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$(X, \mathscr T )$ be a topological space and $A \subset X$. $\{ U_i \mid i \in I \}$ is said to be an open cover of $A$ if $A \subset \cup_{i \in I} U_i$. $A$ is said to be compact if there exists finite $J \subset I$ such that $A \subset \cup_{i \in J} U_i$. $\{ U_i \mid i \in J \}$ is then called finite subcover of $A$.

My doubt:

Let set of open sets which cover $A \subset X$ be $\{U_1,U_2, U_3, \dots\}$. If we add $X$ in this collection of open sets, then also it would cover $A$ i.e. the new covering set be $\{X, U_1,U_2, U_3, \dots\}$. Now we need to take a finite subset of the above "covering set". Let that subset be $\{X, \phi \} $ i.e $\{X, \phi \} \subset \{X, U_1,U_2, U_3, \dots\}$. This subset would cover $A$ since $A \subset X$. So $A$ is compact. But we can virtually do this with any subset of $X$. I know this is wrong but where am I making a mistake?

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    $\begingroup$ The definition of compact is that any open cover has a finite subcover. $\endgroup$ – Quinn Culver Mar 3 '14 at 19:11
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I think the problem in understanding this is that you have combined several definitions into one statement, and the other definitions obscure the one you should focus on. You should separate the definitions so they aren't so intertwined. For example:

(1) Definition of cover: Let $X$ be a set. A cover of $X$ is a set $\mathscr A \subset \mathscr P(X)$ such that $\bigcup\mathscr A = X$.

(2) Definition of open cover: Let $(X,\mathscr T)$ be a topological space. An open cover of $X$ is a cover $\mathscr A$ of $X$ such that $\mathscr A \subset \mathscr T$.

(3) Definition of subcover: Let $\mathscr A$ be a cover of a set $X$. A (finite) subcover is a (finite) set $\mathscr B \subset \mathscr A$ which is also a cover of $X$.

(4) Definition of compactness: Let $(X,\mathscr T)$ be a topological space. $X$ is compact if every open cover $\mathscr A$ of $X$ contains a finite subcover of $X$.

I point out that in all of this, the qualifying phrase "of $X$" is frequently omitted if it is clear from context.

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  • $\begingroup$ If it is given that $X$ is compact then can it be said that every $A \subset X$ is also compact? $\endgroup$ – Richard J Mar 3 '14 at 19:58
  • $\begingroup$ got it!! Closed subsets of compact sets are compact (and not always as I wrote above). thanx for giving a detailed definition of compactness. It cleared many of my doubts. $\endgroup$ – Richard J Mar 3 '14 at 20:21
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The definition is that EVERY open cover has finite subcover. So those open covers which have X as their element can be written on that way, but there are many other open covers that do not have X! If for all those covers we can find finite subcovers, than that set would be called a compact set.

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The definition of compactness means that for any open cover $\mathcal{U}$ of the space $X$ there are finitely many $U_1 , \ldots , U_n$ from that collection $\mathcal{U}$ which also covers $X$. When you "add $X$" to $\mathcal{U}$, you are changing the open cover into a different open cover, let's call it $\mathcal{U}^\prime$. While the collection $\{ X , \varnothing \}$ (or, really, just $\{ X \}$), is a finite subcover of this new open cover $\mathcal{U}^\prime$, but will not (in general) be a finite subcover of the original open cover.

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  • $\begingroup$ Yeah! I should have noticed that for all (any/every) open covers. thanx :) $\endgroup$ – Richard J Mar 3 '14 at 19:22

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