Timeline for Proofs of the uncountability of the reals
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2010 at 10:43 | comment | added | Joel David Hamkins | Bill, the point is that to get the final contradiction, if $f:\mathbb{Q}\to\mathbb{R}$ is the order isomorphism, then the resulting disconnection of $\mathbb{R}$ is the cut determined by $f[\{q\mid q\lt\sqrt{2}\}]$, say, which has least upper bound $z$, but no real works, since the $n$-th real in the enumeration was placed into the range of $f$ at stage $n$. So this is diagonalization. | |
| Nov 23, 2010 at 6:11 | comment | added | Gerry Myerson | If $a\lt b$ are rational then $a+(b-a)/\sqrt2$ is an irrational between them. | |
| Nov 23, 2010 at 1:10 | comment | added | Bill Johnson | I did not understand what point Tim was trying to make in (1) since you can write down an explicit ordering of the rationals. It does address Tim's number (2). | |
| Nov 23, 2010 at 0:52 | comment | added | Joel David Hamkins | +1. I voted this up, but it doesn't satisfy Gowers criterion (1), since the proof of Cantor's theorem that every countable dense subset of $\mathbb{R}$ is order isomorphic to $\mathbb{Q}$ involves enumerating $\mathbb{Q}$. | |
| Nov 23, 2010 at 0:41 | comment | added | George Lowther | Why the downvotes? This looks like an interesting method. | |
| Nov 23, 2010 at 0:40 | comment | added | Michael Renardy | All you need to do is prove that between two rationals is an irrational. A variant of the well known proof that sqrt(2) is irrational should do the trick here. Just exploit the sparsity of squares among "large" integers. | |
| Nov 23, 2010 at 0:38 | comment | added | George Lowther | The rationals are clearly not connected. Partition then into the open sets of rationals less than $\sqrt{2}$ and the rationals greater than $\sqrt{2}$. | |
| Nov 23, 2010 at 0:37 | comment | added | Todd Trimble | Actually, I thought this held up and looked like an interesting possibility. You just need the existence of an irrational, don't you? | |
| Nov 23, 2010 at 0:03 | comment | added | Andrés E. Caicedo | @Bill : Doesn't the non-connectedness of ${\mathbb Q}$ rely on the uncountability of ${\mathbb R}$? How do you argue about it directly? | |
| Nov 23, 2010 at 0:00 | history | answered | Bill Johnson | CC BY-SA 2.5 |