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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?

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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
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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
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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
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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
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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
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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!)

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