Mathematics Prelims

January 25, 2009

Connected Spaces

Filed under: Topology — cjohnson @ 5:06 pm

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 \mathbb{R} to the set S^0 = \{-1, 1\} with the discrete topology.  If such a surjection f existed, then f^{-1}(\{-1\}) = U and f^{-1}(\{1\}) = V would be open subsets of \mathbb{R}.  Also note that as S is partitioned by these sets, \mathbb{R} must be partitioned by U and V; i.e., U \cup V = \mathbb{R} with U \cap V = \emptyset.  We further require that neither U nor V may be empty as f is a surjection.  If we can show that no such U and V can exist, then we will have shown there is no continuous surjection from \mathbb{R} to S^0.

Let u \in U and v \in V and assume, wlog, that u < v and consider the interval {}[u, v]: one endpoint is in U and the other is in V.  Split the interval into two halves of equal size, {}[u, u + (v-u)/2] and {}[v - (v-u)/2, v].  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 U and V are open sets that partition \mathbb{R}.  Call the interval with an endpoint in each of the sets S_1.  Notice if we split S_1 into two closed intervals of equal length, we will again have an interval with an endpoint in each of U and V.  Call this interval S_2.  In general, we will let S_n denote the interval obtained if we split S_{n-1} into two halves and pick the half with an endpoint in U and the other in V.  Notice (S_n)_{n \in \mathbb{N}} is a decreasing sequence of closed intervals.  Furthermore, because of the way we constructed the S_n, 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 U and V.  This contradicts the fact that U and V are non-empty disjoint sets, however.  This means we can not find open sets U and V that are disjoint, both non-empty, and whose union is \mathbb{R}. So we conclude that there can be no continuous surjection from \mathbb{R} to S^0.

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 S mentioned above is a disconnected space.  Another example is \mathbb{Q} with the subspace topology from \mathbb{R}.  Note that this is not simply the discrete topology on \mathbb{Q}, as it was with \mathbb{Z}.  (The set {0} will not be open since any open subset of \mathbb{R} 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 U = \mathbb{Q} \cap (-\infty, \pi) and V = \mathbb{Q} \cap (\pi, \infty) gives us two non-empty open sets whose intersection is empty, but whose union is all of \mathbb{Q}.  Similarly, \mathbb{Z} is disconnected; let U = \mathbb{Z} \cap (-\infty, 1/2) and V = \mathbb{Z} \cap (1/2, \infty).

In fact, if we have any connected space T, there can be no continuous surjection to S^0.  The proof of this follows the same outline for \mathbb{R} above.  We can also show that if X is any disconnected space there is a continuous surjection from X to S^0: we just split X into two disjoint, non-empty open sets that partition X (by disconnectedness), map everything in one set to 1 and everything in the other set to -1.  This tells us that if T is connected and X is disconnected, there is no continuous surjection from T to X.  If there were, we’d just compose that surjection with the surjection from X to S^0 then we’d have a map from T to S^0.

Now we’ll show that any continuous function from \mathbb{R} to \mathbb{Z} must be constant.  Suppose that f : \mathbb{R} \to \mathbb{Z} is continuous and let z be a point in the image of f.  Let U = f^{-1}(\{z\}) and V = f^{-1}(\{z\}^\complement) = f^{-1}(\{z\})^\complement = U^\complement.  Clearly U and V are disjoint sets whose union is \mathbb{R}  Since f is continuous, both U and V must be open.  However, \mathbb{R} is a connected space, so one of U and V must be empty.  We know that U is not empty, since we specifically chose it to be the preimage of a point in the image of f.  This means that V = \emptyset, so \mathbb{R} = U and everything maps to z under f, and so f must be constant.

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2 Comments »

  1. […] 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 […]

    Pingback by Continuous Functions from the Reals to the Rationals « Mathematics Prelims — January 25, 2009 @ 6:45 pm | Reply

  2. […] Connected Spaces […]

    Pingback by Topology, Geometry & Dynamics | Topology, Geometry & Dynamics — September 24, 2010 @ 4:56 pm | Reply


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