It's said a set $A$ is compact if for every finite cover $U$ of $A$ there exists a subset of $U$ which also covers $A$, let's say $U_1$.
Assuming $A$ is a compact set, we must be able to find a subcover of $U_1$, $U_2$ where $U\supset U_1\supset U_2$.
With this in mind, my questions are:

Can this procedure can be done $n$-times?

If so, does some "infimum set" $U_f$ exists, such that from this set there cannot be obtained a subcover?
Sorry for the dull question, but I've been troubling with this for a long time.
Thanks, hope you all have a wonderful day

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    $\begingroup$ Compact is every open cover has a finite subcover, not the condition you stated. $\endgroup$ – Batman Jun 20 '16 at 21:12
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    $\begingroup$ Your definition is incorrect. A set is compact if every cover (in particular every infinite cover) has a finite subcover. $\endgroup$ – Bungo Jun 20 '16 at 21:12
  • $\begingroup$ I can understand the difference between the definition I wrote and the correct definition, but still my doubts are unclear. Thanks both. $\endgroup$ – GBes Jun 20 '16 at 21:18

Once you fix your definition of compact, I think what you are asking is:

Does every compact space satisfy, "Every cover of me has a minimal finite subcover?"

Here "minimal" means "containing no proper subcover", not "contained in every subcover."

The answer is yes: let $U$ be a cover of a compact space $X$. By compactness, $U$ has a finite subcover, $U_1=\{V_1, V_2, . . . , V_n\}$. Now let $U_i=\{V_i, V_{i+1}, V_{i+2}, ... , V_n\}$, and let $k$ be the largest number such that $U_k$ is an open cover of $X$ (there is such a largest number since $U_1$ is a finite cover); then $U_k$ is a minimal subcover of $U$.

Note that there is no reason for every open cover to have a unique minimal subcover.

This property can fail for noncompact spaces, even if we drop the word "finite." For example, take $\mathbb{R}$ with the usual topology and let $U=\{(-n, n): n\in\mathbb{N}\}$. Then $U$ has no minimal subcover.

Interestingly, it is not clear to me whether the property "Every open cover has a minimal subcover" is equivalent to compactness . . .

EDIT: OK, this is neat: it turns out the property "Every open cover has a minimal subcover" is indeed equivalent to compactness! The proof is cute, so I can't resist putting it here:

  • Suppose $X$ is a topological space such that every open cover of $X$ has a minimal subcover, and suppose $X$ is not compact. Let $\kappa$ be the least cardinality of some counterexample to compactness, and let $U=\{V_\eta: \eta<\kappa\}$ be a counterexample to compactness.

  • Let $W_\eta=\bigcup_{\beta<\eta}V_\beta$. Each $W_\eta$ is open, and clearly $U'=\{W_\eta: \eta<\kappa\}$ is an open cover of $X$. So $U'$ has a minimal subcover, $C$.

  • What could $C$ be? Well, it's easy to see that $C$ must have exactly one element: if $W_\eta, W_\theta\in C$ with $\eta<\theta$, then $C\setminus\{W_\eta\}$ would be a strictly smaller subcover.

  • OK, so $C=\{W_\eta\}$ for some $\eta$. Then - by definition of $W_\eta$ - we have that $U''=\{V_\beta: \beta<\eta\}$ is a subcover of $U$. But $\vert\eta\vert<\kappa$, so $U''$ has size $<\kappa$ - so by assumption on $\kappa$, $U''$ has a finite subcover.

  • $\begingroup$ Can this minimal subcover be useful for something? $\endgroup$ – GBes Jun 20 '16 at 21:31
  • $\begingroup$ @SebastianGutierrez Not to the best of my knowledge, but that's not to say that the answer is no. I'm sure there are indeed a few occasions where it's useful to look at a minimal subcover, just because everything's useful somewhere. $\endgroup$ – Noah Schweber Jun 20 '16 at 21:34
  • $\begingroup$ See 17.1 in susanka.org/Notes/Topology.pdf for an independent observation of this fact. $\endgroup$ – Henno Brandsma Jun 20 '16 at 21:36
  • $\begingroup$ @Henrik but there are covers for $\mathbb{N}$ without a minimal subcover. The statement is for all covers. $\endgroup$ – Henno Brandsma Jun 20 '16 at 21:38
  • $\begingroup$ @Henrik Yes, some covers have minimal subcovers - the question is whether there is a noncompact space where every cover has a minimal subcover! $\endgroup$ – Noah Schweber Jun 20 '16 at 21:38

As mentioned in the comments your definition of compactness is wrong. But there is some validity to your questions.

But if $A$ is a compact set and $U$ is an infinite cover, the by the real definition of compactness there exists a finite $U_f\subsetneq U$, and obviously we can add sets from $U\setminus U_f$ to form a sequence of finite covers that are proper subsets of each other.

Some covers might have several (I think it will be easy to construct an example with infinitely many) different minimal (in the sense that no subset is a cover) covers.

  • $\begingroup$ Re: your last paragraph, take $\mathbb{R}$ with the usual topology and let $V_r=\mathbb{R}\setminus\{r\}$ for $r\in\mathbb{R}$ . . . $\endgroup$ – Noah Schweber Jun 20 '16 at 21:47

Some examples to illustrate the idea of compactness.

Firstly with the way you've written it, you could always choose the subset $U_1\subset U$ such that $U_1=U$. If you mean subset in the strict sense, then for example take the unit interval $[0,1]\subset \Bbb R$ and cover it with the sets $(-\frac12,\frac12)$ and $(\frac13,\frac43)$. This certainly covers $[0,1]$ but we certainly can't refine it even once, and $[0,1]$ is compact, so this can't be the right idea.

An enlightening example of compactness can be seen by comparing the compactness of $[0,1]$ and the non-compactness of $(0,1)$.

Take for example the cover $\mathfrak{U}$ of $(0,1)$ defined by $\mathfrak{U}=\{(\frac1n,1-\frac1n)\mid n\in \Bbb{N}-\{1,2\}\}$. We have a countable cover of $(0,1)$ but we cannot choose a finite subcover because even though the intervals are nested, we need an a countably infinite amount of them to cover all of the points arbitrarily close to $0$ and $1$.

$[0,1]$ doesn't have this problem, as some set must contain $0$ and another set (possibly the same set) must contain $1$, and hence some numbers close either side. Of course this requires rigorous proof, but it is part of the idea of the rigorous proof in analysis.


"It's said a set A is compact if for every finite cover U of A there exists a subset of U which also covers A, let's say U1."

Well, as $U \subset U$ that is trivial. This idea that a finite subcover must have strictly smaller subcover is absurd.

Consider that $[0,1]$ is compact. Consider {$(-1,2)$} is a finite cover; it only has one set it in. Can't get much more finite than that. But it has no smaller subcover.

Your definition is wrong. For every open cover $U = $ {$U_{\alpha}$} of A [A cover $U = ${$U_{\alpha}$} is a collection of sets so that $A \subset \cup_{\alpha} U_{\alpha}$] has a finite subcover.

Example: $[0,1]$ is compact. let $U = \{(a,b)| a \in \mathbb R; b \in \mathbb R; a < b\}$. As $[0,1] \subset \cup_{a,b} (a,b)$, $U$ is a an open cover. $U$ has uncountably many intervals $(a,b)$. But as $[0,1]$ is compact there must be a finite subcollection that covers them. And indeed $[0,1] \subset (-1,1) \cup (0,2)$ is a finite subcollection. Indeed $[0,1] \subset (-1,2)$ is a subcollection with only one set.

"Assuming A is a compact set, we must be able to find a subcover of U1,U2. Where U>U1>U2"

Not at all! For example, for $[0,1]$ the subcovers $\{(-1,1),(0,2)\}$ or $\{(-1,2)\}$ don't have any strictly sub covers. But any infinite open cover must have a finite subcover.

"If so, does some "infimum set" Uf exists, such that from this set there cannot be obtained a subcover?"

Obviously. But it's not accurate to call it "infimum". Theres nothing unique or infinimum about it. I've given several examples above. For $[0,1]$. {(-1,2)} is a minimal subcover. As is {(-1,1),(0,2)}. As is {(-.1,1/3),(1/4,2/3),(1/2,1.1)} etc.


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