My Topology textbook defines components of a topological space X as the equivalence classes under the equivalence relation ~ on X defined by: x ~ y iff x and y are in some connected subset of X.
I have been reading about what is called the specialization preorder. The specialization preorder is a preorder on a set X defined by x < y iff every open set containing x also contains y. It appears that for finite topologies the connected components (viewing the preorder as a directed graph) are precisely the components of X defined in my text book.
How do I prove this? Does it suffice to prove the following statement?: If every open set containing x also contains y then x and y are in some connected set. Here is a sketch of my proof.
Suppose BWOC that there are no connected sets in X containing both x and y. Let U be an open set containing both x and y. Since U is disconnected then there are nonempty open sets V and W that form a separation of U. Note that both x and y are in W exclusively or they are in V. WLOG assume x and y are in V. Now V is disconnected so there are nonempty open sets A and B that form a separation of V. Again, x and y must both be in A or must both be in B. But this process can be repeated indefinitely and must terminate (since X is finite) with x and y being the only elements in some open set C. But C is connected so we have a contradiction.
Is this a valid proof? Is there a reference that gives a "better" proof?
I found this introduction by May quite helpful myself. Section 4 discusses connectedness (and it turns out connected is even equivalent to path-conected (in the classical sense) as well.