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I am trying to understand if there is an error in this question, or if the answer is trivial.

Define a topology on $[0,\infty)$ with open sets

  • $(a,\infty), a \in (0,\infty)$

  • $[0,\infty)$ and $\emptyset$.

Show $[0,\infty)$ is compact in this topology.

Since we can only build an open cover out of open sets, surely we can only take $\cup_{n > 0} (n,\infty)$ which does not cover $[0,\infty)$.

Can we either:

  1. Trivially take $[0,\infty)$ as an open finite cover, meaning whenever we take the whole space of any topology it is automatically compact.

  2. The question should read $[a,\infty), a \in [0,\infty)$ are open in this topology

Any help would be greatly appreciated.

Many thanks,

Ash

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1  
Compactness does not mean that you cannot find a finite open covering for your space. That is always true. Compactness means that any open covering has a finite subcovering. –  Rasmus Mar 15 '11 at 8:16
1  
the space is compact (for trivial reasons, i.e. there being very few open covers of the space). the question is probably correct as written, forcing you to understand the definition of compactness (every open cover has a finite subcover). –  yoyo Mar 15 '11 at 15:20
    
The question is correct as written because your proposed change in point 2 does not even define a topology! –  Florian Mar 15 '11 at 16:29
    
As you describe it, the so-called topology is not a topology: $(0,\infty)$ is not one of the sets under your first bullet (as 0 is not greater than 0), nor under the second, but as you say, it is a union of all sets of the form $(a,\infty)$ for $a>0$, so the collection you describe is not closed under unions. It does describe a base for the topology. –  Henno Brandsma Mar 15 '11 at 17:19
    
What Henno means is that you should write $a\in [0,\infty)$ in your first bullet. –  Rasmus Mar 16 '11 at 15:58

1 Answer 1

up vote 5 down vote accepted

The first variant is correct one. The only possible cover of $[0,\infty)$ in this topology is a cover which has $[0,\infty)$ in it. This follows since $\cup_{a}(a,\infty)=(0,\infty)$, so it is not possible to construct cover of $[0,\infty)$ using only sets $(a,\infty)$ with $a>0$.

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Does this then imply that the whole space on any topology is compact, for example $\mathbb{R}$ on $\mathbb{R}$ is compact? –  Ash Mar 15 '11 at 8:08
3  
@Ash, no. Take $\mathbb{R}$ and usual open sets. Then $\mathbb{R}=\cup_n (-n,n)$ and you cannot take finite subcover from this covering. The catch in this example is that the whole space is "isolated", i.e. you cannot construct a cover of it without using itself. –  mpiktas Mar 15 '11 at 8:11
    
Actually every set of open subsets containing $[0,\infty)$ is also a covering. –  Rasmus Mar 15 '11 at 8:18
    
@Rasmus, I am not saying that it is not, only that it is not possible to construct covering without $[0,\infty)$. I revised my answer to stress that. –  mpiktas Mar 15 '11 at 8:22
    
I referred more to the statement "The only possible cover of $[0,\infty)$ in this topology is itself"... –  Rasmus Mar 15 '11 at 12:09

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