## Connectedness

Tied in with the fundamental notion of continuity for studying topology is the notion of connectedness. In fact, once two parts of a space are disconnected, there’s almost no topological influence of one on the other, which should be clear from an intuitive idea of what it might mean for a space to be connected or disconnected. This intuitive notion can be illustrated by considering subspaces of the real plane .

First, just so we’re clear, a subset of the plane is closed if it contains its boundary and open if it contains no boundary points. Here there’s a lot more between open and closed than there is for intervals with just two boundary points. Anyhow, you should be able to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same answer, but it’s interesting to generalize this to find a metric on the product of two arbitrary metric spaces.

Anyhow, back to connectedness. Take a sheet of paper to be your plane, and draw a bunch of blobs on it. Use dotted lines sometimes to say you’re leaving out that section of the blob’s border. Have fun with it.

Now we’ll consider that collection of blobs as a subspace , and thus it inherits the subspace topology. We can take one blob and throw an open set around it that doesn’t hit any other blobs (draw a dotted curve around the blob that doesn’t touch any other). Thus the blob is an open subset because it’s the intersection of the open set we drew in the plane and the subspace . But we could also draw a solid curve instead of a dotted one and get a closed set in the plane whose intersection with is . Thus is also a closed subset of . Some people like to call such a subset in a topological space “clopen”.

In general, given any topological space we can break it into clopen sets. If the only clopen sets are the whole space and the empty subspace, then we’re done. Otherwise, given a nontrivial clopen subset, its complement must also be clopen (why?), and so we can break it apart into those pieces. We call a space with no nontrivial clopen sets “connected”, and a maximal connected subspace of a topological space we call a “connected component”. That is, if we add any other points from to , it will be disconnected.

An important property of connected spaces is that we cannot divide them into two disjoint nonempty closed subsets. Indeed, if we could then the complement of one closed subset would be the other. It would be open (as the complement of a closed subset) and closed (by assumption) and nontrivial since neither it nor its complement could be empty. Thus we would have a nontrivial clopen subset, contrary to our assumptions.

If we have a bunch of connected spaces, we can take their coproduct — their disjoint union — to get a disconnected space with the original family of spaces as its connected components. ~~Conversely, any space can be broken into its connected components and thus written as a coproduct of connected spaces. In general, morphisms from the coproduct of a collection of objects exactly correspond to collections of morphisms from the objects themselves. Here this tells us that a continuous function from any space is exactly determined by a collection of continuous functions, one for each connected component. So we don’t really lose much at all by just talking about connected spaces and trying to really understand them.~~

Sometimes we’re just looking near one point or another, like we’ve done for continuity or differentiability of functions on the real line. In this case we don’t really care whether the space is connected in general, but just that it looks like it’s connected near the point we care about. We say that a space is “locally connected” if every point has a connected neighborhood.

Sometimes just being connected isn’t quite strong enough. Take the plane again and mark axes. Then draw the graph of the function defined by on the interval , and by . We call this the “topologist’s sine curve”. It’s connected because any open set we draw containing the wiggly sine bit gets the point too. The problem is, we might want to draw paths between points in the space, and we can’t do that here. For two points in the sine part, we just follow the curve, but we can never quite get to or away from the origin. Incidentally, it’s also not locally connected because any small ball around the origin contains a bunch of arcs from the sine part that aren’t connected to each other.

So when we want to draw paths, we ask that a space be “path-connected”. That is, given points and in our space , there is a function with and . Slightly stronger, we might want to require that we can choose this function to be a homeomorphism from the closed interval onto its image in . In this case we say that the space is “arc-connected”.

Arc-connectedness clearly implies path-connectedness, and we’ll see in more detail later that path-connectedness implies connectedness. However, the converses do not hold. The topologist’s sine curve gives a counterexample where connectedness doesn’t imply path-connectedness, and I’ll let you try to find a counterexample for the other converse.

Just like we had local connectedness, we say that a space is locally path- or arc-connected if every point has a neighborhood which is path- or arc-connected. We also have path-components and arc-components defined as for connected components. ~~Unfortunately, we don’t have as nice a characterization as we did before for a space in terms of its path- or arc-components. In the topologist’s sine curve, for example, the wiggly bit and the point at the origin are the two path components, but they aren’t put together with anything so nice as a coproduct.~~

*[UPDATE]:* As discussed below in the comments, I made a mistake here, implicitly assuming the same thing edriv said explicitly. As I say below, point-set topology and analysis really live on the edge of validity, and there’s a cottage industry of crafting counterexamples to all sorts of theorems if you weaken the hypotheses just slightly.

When you defined clopen sets, I immediately realized that they’re closed under finite intersection and union, but in fact they’re colesed under arbitrary intersection and union, arent’t they?

Comment by edriv | January 2, 2008 |

I don’t think that’s necessarily so, but I don’t have a counterexample offhand. Point-set topology (like analysis) has a way of breaking when you give it edge cases. There’s even a book (which might have this answer) called

Counterexamples in Topology.Comment by John Armstrong | January 2, 2008 |

Oops in fact we have an easy counterexample: (or . I thought that the connected components were always clopen, but this is not the case (they’re just closed).

Comment by edriv | January 2, 2008 |

Two comments. First: one way to get interesting counterexamples to edriv’s first comment is via Stone duality, which (loosely) says that the category of Boolean algebras is equivalent to the opposite of so-called Stone spaces (compact, Hausdorff, totally disconnected spaces). Take a Boolean algebra A; there’s a way of cooking up a space Spec(A) whose points are Boolean algebra maps A –> 2 (the topology is the Zariski topology). In the other direction, given a Stone space X, we can look at continuous maps X –> 2 where 2 has the discrete topology (notice that these are tantamount to clopen sets of X). Clopen(X) forms a Boolean algebra. These two functors form part of a contravariant adjunction which is in fact a contravariant equivalence.

The point is that Clopen(X) can be

anyBoolean algebra; therefore we shouldn’t expect Clopen(X) to be necessarily complete (i.e., closed under unions and intersections). For a concrete example, take A to be the free Boolean algebra on the countable set N. By definition of “free”, the corresponding Stone space is the space of functions N –> 2, better known as Cantor space C. By Stone duality, we should be able to retrieve A as Clopen(C). We see from this example that the clopens of Cantor space are not closed under arbitrary unions or intersections (since A is not complete).Second comment: it’s not true that every space is the coproduct of its connected components. An easy example is the space of rationals Q, which is totally disconnected (i.e., the connected components are just points). Come to think of it, just about any nontrivial Stone space would also provide a counterexample, by the same reasoning.

Comment by Todd Trimble | January 3, 2008 |

Todd, I think you’re right.. I may have to go back and tweak a bit. In my head I think I might have been making edriv’s mistake, but not consciously. It’s a bit late, though, so I’ll try getting to that in the morning.

Comment by John Armstrong | January 3, 2008 |

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