I follow up on the previous post to report that the question stated there was answered by Dejan Govc. I’ll try to put together a self-contained version of his proof.
Let be a connected topological space. Fix a base point
and define the separation order on
as follows:
iff the points
and
lies in different components of
.
Theorem: at most one of and
holds.
Proof. Suppose . Let
be the connected component of
in
. Note that
. If we can show that
is connected, then
and
are in the same component of
, which is to be proved. Suppose to the contrary that
has a nonempty proper closed-open subset
. We may assume
. Note two things about
:
, because
is closed in
is open in
. Indeed,
open in
. The closure of
is contained in
and therefore does not meet
. Hence
is open in
, and therefore in
.
Since , the set
is not connected. So it has a nonempty proper closed-open subset
. Then either
or
; by replacing
with its complement in
we may assume the latter holds: that is,
. We claim that
is both closed and open in
, contradicting the connectedness of
. Indeed,
is closed in
and
. Hence
is closed in
, and in
.
is open in
and
, hence
is open in
. Since
is open in
, it follows that
is open in
.
Nice exposition! The
notation gives is very suggestive. I think there's a slight problem, though: I'm not sure
is necessarily closed in
.
I think it is quite easy to get rid of this problem though:
is closed in
which is closed in
; and since
, we have that
is open in
. I hope this helps. Best regards, Dejan