1
$\begingroup$

Is there a general method one can follow given a metric space $(X,d)$ and a set $A \subset X$ to show that the set A is

1) open

2)closed

3)compact

I can figure these properties out for some specific sets and I know it depends on what metric we're using , and what set and subset we use but is there any general road marks I can follow when solving questions like this ?

$\endgroup$

closed as too broad by user491874, José Carlos Santos, Anthony Carapetis, Rebellos, Sahiba Arora Jan 10 '18 at 0:33

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I believe this is a too broad question... You have a bunch of theorems such as "inverse image of an open set in a continuous map is an open set", or "the image of a compact set in a continuous map is compact", surely "the complement of an open set is closed and vice versa" etc. but I don't feel there is much more to it than saying "just use the theorems as appropriate". Usually the devil is in detail (of the specific problem setup) IMHO. Happy to be proven wrong by other contributors to Math.SE... $\endgroup$ – user491874 Jan 9 '18 at 21:08
  • $\begingroup$ You usually just prove it from the definition or prove that it has certain properties equivalent to the definition. There's not a general recipe that always works for these though. $\endgroup$ – Alexis Olson Jan 9 '18 at 21:08
  • $\begingroup$ (I should've said "theorems and definitions" rather than just "theorems"...) $\endgroup$ – user491874 Jan 9 '18 at 21:10
2
$\begingroup$

Set A is open, if for every $a\in A$ there exist some open ball containing $a$ with every point in that ball in $A.$

The real numbers is an open set (it is also a closed set). Open intervals i.e. $(0,1)$ is an open set.

Open sets do not have "hard edges"

A set is closed if its compliment is open.

The integers is a an example of a closed set.

The rational numbers are neither open nor closed.

Closed sets have hard edges. i.e. They have all of their limit points.

A set is compact if "every open cover has a finite sub-cover." But what the heck does that mean?

I don't think it was for a year or to after I passed Real Analysis did I feel like I had a solid grip on what compactness really meant. (Not sure if I still do)

Maybe it is easier to talk about a set that not compact.

Consider the set $(-1,1).$ We can establish a sequence of open covers $(-1+\frac {1}n, 1-\frac 1n).$ As an infinite collection it will cover the interval, but there is no finite subset that will cover the entire interval. The set is not compact.

Compact sets are closed and bounded. But sometimes that gets you into a little bit of trouble as not every set that is closed and bounded is compact.

Every sequence in a compact set has a convergent sub-sequence. And every convergent sequence converges to a point in the set.

The set $\{\frac 1n: n\in \mathbb N\}$ is not compact as the sequence $\{\frac1 n\}$ converges to $0$ and $0$ is not in the set.

However the set $\{\frac 1n: n\in \mathbb N\}\cup 0$ is compact. Now thinking about the definition and open covers. Whatever covers $0$ will also cover infinitely many other points in the set, leaving finitely many points left to cover. So, every open cover has a finite subcover.

$k-$cells (closed and bounded intervals in Euclidean space) are compact. And a finite collection of compact spaces is compact.

$\endgroup$
1
$\begingroup$

There is one guideline that I feel I need to give here, even if it is not the complete answer.

If a subset of a metric space is compact, it can be proven that it is closed and bounded.

The converse: I've seen many instances where people have fallen into a trap to believe that the converse is true. The converse is generally not true. It is true in $\mathbb R^n$, but not in general metric spaces. For some examples where it is not true, see e.g. this post: A example of closed and bounded does not imply compactnesss in metric Space

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.