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Consider the real line $\mathbb{R}$ with the half-open interval topology, with a base of open sets given by $[a, b)$ with $a<b$ in $\mathbb{R}$; show that the closed interval $[a, b]$ is not compact.

I saw a similar question asked but no one answered the part that's confusing me. We want to find an open cover with no finite subcover, but I don't even understand how we can have an open cover of $[a,b]$ if our open sets are of the form $[a,b)$.

An open cover is a union of half-open intervals, but I don't see how a union of intervals open on the right could ever give an interval closed on the right. For example, $\bigcup_{n \in \mathbb N}[a,b-\frac1n)$ is never going to include $b$. Can we have $[b,b)=\{b\}$?

I'm not really asking for a solution to the above question, but rather for someone to explain to me what such an open cover will look like or what I'm misunderstanding.

(Also we have $[a,b]$ is compact in the standard topology, right? Again, I don't understand what this would look like. Presumably I've fundamentally misunderstood something.)

Thanks.

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    $\begingroup$ $[a,b)$ generate the open sets in that topology, but they are not the only open sets. $\endgroup$ – user107952 Sep 15 '17 at 16:17
  • $\begingroup$ @user296602 but it is not an open cover of $[a,b]$, right? $\endgroup$ – Aka_aka_aka_ak Sep 15 '17 at 16:18
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    $\begingroup$ You're right, it isn't an open cover. But this is easy to fix; just take $[b, b + 1)$ as an additional element of your cover. $\endgroup$ – T. Bongers Sep 15 '17 at 16:18
  • $\begingroup$ @user296602 but does an open cover not have to give us $[a,b]$ exactly? $\endgroup$ – Aka_aka_aka_ak Sep 15 '17 at 16:19
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    $\begingroup$ @Aka_aka_aka_ak I think you need to check the definition of open cover again. There is no requirement that the union of the cover is equal to the original set, just that it contains the original set. $\endgroup$ – T. Bongers Sep 15 '17 at 17:20
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You can use relatively open subsets of $[a,b]$ (so that the cover consists of subsets of $[a,b]$ that are open in the subspace topology and whose union is exactly $[a,b]$), in which case note that $\{b\}$ is open in $[a,b]$ because $[a,b] \cap [b, b+1) = \{b\}$ and $[b, b+1)$ is open in the total space.

Or you can use open sets of the whole space and their union contains $[a,b]$. (and use $[b,b+1)$ as one of the sets in the cover).

Both ways of defining compactness for a subspace (intrinsic and extrinsic) are equivalent.

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