# Connectedness and the real line

It is fundamental to topology that $\mathbb{R}$ is a connected topological space. However, all the topology books that I have ever looked in give the same proof. (the proof I am thinking of can be seen in Munkres's topology or Lee's Introduction to topological manifolds)

This seems strange to me, because for other fundamental results such as the Compactness of $[0,1]$, I can think of several proofs.

Does anyone know any different proofs of the connectedness of $\mathbb{R}$?

• I guess that all proofs will mainly rely on the least upper bound property, so that they shall be the same proof possibly under different disguise. May 31, 2010 at 8:05
• The least upper bound property is not the only property which characterises the real line. Am I correct in saying that the least upper bound property if equivalent to the completeness of the reals? Could there possibly be a proof of connectedness which relies on the convergence of Cauchy sequences? May 31, 2010 at 8:21
• You might as well ask about the intermediate value theorem, which is formally equivalent. May 31, 2010 at 9:11

If you've already developed basic facts about compactness you can prove it this way:

Let $[0,1] = A \cup B$ with $A$ and $B$ closed and disjoint. Then since $A \times B$ is compact and the distance function is continuous, there is a pair $(a, b) \in A \times B$ at minimum distance. If that distance is zero, $A$ and $B$ intersect. If not, you get a contradiction by taking any point in the interval from $a$ to $b$: it can't be in either $A$ or $B$ because its distance from $b$ or $a$ is smaller than the minimum.

That shows a compact interval in $\mathbb{R}$ is connected. If $\mathbb{R} = A \cup B$ with $A$ and $B$ closed and disjoint, then for any closed interval $I$ with one endpoint in $A$ and one in $B$, $I = (A \cap I) \cup (B \cap I)$ is disconnection of $I$. Alternatively, you could write $\mathbb{R}$; as a union of closed intervals with a common point.

• You can also prove compactness of [0,1] from its connectedness: For a fixed open cover of [0,1] consider the set A of points t such that [0,t] is covered by finitely many sets from the cover. Then A is easily seen to be both open (if t is in A, then the finite cover of [0,t] actually covers a little bit beyond t) and closed (if t is not in A and U is a set from the open cover containing t and thus also, say, (t-x, t+x), then no point of (t-x, t+x) can be in A either). May 31, 2010 at 16:26

If you want to prove that 'complete plus densely ordered' implies connected you are almost forced to use the 'standard' proof. For the real line you could also use the bisection method: if $I$ is convex and the union of two closed sets $A$ and $B$ take $a\in A$ and $b\in B$, with $a < b$, say. Now create two sequences $(a_n)_n$ (increasing) and $(b_n)_n$ (decreasing) with $a_n\in A$, $b_n\in B$ and $b_n-a_n=(b-a)2^{-n}$; then the common limit of these sequences belongs to $I\cap A\cap B$.

• Indeed, if you use the order topology, we have that connected order topology--> "almost compact" (we only need add a minimum and a maximum element, if they do not already exist, to have a compact space). So the proof then has to use similar elements to compactness proofs for such spaces (i.e. using completeness properties). Apr 25, 2011 at 12:56

It's a notoriously thorny matter to decide whether two proofs of a given theorem are "really different". But...a proof of the connectedness of the real line using real induction is given in Theorem 9 of this note of mine. This proof (to me) feels moderately different from the usual LUB proof, and I think I like it a little better.

1. Actually what is proved is that any closed, bounded interval $$[a,b]$$ is connected. But you can get from here to the connectedness of $$\mathbb{R}$$ with no trouble at all: e.g. the union of a chain of connected subspaces is connected.
If $\mathbb{R} = A \uplus B$, for $A,B$ nonempty open subsets, then each of them is a countable union of disjoint open intervals (to prove this, one has to use the completeness of $\mathbb{R}$). Let $(a_1,a_2)$ be an interval which appears in this decomposition of $A$. Then $a_2$ can't be in $A$, but also can't be in $B$ since $B$ is open. Contradiction.