Subset of Real Numbers is Interval iff Connected

Theorem
Let the real number line $$\R$$ be considered as a topological space.

Let $$S$$ be a subspace of $$R$$.

Then $$S$$ is connected iff $$S$$ is an interval of $$\R$$.

That is, the only subspaces of $$\R$$ that are connected are intervals.

Proof

 * Suppose $$S \subseteq \R$$ is not an interval.

Then by Interval Defined by Betweenness, $$\exists x, y \in S$$ and $$z \in \R - S$$ such that $$x < z < y$$.

Consider the sets $$S \cap \left({-\infty \, . \, . \, z}\right)$$ and $$S \cap \left({z \, . \, . \, +\infty}\right)$$.

Then $$S \cap \left({-\infty \, . \, . \, z}\right)$$ and $$S \cap \left({z \, . \, . \, +\infty}\right)$$ are open by definition of the subspace topology on $$S$$.

Neither is empty because they contain $$x$$ and $$y$$ respectively.

They are disjoint, and their union is $$S$$, since $$z \notin S$$.

Therefore $$S \cap \left({-\infty \, . \, . \, z}\right) | S \cap \left({z \, . \, . \, +\infty}\right)$$ is a partition of $$S$$.

It follows by definition that $$S$$ is disconnected.


 * Now suppose $$S \subseteq \R$$ is an interval.

Suppose $$A | B$$ partitions $$S$$.

Let $$a \in A, b \in B$$, and suppose WLOG that $$a < b$$.

Since $$a, b \in S$$ and $$S$$ is an interval, we have that $$\left[{a \,. \, . \, b}\right] \subseteq S$$.

Let $$A' = A \cap \left[{a \,. \, . \, b}\right]$$ and $$B' = B \cap \left[{a \,. \, . \, b}\right]$$.

By the definition of a partition, both $$A$$ and $$B$$ are closed in $$S$$.

Hence by Closed Sets in Topological Subspace, $$A'$$ and $$B'$$ are also closed in $$\left[{a \,. \, . \, b}\right]$$.

From its corollary, $$A'$$ and $$B'$$ are closed in $$\R$$.

Now, since $$B' \ne \varnothing$$, and $$B$$ is bounded below (by, for example, $$a$$), by the Continuum Property $$b' = \inf \left({B'}\right)$$ and $$b' \ge a$$.

Since $$B'$$ is closed in $$\R$$, by Closure of Real Interval $$b' \in B'$$.

Since $$a \in A'$$ and $$A \cup B = \varnothing$$, it follows that $$b' > a$$.

Now let $$A'' = A' \cap \left[{a \,. \, . \, b'}\right]$$.

Using the same argument as for $$B'$$, we have that $$a = \sup \left({A}\right)$$ exists, that $$a \in A$$ and $$a'' < b'$$.

Now $$\left({a \, . \, . \, b'}\right) \cap A' = \varnothing$$ or else $$a$$ would not be an upper bound for $$A''$$.

Similarly, $$\left({a' \, . \, . \, b}\right) \cap B' = \varnothing$$ or else $$b$$ would not be a lower bound for $$B''$$.

But $$A' \cup B' = \left[{a \,. \, . \, b}\right]$$ and $$\left({a'' \, . \, . \, b'}\right) \subseteq \left[{a \,. \, . \, b}\right]$$.

Thus we have deduced a contradiction, and hence there can be no such partition $$A | B$$ on the interval $$S$$.