Separation order

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 X be a connected topological space. Fix a base point 0 and define the separation order on X as follows: x<y iff the points 0 and y lies in different components of X\setminus \{x\}.

Theorem: at most one of x<y and y<x holds.

Proof. Suppose x<y. Let Y be the connected component of y in X\setminus \{x\}. Note that 0,x\in X\setminus Y\subset X\setminus \{y\}. If we can show that X\setminus Y is connected, then 0 and x are in the same component of X\setminus \{y\}, which is to be proved. Suppose to the contrary that X\setminus Y has a nonempty proper closed-open subset A. We may assume x\notin A. Note two things about A:

  • \overline{A}\subset A\cup Y, because A is closed in X\setminus Y
  • A is open in X. Indeed, A open in X\setminus Y. The closure of Y is contained in Y\cup \{x\} and therefore does not meet A. Hence A is open in X\setminus \overline{Y}, and therefore in X.

Since Y\subsetneq A\cup Y\subset X\setminus \{x\}, the set A\cup Y is not connected. So it has a nonempty proper closed-open subset Z. Then either Y\subset Z or Y\cap Z=\varnothing; by replacing Z with its complement in A\cup Y we may assume the latter holds: that is, Z\subset A. We claim that Z is both closed and open in X, contradicting the connectedness of X. Indeed,

  • Z is closed in A\cup Y and Z\subset \overline{A}\subset A\cup Y. Hence Z is closed in \overline{A}, and in X.
  • Z is open in A\cup Y and Z\subset A, hence Z is open in A. Since A is open in X, it follows that Z is open in X.

Quasicomponents

Every topological space X comes with a natural equivalence relation: x\sim y iff there is a connected subset C\subset X such that x,y\in C. The equivalence classes are connected components of X.

How do we know if C\subset X is connected? We must check that there is no way to write it as C=U\cup V where U and V are nonempty, open in C, and disjoint. This must be done for every candidate for C (actually, we can save time by considering only closed C). 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
    return('equivalent')
return('not equivalent')

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')
return('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 X.

Let’s compare these two notions of equivalence, denoting the first one x\sim y and the second x\approx y. If x\not\approx y, then there exists a closed-open set A such that x\in A and y\notin A. Hence, any set C containing \{x,y\} is disconnected by C\cap A and C\setminus A. We conclude that x\not\sim y. To summarize: x\sim y\implies x\approx y.

The converse is not true in general. Here is an example adapted from General Topology by Willard (and made in Sage)

Quasicomponents

The set X lives in the plane and contains of infinitely many concentric squares that converge to the square with endpoints (\pm1, \pm1). The vertical sides of the unit square are included in X, but the horizontal ones are not. So, the connected components of X are: left vertical line (in red), right vertical line (in red), and each of the blue squares. In particular, (-1,0)\not\sim (1,0).

However, (-1,0)\approx (1,0). Indeed, let A be a closed-open set that contains (-1,0). Since A is open, it intersects (hence, contains) infinitely many blue squares. The union of these squares has (1,0) as a limit point, and since A is closed, (1,0)\in A.

The equivalence classes of \approx are called quasicomponents. The quasicomponent of x is the intersection of all closed-open sets containing x. Quasicomponents are closed but not necessarily open; in this they are similar to connected components. Here is another similarity: X 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 (1,0) and (-1,0) will be in the same component. Perhaps quasicomponents see the space as it should be, not as it is.

One of These Days

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 X be a connected topological space (with more than 1 point) which we “cut” by removing a point a\in X. Let C be a connected component of X\setminus \{a\}; this is one of the little pieces. Is it true that a\in \overline{C}?

True if we replace “connected” by “path-connected”. Indeed, let \gamma\colon [0,1]\to X be a continuous function such that \gamma(0)\in C and \gamma(1)=a. If necessary, we truncate \gamma so that \gamma(t)\ne a for 0\le t<1. Now the set \gamma([0,1)) is a path-connected subset of X\setminus\{a\}, hence \gamma([0,1))\subset C. Since \gamma(t)\to a as t\to 1-, we have a\in \overline{C}.

False as stated, even if X is a subset of \mathbb R^2. Here is a counterexample given by Niels Diepeveen on Math.StackExchange:

Counterexample by Niels Diepeveen

The set X consists of the point (0,0) and of closed line segments from (1/n,0) to a=(0,1). The union of line segments is connected, and since it is dense in X, X is also connected. However, once a is removed, each line segment becomes its own connected component, and so does the point (0,0). Clearly, the closure of \{(0,0)\} does not contain a.

In this example X is not compact. It turns out that the statement is true when X is a compact connected Hausdorff space (i.e., a continuum), but the proof (also given by Niels Diepeveen) is not easy.