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Joseph Van Name
Joseph Van Name
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The order topology provides many examples of compact connected ordered sets which are totally path disconnected. A totally ordered set is said to be complete if it complete as a lattice. A totally ordered set $X$ is said to be dense if whenever $x<y$ then there is some $z$ with $x<z<y$. One can get a totally ordered set that is both complete and dense by taking the Dedekind-MacNielle completion of a dense totally ordered set. A totally ordered set is compact in the order topology if and only if it is complete, and a totally ordered set is connected in the order topology if and only if it is dense and every bounded nonempty subset has a least upper bound. Therefore the dense complete total orders are precisely the compact connected total orders. On the other hand, except for the trivial cases, a compact totally ordered set need not have any paths. To be more precise, if $X$ is a compact connected totally ordered set, then there is a path from a point $x$ to a point $y$ with $y>x$ if and only if the interval $[x,y]$ is order isomorphic to the unit interval in the real numbers.

The order topology provides many examples of compact connected ordered sets which are totally path disconnected. A totally ordered set is said to be complete if it complete as a lattice. A totally ordered set $X$ is said to be dense if whenever $x<y$ then there is some $z$ with $x<z<y$. One can get a totally ordered set that is both complete and dense by taking the Dedekind-MacNielle completion of a dense totally ordered set. A totally ordered set is compact in the order topology if and only if it is complete, and a totally ordered set is connected in the order topology if and only if every bounded nonempty subset has a least upper bound. Therefore the dense complete total orders are precisely the compact connected total orders. On the other hand, except for the trivial cases, a compact totally ordered set need not have any paths. To be more precise, if $X$ is a compact connected totally ordered set, then there is a path from a point $x$ to a point $y$ with $y>x$ if and only if the interval $[x,y]$ is order isomorphic to the unit interval in the real numbers.

The order topology provides many examples of compact connected ordered sets which are totally path disconnected. A totally ordered set is said to be complete if it complete as a lattice. A totally ordered set $X$ is said to be dense if whenever $x<y$ then there is some $z$ with $x<z<y$. One can get a totally ordered set that is both complete and dense by taking the Dedekind-MacNielle completion of a dense totally ordered set. A totally ordered set is compact in the order topology if and only if it is complete, and a totally ordered set is connected in the order topology if and only if it is dense and every bounded nonempty subset has a least upper bound. Therefore the dense complete total orders are precisely the compact connected total orders. On the other hand, except for the trivial cases, a compact totally ordered set need not have any paths. To be more precise, if $X$ is a compact connected totally ordered set, then there is a path from a point $x$ to a point $y$ with $y>x$ if and only if the interval $[x,y]$ is order isomorphic to the unit interval in the real numbers.

Source Link
Joseph Van Name
Joseph Van Name
  • 1
  • 4
  • 75
  • 118

The order topology provides many examples of compact connected ordered sets which are totally path disconnected. A totally ordered set is said to be complete if it complete as a lattice. A totally ordered set $X$ is said to be dense if whenever $x<y$ then there is some $z$ with $x<z<y$. One can get a totally ordered set that is both complete and dense by taking the Dedekind-MacNielle completion of a dense totally ordered set. A totally ordered set is compact in the order topology if and only if it is complete, and a totally ordered set is connected in the order topology if and only if every bounded nonempty subset has a least upper bound. Therefore the dense complete total orders are precisely the compact connected total orders. On the other hand, except for the trivial cases, a compact totally ordered set need not have any paths. To be more precise, if $X$ is a compact connected totally ordered set, then there is a path from a point $x$ to a point $y$ with $y>x$ if and only if the interval $[x,y]$ is order isomorphic to the unit interval in the real numbers.