# Prove: [0,1] is supercompact

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.

• 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 $a_{i}<c$. – Floris Claassens Feb 1 at 21:28
• 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. – DanielWainfleet Feb 2 at 10:38

## 1 Answer

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.