# Total, transitive, not anti-symmetric orderings

1. Apr 8, 2013

### stlukits

I have a total order on a multi-dimensional, real-numbered vector space $X$. This means that for any vectors $x,y\in{}X$ either $xRy$ or $yRx$. Total orders are usually transitive, total, and antisymmetric (if $xRy$ and $yRx$ then $x=y$), 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. $xRy$ and $yRx$ imply $x=y$ 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 ($\geq$) or greater than ($\leq$) relations: $xRy$ if and only if $x_{i}>y_{i}$ for some fixed $i$, and we ignore all other coordinates, all $x_{j}$ for $j\neq{}i$. 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 $R$ of a $n$-dimensional, real-numbered vector space are those which only consider one dimension and disregard the others. Let $R$ be a such an ordering. Then there exists a transitive, total, antisymmetric ordering $R'$ of the real numbers such that for a fixed $k\in\{1,\ldots,n\}$ $xRy$ if and only if $x_{k}R'y_{k}$. How would I prove this (and is it correct)?

Last edited: Apr 8, 2013
2. Apr 9, 2013

### stlukits

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.