Thoughts About Connectedness (Components)
It is not surprising that if one cannot speak of a connected space that they would like to at least speak of the “biggest” connected subspaces. To motivate our point let us consider our example we used in the first post. . It is intuitively clear once again what a component of this space would be, but defining it is a little trickier than may seem. We clearly don’t want to say a component is a connected subspace, for clearly is a connected subspace but not a “component”. It is clear once again that our minds are drawn to maximality. We want to describe our space by as few connected subspaces as possible. Defining this maximality is what requires finesse though. Usually in math when we think of biggest thing that contains something we usually think of the intersection of all things containing it. That is a problem here though, since the intersection of two connected sets need not be connected (think about the intersection of a line and a circle)….
With this in our noodles as motivation we begin the actual work.
Component: If is a topological space we define to be a component if it is connected and it is not contained in any other connected subspace.
Before we get into the meat of the argument we prove some facts regarding unions of connected sets.
Theorem: Let be a topological space and and arbitrary collection of non-empty connected subspaces. Then if is non-empty then is connected.
Proof: Suppose not, then there exists some which are open and non-empty and such that and and . Now, clearly no one particular may be in both otherwise so by assumption that we may conclude in particular that there exist some such that , and there is some . But, since we see that is a disconnection of and this is a contradiction.
We can actually prove some pretty cool theorems using this. For example:
Theorem: Let be a normed vector space over . Then is connected.
Proof: Fix and define by .
Lemma: is continuous.
Proof: Since these are both metric spaces we use the more metric space oriented definition of continuity.If this is trivial, so assume not. Let be given, then choosing such that we see that . The conclusion follows.
Now clearly is a connected subspace of since it is the continuous image of a connected space. But, seeing that and appealing to the previous theorems finishes the argument.
Corollary: Every Banach space is connected.
Corollary: and thus are connected.
Corollary: with the norm is connected for any topological space .
We next wish to prove something that might have been intuitively obvious.
Theorem: Let be a topological space and be a connected subspace then given any we have that is connected, in particular is connected.
Proof: Suppose that is a disconnection of in (that is there restrictions to is a disconnection of in the normal sense). Since is connected we clearly must have that either or . Now, it is relatively easy to show that this implies that and thus which contradicts our choice of .
Remark: There is a particularly nice way of proving that is connected implies that is connected. To see this consider then from basic topology we have that and since must be either or which are both closed in it follows that . From where the conclusion is immediate.
Corollary: If is a topological space with a connected dense subset then is connected.
We can now prove the main properties of components which interest us.
Theorem: If is a topological space and then it is contained in exactly one component.
Proof: Define to be the set of all connected subsets of which contain . Clearly then and so by a previous theorem we see that is connected. This is clearly a component which contains Also, it is clearly the only one. For, suppose that is a component which contains then and so and by ‘s maximality it follows that . The conclusion follows. .
Theorem: Any connected subspace of is contained in exactly one component.
Proof: The fact that if they are contained in one it is unique is obvious. To see that it is contained in at least one merely note that it is contained in the component constructed in the above theorem. .
Theorem: The components of are closed.
Proof: Let be a component of , then by previous theorem we have that is a connected subspace of which contains , and by ‘s maximality it follows that . .
Theorem: A connected subspace of which is both open and closed is a component of .
Proof: Suppose that is such a subspace and where is another connected subspace of . Then is clearly a disconnection of . It follows that no connected subspace of may properly contain . The conclusion follows. .
One of the important facts about components is the following.
Theorem: Let be a topological space and the set of components of , then forms a partition of .
Proof: Since every point of belongs to at least one component it’s clear that . Also, suppose that were such that then by a previous theorem we have that is a connected subspace of which properly contains and , which of course contradicts their maximality. The conclusion follows.
Corollary: If is a component of then
From this we can prove the following theorem.
Theorem: Let have finitely many components, then they are open subsets of .
Proof: Let be as above and let then which is the finite union of closed subspaces of and thus a closed subspace. The conclusion follows.
This finished up what I wanted to say about components. So, I am just going to prove some related theorems. Most of them relate to union of connected subspaces.
Theorem: Let be a topological space and a sequence of closed subspaces of such that . Then is a connected subspace of .
Proof: Suppose there existed non-empty open subsets of such that their restrictions to form a disconnection. Let . Clearly we must have that either or (it obviously can’t have some in one and some in the other or the restrictions of to would be a disconnection). WLOG assume that and let . If we are done, so assume not. Then by the well-ordering principle exists. Clearly we must have that .
Now, assume that then since we have that are both non-empty open subspaces of and their union is clearly . This of course contradicts ‘s connectedness. It follows that which contradicts the minimality of . It follows that . The conclusion follows.
Theorem: Let be a topological space and a collection of connected subspaces of such that . Then, is a connected subspace of .
Proof: Suppose not, and there existed some were are non-empty and open in and is a disconnection of . Then, since are non-empty and each connected we must have that and for some but since it is pretty plain that is a disconnection of . The fact that is disconnected is therefore untenable. .
We now give an alternate theorem that the product of finitely many connected spaces is connected.
Theorem: Let and be connected topological spaces, then is connected under the product topology.
Proof: Fix and . Clearly for any and . Thus, given any we have that is the union of two interesting connected sets, thus connected. Also, we have that for any . It follows that is connected. .
Corollary: It follows from induction that the product of finitely many connected spaces is connected.
We now need a definition. Note though that the following may be extended considerably.
Convex set: Let be a normed vector space over . We then call convex if for every .
Theorem: Let be a normed vector space and a convex subset, then is connected.
Lemma: Let and define . Then if the mapping given by is continuous. (it is assumed that is a subspace of under the usual topology)
Proof: Notice that if , that
Now it is assume that we see that . So, letting we see that
The conclusion follows.
Now that would help us if we knew that is connected, so we must prove it. Don’t fret though, we must merely note that if is given by then it is evident that is continuous and . It follows from an earlier question that and thus is connected.
So, assume that is continuous and surjective. Let then clearly the mapping is also continuous and surjective. This contradicts that is connected. The conclusion follows.
From this we can derive a theorem which will become mainly interesting to us in the next post.
Theorem: Let be a normed vector space. Then the open (or closed) ball is convex for every and for every .
Proof: Let then
it follows that . The conclusion follows. .
Corollary: Every open (or closed) ball in a normed vector space is connected.
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