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i) Show that the intervals $(a, \infty)$, $a \in (0, \infty)$ together with $\emptyset$ and $[0, \infty)$ form a topology on $[0, \infty)$.

ii) Show that in this topology $[0, \infty)$ is compact. Show that for each $a \in [0, \infty)$, the subspace $[a, \infty)$ is also compact.

iii) Notice that intersection $\bigcap_{k=1}^\infty [k,\infty)= \emptyset$,why doesn’t this contradict the statement that “the intersection of a decreasing sequence of nonempty compact sets is nonempty.”

this is not homework, it is from a book with no solutions.

i is easy from definition and checking union and intersection.

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What topology are you working with? If you are working in the Euclidean metric space, then $[0, \infty)$ is not compact since it is not bounded. – Dávid Tóth Feb 24 '13 at 17:22
He is defining the topology. – copper.hat Feb 24 '13 at 17:23
Please use actual words, not "defn, soln" and so on. – Asaf Karagila Feb 24 '13 at 17:37
Your topology is known as the right order topology:… – Seirios Feb 24 '13 at 18:27
up vote 2 down vote accepted

i) is straightforward.

For ii), note that if $\cal O$ is an open cover of $[a,\infty)$, then at least one element of $\cal O$ must contain the point $a$. If $a=0$, then the only set satisfying this is $[0,\infty)$, and if $a>0$, then there must exists a set of the form $[b,\infty) \in {\cal O}$ where $b < a$ (otherwise $\cup_{U \in {\cal O}} U = (a, \infty)$). Hence there exists a finite cover.

For iii), the statement should be in terms of closed, not compact sets. In this particular topology, the only non-empty compact sets that is closed is $[0,\infty)$.

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Just to add a little bit. This kind of problem (I'm refering to iii)) happens when the space is not Hausdorff. – Tomás Feb 24 '13 at 17:39
@Tomás: Good point. Without textbook at hand I forget the various separation axioms... – copper.hat Feb 24 '13 at 18:03

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