Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The definition of compact is every open cover has a finite subcover; I want to prove $S^n$ is compact directly, i.e. choose any infinite open cover, $\{a_i\}$, how to find a finite subcover?

What spaces can we get if we only require that at least one infinite open cover has a finite subcover,by proper open subsets., is this sufficient for compactness?

Also, how to show (closed and bounded) $\implies$ compact?

How to prove that every open cover of [0,1] has a finite subcover?

share|improve this question
The latter half of your question is part of the famous Heine-Borel theorem, which you should be able to apply to $S^n$ to show it is compact. –  Sid Raval Jan 29 '12 at 19:27
Every space $X$ with an infinite open cover $\mathscr{U}$ has an infinite open cover with a finite subcover, namely, $\mathscr{U}\cup\{X\}$. Perhaps you should revise the question to say at least one infinite cover by proper open subsets. –  Brian M. Scott Jan 29 '12 at 19:32
If I may ask, why is it that you want to show the compactness of $S^n$ from the open cover definition? There are at least two easier ways to do this that I can think of - first of all, the complement of $S^n$ in $\mathbb{R}^{n+1}$ is open, and $S^n$ is bounded, so as Sid says, we can use our criterion for compactness in $\mathbb{R}^N$. Alternatively, $S^n$ is the quotient of the closed disk $D^n$ by identifying points on the boundary, and the image of a compact set is compact. –  NKS Jan 29 '12 at 19:38
Probably the easiest proof from scratch is to prove that every open cover of $[0,1]$ has a finite subcover, then prove that the Cartesian product of two compact spaces is compact, from which it follows by induction that $[0,1]^n$ is compact for every $n$, and finally show that compactness is preserved by closed subsets. –  Brian M. Scott Jan 29 '12 at 19:46
By the way, note that Bounded + Closed $\Rightarrow$ Compact is not true in general, but e.g. in $\mathbb{R}^{n}$ it is true. Also, when dealing with metric spaces it can sometimes be useful to work with sequential compactness instead of compactness, as they are equivalent. Proving the Heine-Borel theorem also turns out quite straight forward once working with sequences. –  Thomas E. Jan 29 '12 at 20:48
show 2 more comments

1 Answer 1

up vote 4 down vote accepted

For the first question, proving $S^n$ is compact directly from the definition could be very long and messy; I would suggest you use Heine Borel (your third question): consider the Euclidean norm $$ | \ \ \ |: \mathbb{R}^n \to [0, +\infty)$$ this is a continuous function, hence the preimage of a closed set of $[0, +\infty)$ is closed in $\mathbb{R}^n$. Since points are closed in $\mathbb{R}^n$ $$S^n = | \ \ \ |^{-1} (1) $$ is closed. Since the $S^n$ is bounded, it is compact.

For the second question the answer is all spaces with infinite open sets. Take a general topological space $(X, \tau)$, and simply take the cover of all open sets, $\tau$: a finite subcover is trivially given by $\{ X \}$. Also, asking that $X$ not be part of the cover still leaves you quite far from compactness: take $(0, 1)$, and consider an open cover $\mathscr{U}$ such that $\left(0, \frac23 \right), \ \left(\frac13, 1 \right) \in \mathscr{U}$. (If you want $\mathscr{U}$ to be infinite, just throw in infinite other open sets). $\{\left(0, \frac23 \right), \ \left(\frac13, 1 \right) \} \subseteq \mathscr{U}$ is a finite subcover of $\mathscr{U}$, and clearly an open interval is far from being compact.

For the third question I suggest you read here , or here. Also note that the " $\Leftarrow $" is also true.

For your last question, I will include a sketch of a proof I am familiar with (the whole proof is rather long, and you can find it, and probably others, in many General Topology textbooks):
Let $I = [0, 1]$, let $\mathscr{U}$ be an open cover of $I$, and let $$X = \{x \in I \ | \ [0, x] \ \text{is covered by finitely many} \ U \ \in \mathscr{U} \}$$ if you prove that $I = X$ you are done. You can do this by proving that $ \emptyset \neq X$ is an interval, and is simulteneously open and closed in $I$. Since the only interval $\subseteq I$ with this property is $I$ itself, you are done.

(sorry, I'm slow with latex and by the time I finished, most of this was already covered in the comments; I hope it helps anyway!)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.