If we remove the diagonal from $X\times X$, is it necessarily disconnected?

Question

If $X$ is a compact, connected Hausdorff space, we know that the diagonal $\Delta_X=\{(x,x)\in X\times X\}$ is closed in $X\times X$ by Hausdorffness. But is $X\times X\setminus\Delta_X$ disconnected in general? I know that, if $X$ has a total order $<$ such that the induced order topology is contained in the original topology (i.e. it is coarser), then we can write $$X\times X\setminus\Delta_X=\{(x,y)\in X\times X:x<y\}\sqcup\{(x,y)\in X\times X:x>y\}$$ Also, if $\Delta_X$ is a zero-set, then we also have the disconnectedness of $X\times X\setminus\Delta_X$ since, if $\Delta_X=f^{-1}[\{0\}]$, then we have the disjoint open sets $f^{-1}[(-\infty,0)],f^{-1}[(0,+\infty)]$. But we don't have that $\Delta_X$ is a zero-set in general as $X$ may not be first countable and $\Delta_X$ being $G_\delta$ implies $X$ first countable for compact Hausdorff spaces.

Answer 1

For a counterexample, take $X=\mathbb R^2$.

The diagonal in $X \times X = \mathbb R^4$ consists of all $(a,b,c,d) \in \mathbb R^4$ such that $a=c$ and $b=d$, which defines a 2-dimensional linear subspace of $\mathbb R^4$.

With a bit of linear algebra one can apply an invertible linear transformation to convert this into the subspace defined by the equations $a=0$ and $b=0$, the complement of which is homeomorphic to $(\mathbb R^2 - \{(0,0\}) \times \mathbb R^2$, which is path connected.

Answer 2

Let $X$ be $\Bbb{R}$ with the cofinite topology (any infinite set would do: I'm just using $\Bbb{R}$ so I can use geometrical language). The closed subsets of $X$ are either empty, finite or equal to the whole real line $\Bbb{R}$. For each point $(x, y)$ in a closed subset $C$ of $X \times X$, either $(x', y) \in C$ for all $x'$ or for only finitely many $x'$ and either $(x, y') \in C$ for all $y'$ or for only finitely many $y'$. It follows that, if the projection of $C$ onto the $x$-axis is infinite, then $C$ contains an entire horizontal line and if the projection onto the $y$-axis its infinite, then $C$ contains an entire vertical line. If $X \times X \setminus \Delta_X = C \cup D$, where $C$ $D$ are closed in the subspace topology on $X \times X \setminus \Delta_X$, then the projections of both $C$ and $D$ onto the coordinate axes are surjective and this implies that $C$ and $D$ cannot be disjoint.

Answer 3

In fact, there is a famous theorem due to Eilenberg, see here or here, Theorem 3.

Definition: A topological space $(X, \tau)$ is weakly orderable, if there is a linear order on $X$, such that $\tau$ is finer than the order topology generated by this linear order.
$(X, \tau)$ is orderable, if there is a linear order on $X$, such that $\tau$ equals the order topology generated by this linear order.
(Note that in the above referenced paper of Kok, weakly orderable spaces are called orderable and in Eilenberg's paper ordered.)

Theorem (Eilenberg). Let $X$ be connected, $T_1$ and $|X| \ge 2$.

1. X is weakly orderable, iff $X^2 \setminus \Delta_X$ is not connected.
2. $X$ is orderable, iff $X^2 \setminus \Delta_X$ is not connected and $X$ is locally connected.