Let $X$ be a totally-ordered set under an order relation $<$ 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 $x<y$ there exists $z\in X$ such that $x<z<y$ ).
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 $\mathbb{R}$ , the set $[0,1]\times[0,1]$ in the dictionary order, and the so-called long line $\Omega\times[0,1)$ in the dictionary topology. (The third example is a special case of a general result on well-ordered sets and linear continua.)

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:

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

Proof. This is essentially a slightly generalized version of the Heine-Borel Theorem for $\mathbb{R}$ , and the proof is almost identical.