Compact set Sorgenfrey's topology.

Question

Be $\mathbb{R}$ with Sorgenfrey's topology where the basis elements are of the form $[a,b)$. If $A\subseteq\mathbb R$ is compact, then $A$ is countable.

I tried to find a set $A$ not countable which has a cover that can not be reduced to a finite one, but I could not find it. Is it possible to do it another way?

Answer 1

You might look at this article in Wikipedia which has a proof.

Answer 2

HINT: This answer shows that if $A$ is uncountable, there is an uncountable $A_0\subseteq A$ such that if $V$ is an open interval in $\Bbb R$, then either $V\cap A_0=\varnothing$, or $V\cap A_0$ is uncountable. Fix $a\in A_0$, and look at the open cover

$$\{(\leftarrow,x):x<a\}\cup\{[a,\to)\}$$

of $A$.

Answer 3

Hints:

1. Any uncountable set of real numbers contains a strictly increasing infinite sequence.

2. Show that if a subspace $X$ of the Sorgenfrey line contains a strictly increasing infinite sequence, then $X$ has an open cover with no finite subcover.

Answer 4

A simple example: $[0,1]$ is not compact in the Sorgenfrey topology because the open cover $U_n = [0, 1-\frac{1}{n}), n \ge 2$ together with $[1,2)$ is an open cover of $[0,1]$ that has no finite subcover: suppose it had: let $k$ be the largest of the $n$ used in the first type of open set; then $1-\frac{1}{k+1}$ is not covered by any of the sets in the cover (as all right hand points of the $U_n$ in the cover are smaller by construction, and $[1,2)$ is only there to cover $1$).