I am struggling to draw this point home:(adsbygoogle = window.adsbygoogle || []).push({});

To prove that R has LUB property, we used the following reasoning:

First we defined R to be set of cuts (having certain properties) where each cut corresponds to a real number and then we proved any subset A of R has LUB (least upper bound) property.

If I use similar approach to define Q* to be the set of all rational cuts.

Then let A be the subset of Q*. Let Υ be the union of all cuts in A. Then this Υ is a cut in Q* and is supremum of A.

(Note, to define Q* above, we didn't need to introduce R, as we are playing with only rational cuts so far.)

Is above reasoning false? and if false, how to prove it.

If the reasoning is true, then the coclusion that Q* has LUB leads contrary to what we have read in the books (i.e. Q does not have LUB). Or the conclusion with the help of cuts that R has LUB has problem?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Q* (the set of rational cuts) has least upper bound property or not?

Loading...

Similar Threads - rational cuts least | Date |
---|---|

I Importance of Archimedean Property and Density of Rationals | Feb 8, 2018 |

I Can a tetrahedron have all dihedral angles rational? | Aug 1, 2017 |

Proof of rational density using Dedekind cuts | Aug 7, 2014 |

Defining Cuts in the Rationals | Dec 2, 2013 |

A proof regarding Rational numbers | Jun 25, 2013 |

**Physics Forums - The Fusion of Science and Community**