Thoughts about connectedness (Locally connected spaces)
Just as with compactness there is a certain local connectedness property that spaces may posses without being themselves connected.
Locally Connected: Let be a topological space, then we call locally connected if for every and every neighborhood of there exists some neighborhood connected of such that .
Remark: It is apparent that this is equivalent that has a connected open base.
We have already said that every normed vector space with the usual topology is locally connected (since every open ball is convex and thus connected). Also, it is clear that our example is locally connected but not connected and that the rationals are neither connected nor locally connected. In fact, you can have a locally connected space which is totally disconnected, take any discrete space.
But, unlike local compactness which is implied by compactness local connectedness is not implies by connectedness as the following example shows.
Example (Topologist’s Sine Curve): Define by . So, define . It is clear that is closed since it is equal to which is the graph of a continuous function on a connected space. But it is fairly easy to see that is not locally connected.
We now prove some facts about locally connected spaces.
Theorem: Let be a locally connected space. If is an open subspace of , then each component of is open in . In particular each component of is open.
Proof: Let be a component in and let . Since is a neighborhood of and is locally connected there exists some connected neighborhood of contained in . But, since is a component we must have that . The conclusion follows. .
Theorem: Let be a topological space where all the components of every open subspace of are open, then is locally connected.
Proof: Let and let be any neighborhood of it. Since is an open subspace of it’s components are also open in . In particular, the component which contains is open in . Thus, we have found a connected neighborhood of contained in . The question is whether or not it is connected with respect to . But, since the topology of as a subspace of is the same as that with respect to there is nothing to prove.
Theorem: Let be a compact locally connected space. Then, has finitely many components.
Proof: Suppose not. Then we have that the class of components of is infinite. Since the formation of components is a partition we see in particular that covers and it admits no finite subcovering. But, is in fact an open cover of by previous theorem. It follows that is an open cover of which admits no finite subcover, which of course contradicts its compactness.
Theorem: Let be locally connected and open, surjective, and continuous. Then, is locally connected.
Proof: Let be arbitrary and any neighborhood of . Since is continuous we have that is a neighborhood of . Now, by ‘s local connectedness there then exists some connected neighborhood of such that . It follows that is connected and open and . The conclusion follows. .
Corollary: Let be an arbitrary collection of topological spaces and locally connected. Then, is locally connected for every .
We know answer the question of whether or not local connectedness is invariant under products.
Theorem: Let be a finite collection of locally connected spaces. Then, is locally connected.
Proof: Let be arbitrary and any neighborhood of . Clearly then is a neighborhood of for each . Using ‘s local connectedness we are furnished with some open connected such that . Clearly then, is a connected open subspace of such that . The conclusion follows. .
The general case for the product of arbitrarily many locally connected spaces need not be locally connected. With a little bit of unjustified knowledge we can see that under the discrete topology is locally connected but where is the Cantor set. Clearly the Cantor set is not locally connected though.
If we strengthen our conditions though we’re right as rain.
Theorem: Let be an arbitrary collection of connected and locally connected spaces. Then, is locally connected under the product topology.
Proof: Let be arbitrary and any neighborhood of . We of course may assume WLOG that is basic open. So, let be the finitely many indices such that . For each we have that is a neighborhood of . And thus, by ‘s local connectedness there exists a connected neighborhood such that . Clearly then if
Then is open and connected and . The conclusion follows. .
Remark: The necessity for each to be connected was used when we were able to conclude that the product of the ‘s was connected.
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