How there are lots of points in $(0,1)\setminus I$?

Question

My doubts: there are only two point in $ (0,1) - [0,1]= \{0,1\}$. But user Ihf said that there are lots of points?

My question How there are lots of points in $(0,1)\setminus I$?

$[0,1]$ is a superset of $(0,1)$ , there are no points left if we remove the larger set. — Peter, Aug 15 '19 at 08:03
$I$ here is defined to be $[a,b]$ with $a$ and $b$ belonging to $(0,1)$; it is not the closed unit interval. — David Mitra, Aug 15 '19 at 08:03

score 4 · Accepted Answer · 2019-08-15T08:13:56.637

4

If it is unclear, $I$ is not the closed unit interval in this proof, but an interval defined earlier in the paragraph.

Let $x=\min\{a,b\}$ and $y=\max\{a,b\}$. Then $(0,1)\setminus I=(0,x)\cup(y,1)$, which is uncountable.

edited Aug 15 '19 at 08:13

answered Aug 15 '19 at 08:02

1 Answers1