4 ms·
Ok, that makes more sense. The part I still don't get is how we can pick a number x such that s2 < x < s3 without directly assuming that [0, 1] is uncountable.
by saiojd 6y ago
Ok, that makes more sense. The part I still don't get is how we can pick a number x such that s2 < x < s3 without directly assuming that [0, 1] is uncountable.
- blake1 6y agoEven the rational numbers have that property (the Archimedean property): for any distinct rationals s2 and s3, there exists a rational x such that s2 < x < s3. Proof: x = (s2 + s3) / 2. This is easily extended to show that there is an arbitrary number of such x.
- deleted 6y ago[deleted]