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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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    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.2011-03-15
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    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).2011-03-15
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    The question is correct as written because your proposed change in point 2 does not even define a topology!2011-03-15
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    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.2011-03-15
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    What Henno means is that you should write $a\in [0,\infty)$ in your first bullet.2011-03-16

1 Answers 1

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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?2011-03-15
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    @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.2011-03-15
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    Actually every set of open subsets *containing* $[0,\infty)$ is also a covering.2011-03-15
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    @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.2011-03-15
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    I referred more to the statement "The only possible cover of $[0,\infty)$ in this topology is itself"...2011-03-15
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    @Rasmus, if cover is the union of the sets in the covering, then yes, this is wrong.2011-03-15
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    Oh, I thought you used "cover" as a synonym for "covering" but apparently you don't. How do you define a cover in this context? A superset?2011-03-15
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    It's more correct to say (if we assume that we have here a description of a base, not a topology) that, if we have a cover for $[0,\infty)$, the only set that could cover the point $0$ is $[0,\infty)$ itself, so this set **must** be present in any open cover of the space, and every cover of the space has a subcover of size 1.2011-03-15
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    @Rasmus, I fixed the wording, you were right, the mathematical statement was not rigorous.2011-03-16
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    @mpiktas: One last nitpicking: There are lots of coverings containing the set $[0,\infty)$, so it does not really make sense to say "The only possible cover of [0,∞) in this topology is **the** cover which has [0,∞) in it". I will stop razzling now. =)2011-03-16
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    @Rasmus, thanks. In my native language there are no articles, so although I know rules about them, I tend to forget how to use them :)2011-03-16