spaces, do you mean the two concepts are eqivalent in all metric spaces? $\endgroup$
    – Physor
    Sep 4, 2020 at 10:40
  • 2
    $\begingroup$ @Physor sorry, what I've meant is that for metric spaces open cover compactness is equivalent to sequential compactness. They are not equivalent to "closed and bounded" property unless we are dealing with subsets of the standard $\mathbb{R}^n$. $\endgroup$
    – freakish
    Sep 4, 2020 at 10:44
0
$\begingroup$

In some commonplace metric spaces such as $\ell^2,$ there are sets that are closed and bounded but NOT compact. In particular, the standard orthonormal basis of $\ell^2$ is an example of such a set. And the closed interval from $0$ to $1$ within the space of rational numbers with the usual metric is another example. These examples are closed and bounded but not compact.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .