Question 7.2.16 from S. Morris's Topology without Tears

I'm just trying to get a grip on how these concepts work. So I've written this up for checking.

Prove that $[0,1]$ with the euclidean topology is supercompact:

That is, we need to find a subbasis such that every subbasis cover of $[0,1]$ has a two-element subcover.

We will use the set $\{(-\infty, a): a\in \mathbb{R}\} \bigcup \{(b, \infty):b\in \mathbb{R}\}$ as a subbasis for the euclidean topology on $\mathbb{R}$.

Letting $i \in I$ be elements of an index set, any subbasis cover $\{O_{i}\}$ of $[0,1]$ contains the point $1$ and the point $0$.

Any cover that consists exclusively of $(-\infty, a)$ sets or $(b, \infty)$ sets clearly has one set that contains the entirety of $[0,1]$. Take that one set and union it with any other set in the cover, and those cases are done.

Assume that a given subbasis cover has a mix of $(-\infty, a)$ sets and $(b, \infty)$ sets. WLOG, consider how to include 1.

Considering all subsets $O_i = (-\infty, a_i)$, let $c = \sup_{i\in I}a_i$

Then, either $c \leq 1$ or $c > 1$.

If $c >1$, then $[0,1]\subset O_i = (-\infty, a_i)$ for some $a_i$ such that $1 < a_i < c$.

Take the union of that one set and any other in the cover, and we're done.

If $c \leq 1$, then $c\in O_j = (b_j,\infty)$ for some $b_j$ such that $b_j < c$

Then, there must be some $O_k = (-\infty, a_k)$ such that $b_j < a_k < c$.

So $[0,1] \subset O_j \cup O_k$, and we're done.

  • $\begingroup$ Looks good to me. Minor technical mistake is that if $c>1$ there is some $a_{i}$ such that $1<a_{i}\leq c$. Suppose your cover is just the set $\{(-\infty,a_{0}):a_{0}=2\}$ then $c>1$ but there is no $a_{i}<c$. $\endgroup$ – Floris Claassens Feb 1 at 21:28
  • $\begingroup$ Minor objection: It should say that every cover of [0,1] by a subset of the sub-base has a sub-cover with $at$ $most$ two members. $\endgroup$ – DanielWainfleet Feb 2 at 10:38

Cleaner and fewer cases:

Take the subbase $\mathcal{S}=\{(a,1], a \in [0,1]\} \cup \{[0,b): b \in [0,1]\}$, which is the standard subbase for an ordered space having a maximum $1$ and a minimum $0$ (they're the intersections of $(a,+\infty)$ with $[0,1]$ with $a\in [0,1]$ (if $a$ is not in that range this intersection is either $\emptyset$ or $[0,1]$ hence useless. Likewise for $(-\infty,a) \cap [0,1]$ etc.)).

Let $\{(a_i, 1]: i \in I\} \cup \{[0,b_j): j \in J\}$ be an arbitrary cover by subbasic elements (so all $a_i, b_j \in [0,1]$). $J \neq \emptyset$ as we need to cover $0$ and $I \neq \emptyset$ as we need to cover $1$.

Let $b=\sup\{b_j: j \in J\}$ which is well-defined. Note that $b$ cannot be covered by a set of the form $[0,b_j), j \in J$ or else we would have $b_j > b$ contradicting that $b$ is an upperbound for $\{b_j : j \in J\}$.

So $b$ must be covered by some set $(a_{i_0}, 1]$ with $i_0 \in I$. As $a_{i_0} < b$, $a_{i_0}$ cannot be an upperbound for $\{b_j: j \in J\}$ (because the sup is the smallest upperbound) so there is some $j_0 \in J$ with $a_{i_0}< b_{j_0}$.

It follows that $[0,1] = [0,b_{j_0}) \cup (a_{i_0}, 1]$ and we have a two-element subcover of our subbasic cover.


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