linear continuum

Let $X$ be a totally-ordered set under an order having at least two distinct points. Then $X$ is said to be a linear continuum if the following two conditions are satisfied:

1. The order relation is a dense total order (i.e., for every $x,y\in X$ with there exists $z\in X$ such that ).

2. Every non-empty subset of $X$ that is bounded above has a least upper bound (i.e., $X$ has the least upper bound property).

Some examples of ordered sets that are linear continua include $ℝ$, the set $[0,1]\times [0,1]$ in the dictionary order, and the so-called long line $\mathrm{\Omega }\times [0,1)$ in the dictionary topology. (The third example is a special case of a general result on well-ordered sets and linear continua.)

Proposition.

If $X$ is a well-ordered set, then the set $X\mathrm{\times }\mathrm{[}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$ is a linear continua in the dictionary order topology.

Linear continua are of special interest when they are made into topological spaces under the order topology, and the following two establish some useful properties of such spaces:

Proposition.

If $X$ is a linear continuum in the order topology, then $X$ is http://planetmath.org/node/941connected and so are intervals in $X$.

As a corollary of the preceding , we obtain the result that $ℝ$ is in its usual topology, as are the intervals $[a,b]$ and $(a,b)$, where .

Proposition.

If $X$ is a linear continuum in the order topology, then every closed interval in $X$ is compact.

Proof.

This is essentially a slightly generalized version of the Heine-Borel Theorem for $ℝ$, and the proof is almost identical.∎