Crossley motivates the idea of connectivity of a topological space by posing the question of whether or not there can be a continuous surjection from to the set with the discrete topology. If such a surjection existed, then and would be open subsets of . Also note that as is partitioned by these sets, must be partitioned by and ; i.e., with . We further require that neither nor may be empty as is a surjection. If we can show that no such and can exist, then we will have shown there is no continuous surjection from to .

Let and and assume, wlog, that and consider the interval : one endpoint is in and the other is in . Split the interval into two halves of equal size, and . One of these intervals will lie entirely in one set, while the other will have an endpoint in each set. This follows from the fact that and are open sets that partition . Call the interval with an endpoint in each of the sets . Notice if we split into two closed intervals of equal length, we will again have an interval with an endpoint in each of and . Call this interval . In general, we will let denote the interval obtained if we split into two halves and pick the half with an endpoint in and the other in . Notice is a decreasing sequence of closed intervals. Furthermore, because of the way we constructed the , their measure is decreasing to zero. These sets must have a non-empty intersection, but because the measure is shrinking to zero, this intersection must consist of a single point. This point, then, must be in both and . This contradicts the fact that and are non-empty disjoint sets, however. This means we can not find open sets and that are disjoint, both non-empty, and whose union is . So we conclude that there can be no continuous surjection from to .

In general a topological space that *does* have two disjoint, non-empty open sets that partition the space is called *disconnected*, while a space where no such sets exist is called *connected*. We say that a subset of a topological space is disconnected if it’s a disconnected topological space when endowed with the subspace topology. The mentioned above is a disconnected space. Another example is with the subspace topology from . Note that this is *not* simply the discrete topology on , as it was with . (The set {0} will not be open since any open subset of will contain an open interval, so if we take some open set of reals and intersect it with the rationals, we’ll get all of the rationals in the intervals making up the open set.) Letting and gives us two non-empty open sets whose intersection is empty, but whose union is all of . Similarly, is disconnected; let and .

In fact, if we have any connected space , there can be no continuous surjection to . The proof of this follows the same outline for above. We can also show that if is any disconnected space there is a continuous surjection from to : we just split into two disjoint, non-empty open sets that partition (by disconnectedness), map everything in one set to 1 and everything in the other set to -1. This tells us that if is connected and is disconnected, there is no continuous surjection from to . If there were, we’d just compose that surjection with the surjection from to then we’d have a map from to .

Now we’ll show that any continuous function from to must be constant. Suppose that is continuous and let be a point in the image of . Let and . Clearly and are disjoint sets whose union is Since is continuous, both and must be open. However, is a connected space, so one of and must be empty. We know that is not empty, since we specifically chose it to be the preimage of a point in the image of . This means that , so and everything maps to under , and so must be constant.

[…] Filed under: Topology — cjohnson @ 6:45 pm Using the ideas of connectedness previously discussed, we can show that a continuous function from to must be constant. We will do this by showing […]

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