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We know the following definition of compactness of a set A in any topological space X.

Definition: In a topological space a subset A of X is said to be compact if every open cover of A has a finite sub cover.

Now my question is following

  1. If it is given that for any topological space X, a subset A of X is compact , can I say every open cover of it has a finite sub cover ? if so , then in the definition the term 'if' would be replaced by 'iff'; if not ,can you kindly provide me a suitable counter example? And What is infact compactness if I don't include the notion of cover? ( Well I am not entering into any other compactness like limit point compactness or sequential compactness )

If there is anything mistake in my question or my understanding, I shall be highly glad if you kindly help me by answering my question.

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    $\begingroup$ Question 1. is a bit broad; what topological properties are connected to compactness is a big topic that'd you'd do better to read up on in a textbook and then ask about specific notions. For 2., what do you mean by "$A$ is compact under any topological space"? This doesn't compile. For 3. If you don't include the notion of cover and at the same time exclude all other notions of compactness then I don't see how this question is supposed to be answerable. Please pick one of these questions to ask and address the respective concerns. $\endgroup$ Commented May 18 at 10:34
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    $\begingroup$ Hi Albert, to be honest I'ld change the way you ask questions cause "if not give me a counterxample" isn't very friendly. Despite this, I don't really get what you mean when you say "a set $A$ is compact under any topological space". Do you mean when a set $A$ of a topological space is compact in any topological space S contained in X equipped with induced topology? $\endgroup$
    – Francesco
    Commented May 18 at 10:39
  • $\begingroup$ @Francesco I think now my question is perfect for understanding. $\endgroup$
    – Albert
    Commented May 18 at 10:50
  • $\begingroup$ For question $1$, you can check $\pi$-base. $\endgroup$ Commented May 18 at 18:42

2 Answers 2

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It is a standard (but sometimes confusing, as here) convention that, in definitions, "if" means "if and only if". So compactness is being defined here to mean exactly that every open cover has a finite subcover.

More precisely, a definition of the form "A if B" really means "A iff B". An occurrence of "if" within B, however, has its normal meaning.

(I've been around for a long time, but not long enough to complain when this convention became standard. So don't blame me.)

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    $\begingroup$ When I teach advanced mathematics, I've switched over to a different expression for such definitions. In the OP's example, I would express it like this: To say that a subset $A$ of $X$ is compact means every open cover of $A$ has a finite subcover. $\endgroup$
    – Lee Mosher
    Commented May 18 at 15:07
  • $\begingroup$ Thank you sir now it is clear to me 😊 $\endgroup$
    – Albert
    Commented May 19 at 4:03
  • $\begingroup$ @LeeMosher Your formulation is good, but I trust you also warn students about the traditional (mis)use of "if",, since they're likely to encounter it elsewhere. $\endgroup$ Commented May 19 at 19:12
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For 2 there is a really simple answer; take $X$ to be any infinite set and let $A=X$ $-$think of $X$ as being the set of integer numbers, if you will$-$.

There are topologies for which $X$ is compact, and other for which $X$ is not; for instance, the discrete topology on $X$ makes it not compact of obvious reasons (for we are choosing $X$ to be an infinite set), whereas if we equip $X$ with the cofinite topology, $X$ becomes compact.

To check that $X$ with the cofinite topology has to be compact, take an open cover of it, say $\mathcal U$, and choose one open set $U\in\mathcal U$. Then, by the definition of the cofinite topology, $X\setminus U$ is a finite set, say $$X\setminus U=\{x_1,\dots,x_n\}\text{.}$$ Then, from $\mathcal U$, we can choose an open set, say $U_{x_1}$, that is such that $x_1\in U_{x_1}$; we can do this because $\mathcal U$ is a covering of $X$, and the elements of $\mathcal U$ have to contain every single element of $X$. Notice that this choice of $U_{x_1}$ is not canonical, and some use of the axiom of choice might have to be applied, depending on the nature of $\mathcal U$.

Now, with $U$ and $U_{x_1}$ at hand, it so happens that $$X\setminus(U\cup U_{x_1})=\{y_1,\dots,y_m\},$$ where the $y_i$ are just the original $x_i$ that don't belong in $U_{x_1}$. Notice that $m$ is strictly less than $n$ because in the list of the $y_i$, $x_1$ does not appear.

Since there are a finite number of $x_i$ to cover, we can repeat this process until one gets a finite subcover $U, U_{x_1}, U_{y_1},...$, for $X$. Therefore, $X$ is compact.

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    $\begingroup$ Notice that with the trivial topology any set equipped with it is compact, but I think my answer is cooler, for it showcases other topologies that also make the trick. $\endgroup$
    – Akerbeltz
    Commented May 18 at 11:07
  • $\begingroup$ Whoops, I answered before you changed your question... my answer showcases that in general you can't find a topological space that is compact independently of its topology... any finite set will do, but other than that... $\endgroup$
    – Akerbeltz
    Commented May 18 at 11:14
  • $\begingroup$ No problem 😊 sir Akerbeltz, I must read your answer for learning anything new $\endgroup$
    – Albert
    Commented May 18 at 11:17
  • $\begingroup$ For the first and third part, there are many characterizations of compactness (check the Wikipedia article on compact spaces for that), that don't depend necesarily on the notion of open covers. $\endgroup$
    – Akerbeltz
    Commented May 18 at 12:40

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