a space is connected under the ordered topology if and only if it is a linear continuum.


Let X be totally orderedPlanetmathPlanetmath by the strict total order <, and let it have the order topology.

Suppose X is not a linear continuum. Then either X is not bounded-complete, or the order on X is not a dense total order.

Suppose X is not bounded-complete. Let A be a nonempty subset of X that is bounded above by b, but has no least upper boundMathworldPlanetmath. Let U be the set of upper bounds of A. If xU, then x is not a least upper bound of A, so there is a zX such that z<x and z is an upper bound of A. Then the set {yXy>z} is open and contains x. Furthermore, all of its elements exceed z, so it is a subset of U. Thus, U is open. U contains b, so it is not empty. Let xXU. Then x is not an upper bound of A, so there is an aA such that x<a. The set {yXy<a} is open, and contains no upper bounds of A, so it is a subset of XU. Thus XU contains a neighborhood of each of its points, and is therefore open. Since U and XU are open, X is not connectedPlanetmathPlanetmath.

Suppose the ordering of X is not dense, so there are a and b in X with a<b so that there is no c in X with a<c<b (there is a gap between a and b). Let U={xXx<b} and let V={xXx>a}. Because there are no elements between a and b, UV=. By transitivity and trichotomy, UV=X. U and V are both open. aU and bV, so neither U nor V is empty. Thus, U and V separate X, so X is not connected.

Therefore, if X is connected, then X is a linear continuum.

Now suppose that X is disconnected and bounded-complete, and that U and V are (nonempty, open and closed) sets separating X. Suppose that aU, and suppose also that there is an element bV such that a<b (if there is none, swap the names of U and V, or reverse the ordering). Z={xVx>a} is open (it is the intersectionDlmfMathworldPlanetmath of two open sets), and contains b (so it is not empty). Z is bounded below by a, so it has a greatest lower boundMathworldPlanetmath z.

If zU, then, since U is open, there is an interval in U containing z, which must, to exclude b, be of the form {xx<k} or of the form {xj<x<k}, for some k and perhaps j. But then k would be a lower bound of Z, contradicting the fact that z is the infimumMathworldPlanetmath of Z.

If zV, then, since V is open, there is an interval in V containing z, which must, to exclude a, be of the form {xj<x} or of the form {xj<x<k} for some j and perhaps k. In either case, a<j<z and the set W={xj<x<z} is a subset of V. If xW, then xV and x>a, so xZ, contradicting the fact that z is the infimum of Z. Thus, there are no elements of X between j and z, so the order on X is not dense. This proves that if X is a linear continuum, then X is connected.

Title a space is connected under the ordered topologyMathworldPlanetmath if and only if it is a linear continuum.
Canonical name ASpaceIsConnectedUnderTheOrderedTopologyIfAndOnlyIfItIsALinearContinuum
Date of creation 2013-03-22 17:38:38
Last modified on 2013-03-22 17:38:38
Owner dfeuer (18434)
Last modified by dfeuer (18434)
Numerical id 7
Author dfeuer (18434)
Entry type Result
Classification msc 54B99
Classification msc 06F30
Related topic LinearContinuum
Related topic OrderTopology