two versions of the nested interval property - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:06:07Z http://mathoverflow.net/feeds/question/91135 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/91135/two-versions-of-the-nested-interval-property two versions of the nested interval property James Propp 2012-03-13T23:20:03Z 2012-03-13T23:20:03Z <p>There appear to be two different nested interval properties for the reals with the punchline "... then the intersection of the intervals is non-empty", and I'd like to know their respective histories (and the appropriate terminology to use for distinguishing between them). Here I am not concerned with the issue of "singleton-icity" of the intersection; only non-emptiness.</p> <p>In one version (which I believe is due to Cantor), the length of the intervals is assumed to shrink down to 0; in the other version, their lengths are not constrained.</p> <p>One sense in which the two properties are inequivalent is that there are non-archimedean ordered fields that satisfy the first but not the second. (Side question: Which is the preferred spelling, "archimedean" or "archimedian"? The number of Google hits for each spelling is about 800,000, so we can't use the Google-hits heuristic to decide, though Google seems to think that the former is preferred.) Specifically, consider the ordered field of formal Laurent series in \$x\$ over the reals, "ordered by size at zero" (so that effectively we're adjoining a formal infinitesimal \$x\$ to the reals). This ordered field satisfies the first nested interval property but not the second. (Consider the closed intervals \$[nx,1/n]\$. Their intersection is empty, but this doesn't contradict the first version of the nested interval property because the lengths of these intervals don't shrink to 0; indeed all the lengths are greater than \$x\$.)</p>