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Total, transitive, not anti-symmetric orderings

  1. Apr 8, 2013 #1
    I have a total order on a multi-dimensional, real-numbered vector space [itex]X[/itex]. This means that for any vectors [itex]x,y\in{}X[/itex] either [itex]xRy[/itex] or [itex]yRx[/itex]. Total orders are usually transitive, total, and antisymmetric (if [itex]xRy[/itex] and [itex]yRx[/itex] then [itex]x=y[/itex]), but this one is not necessarily antisymmetric, it is only transitive and total. Which further conditions do I need to prove that it is antisymmetric, ie. [itex]xRy[/itex] and [itex]yRx[/itex] imply [itex]x=y[/itex] just based on the fact that the ordering is total and transitive? One example of a transitive, total ordering which is not antisymmetric is one that uses only one coordinate of the vector and the usual less than ([itex]\geq[/itex]) or greater than ([itex]\leq[/itex]) relations: [itex]xRy[/itex] if and only if [itex]x_{i}>y_{i}[/itex] for some fixed [itex]i[/itex], and we ignore all other coordinates, all [itex]x_{j}[/itex] for [itex]j\neq{}i[/itex]. I have a hunch that ALL transitive, total orderings that are not antisymmetric are of this type. How could I formalize and prove this claim?

    The claim, in other words, is that the only transitive, total, not-antisymmetric orderings [itex]R[/itex] of a [itex]n[/itex]-dimensional, real-numbered vector space are those which only consider one dimension and disregard the others. Let [itex]R[/itex] be a such an ordering. Then there exists a transitive, total, antisymmetric ordering [itex]R'[/itex] of the real numbers such that for a fixed [itex]k\in\{1,\ldots,n\}[/itex] [itex]xRy[/itex] if and only if [itex]x_{k}R'y_{k}[/itex]. How would I prove this (and is it correct)?
    Last edited: Apr 8, 2013
  2. jcsd
  3. Apr 9, 2013 #2
    Taking myself out of the NR category. Don't bother answering this one. The claim is patently false and needs to be re-articulated. Sorry.
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