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 .
Every topological space comes with a natural equivalence relation: iff there is a connected subset such that . The equivalence classes are connected components of .
How do we know if is connected? We must check that there is no way to write it as where and are nonempty, open in , and disjoint. This must be done for every candidate for (actually, we can save time by considering only closed ). So, we run in a nested loop:
for C in [closed sets]:
if either x or y are not in C, continue with next C
for U in [open sets]:
for V in [open sets]:
if U and V disconnect C, continue with next C
Perhaps this can be optimized, but it looks like a lot of work to me.
Here is what could be done instead.
for U in [open sets]:
if X\U is also in [open sets], add U to [clopen sets]
for A in [clopen sets]
if x is in A and y is not in A, return('not equivalent')
There are no nested loops here: the algorithm is linear in the size of the topology. Also, the first loop should run only once for the space, creating a catalog of closed-open subsets in .
Let’s compare these two notions of equivalence, denoting the first one and the second . If , then there exists a closed-open set such that and . Hence, any set containing is disconnected by and . We conclude that . To summarize: .
The converse is not true in general. Here is an example adapted from General Topology by Willard (and made in Sage)
The set lives in the plane and contains of infinitely many concentric squares that converge to the square with endpoints . The vertical sides of the unit square are included in , but the horizontal ones are not. So, the connected components of are: left vertical line (in red), right vertical line (in red), and each of the blue squares. In particular, .
However, . Indeed, let be a closed-open set that contains . Since is open, it intersects (hence, contains) infinitely many blue squares. The union of these squares has as a limit point, and since is closed, .
The equivalence classes of are called quasicomponents. The quasicomponent of is the intersection of all closed-open sets containing . Quasicomponents are closed but not necessarily open; in this they are similar to connected components. Here is another similarity: is connected if and only if it has one quasicomponent.
In compact Hausdorff spaces quasicomponents coincide with components (this is not trivial). If we compactify the example above by adding horizontal sides of the unit square, the points and will be in the same component. Perhaps quasicomponents see the space as it should be, not as it is.
When you cut something with a knife, does the blade come into contact with each of the little pieces? I think so.
Now consider a topological version of this question. Let be a connected topological space (with more than 1 point) which we “cut” by removing a point . Let be a connected component of ; this is one of the little pieces. Is it true that ?
True if we replace “connected” by “path-connected”. Indeed, let be a continuous function such that and . If necessary, we truncate so that for . Now the set is a path-connected subset of , hence . Since as , we have .
False as stated, even if is a subset of . Here is a counterexample given by Niels Diepeveen on Math.StackExchange:
The set consists of the point and of closed line segments from to . The union of line segments is connected, and since it is dense in , is also connected. However, once is removed, each line segment becomes its own connected component, and so does the point . Clearly, the closure of does not contain .
In this example is not compact. It turns out that the statement is true when is a compact connected Hausdorff space (i.e., a continuum), but the proof (also given by Niels Diepeveen) is not easy.