3
$\begingroup$

Definition: A Space $\mathbb{X}$ is said to be Compact if every Open Cover has a Finite Subcover.

My Question:
Suppose, Space $\mathbb{X}$ has an Open Cover $A$ whose finite subcover is $A_{1}$.
Now, $A_{1}$ becomes a open cover of space $\mathbb{X}$. Suppose that $A_{1}$ has a finite subcover $A_{2}$. Again, Repeat the process.
After fintely many steps of this process, we will arrive at a stage when, there will be no finite subcover for the previously made cover. (Say, the process ends at Nth step. So, there is no finite subcover at (N+1)th step. )
So, By Definition, I conclude that: There is No space $\mathbb{X}$ which is compact.

$\endgroup$
11
  • 4
    $\begingroup$ Well the cover $A_1$ is itself a finite cover, so $A_2$ could be taken equal to $A_1$. $\endgroup$ Mar 9, 2019 at 13:21
  • 2
    $\begingroup$ It is not forbidden for subcovers to coincide with the original cover. $\endgroup$
    – drhab
    Mar 9, 2019 at 13:23
  • 1
    $\begingroup$ It is just the same principle as used for sets. If $A$ is a set then it has subsets and $A$ itself (so coinciding with the original set) is one of those subsets. So also subsets of a set are not forbidden to coincide with the original set. $\endgroup$
    – drhab
    Mar 9, 2019 at 13:33
  • 1
    $\begingroup$ @drhab In the back of my mind, I had the notion that the subcover needs to be a proper subcover. The reason, for this is that I have only encountered examples in which, the author and instructors, construct an infinite cover and then, a finite subcover. So, I assumed that the containment is always proper. $\endgroup$
    – Kumar
    Mar 9, 2019 at 13:46
  • 1
    $\begingroup$ The reason that authors and instructors have used only examples where the original cover is infinite (and so a finite subcover is necessarily proper) is that the other possibility, where the original cover is finite, was considered too trivial to mention, since the original cover will itself serve as the desired finite subcover. $\endgroup$ Mar 9, 2019 at 14:46

1 Answer 1

2
$\begingroup$

This community wiki solution is intended to clear the question from the unanswered queue.

A cover $\mathfrak{C}$ of $X$ is a family $(C_\alpha)_{\alpha \in A}$ of subsets $C_\alpha \subset X$ such that $\bigcup_{\alpha \in A} C_\alpha = X$. A subcover $\mathfrak{C}'$ of $\mathfrak{C}$ is any subfamily $(C_\alpha)_{\alpha \in A'}$ with $A' \subset A$ such that $\bigcup_{\alpha \in A'} C_\alpha = X$. In particular, $A' = A$ is allowed. This is equivalent to $\mathfrak{C}'= \mathfrak{C}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .