**Exercise**
Prove: Let *X* be connected, and *C* a connected subset,
and *Q* either open-and-closed, or a component, or a quasicomponent
in *X\C*. Then *X\Q* is connected.

**Proof**
Let us use *+* here to denote a separation: a decomposition
into the union of two open-and-closed subsets.
A disconnected space has a separation where both parts are nonempty.

(i) Let *Q* be nonempty and open-and-closed in *X\C*,
say *X\C = Q + R*. If *X\Q = Y + Z*, where *C* is
contained in *Y*, then *Z* is open-and-closed in *X*,
hence empty.

(ii) Let *Q* be a component of *X\C*.
If *X\Q = Y + Z*, where *C* is contained in *Y*,
then by (i) *X\Y* is connected, contains *Q*
and is contained in *X\C*. Hence *X\Y = Q* and
*Z* is empty.

(iii) Let *Q* be a quasicomponent of *X\C*.
Since *Q* is the intersection of all open-and-closed
subsets *S* of *X\C* containing it, *X\Q*
is the union of all subsets *X\S*. Each *X\S*
contains *C* and is connected by (i), so their
union is connected as well.