Doubt in the definition of a compact set 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
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 $U=\{V_\eta: \eta&lt;\kappa\}$ be a counterexample to compactness.
Let $W_\eta=\bigcup_{\beta&lt;\eta}V_\beta$. Each $W_\eta$ is open, and clearly $U'=\{W_\eta: \eta&lt;\kappa\}$ 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 $\eta&lt;\theta$, 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 $U''=\{V_\beta: \beta&lt;\eta\}$ is a subcover of $U$. But $\vert\eta\vert&lt;\kappa$, so $U''$ has size $&lt;\kappa$ - so by assumption on $\kappa$, $U''$ has a finite subcover.
"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 $U = \{(a,b)| a \in \mathbb R; b \in \mathbb R; a &lt; b\}$.  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.