# Show that for each $a\in [0, \infty)$, the subspace $[a, \infty)$ is also compact

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: en.wikipedia.org/wiki/… –  Seirios Feb 24 '13 at 18:27

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 $[0,\infty)$.