## Path Connectedness

During our discussion of connectedness of topological groups I alluded to the fact that I would not be talking about path connectedness, but why? It seems natural that the two concepts, ostensibly so connected (), should be discussed at the same time. While I cannot refute that logic, my reason for separating (I’m on fire) them was purely a personal matter of connotation. To me connectedness while very geometric is much more aligned in the depths of point-set topology. It is a fundamental property of spaces. Path connectedness (while admittedly possessing all of the aforementioned properties) on the other hand has a *very* geometric flavor to it. Also, path connectedness will be an integral part of our discussion of manifolds. Thus, without further justification to a non-existent audience let us begin:

**Path:** Let be a topological space, then we call a continuous map such that and a *path* from to .

**Path Connected:** We call a topological space *path connected* if for any there exists a path between them.

Intuitively we can think about a path between two points as just connecting them with a “squiggly line”. So, it is natural to assume that every path connected space is connected but as some wise old guy once said “What is asserted without proof may be dismissed without proof”:

**Theorem:** Let be path connected, then is path connected.

**Proof:** Suppose that is a continuous surjection (where as always is the two-point discrete space). Then, choose . By assumption there exists a path between them. Clearly then, is a continuous surjection and so it follows that is also a continuous surjection, but this clearly contradicts the connectedness of .

So, what are some examples of path connected sets you may ask? We can find a plethora, but first we need a little itsy bitsy lemma (barely worth saying).

**Theorem:** Every convex subset of (a normed vector space over ) is path connected.

**Proof:** Clearly given any where is the convex set the map is continuous obviously and by assumption , from where the conclusion follows.

So, from this we can derive zillions and zillions (technical term of course) of path connected sets, for if you’ll remember we proved earlier that every open ball in a normed vector space is convex. So, in particular every open ball in is path connected. Also, realizing that themselves are convex (obviously) and in fact so is the punctured plane (just take a polygonal line if needed). Slightly less obvious is that for some arbitrary (once again take a polygonal line)

So, we begin to prove some pretty obvious theorems.

**Theorem:** Let be path connected and a continuous map. Then, is a path connected subspace.

**Proof:** Given it is evident that where is the path from to in is the required path.

**Theorem:** Let be a collection of path connected topological spaces, then with the product topology is path connected. The converse is also true.

**Proof:** Recall that if is a collection of continuous maps then the map

is connected (since each of the projections is connected). So, for any we merely take the guaranteed guaranteed path from to and note that

is the desired path.

Conversely, if is path connected then so are each of it’s coordinate spaces since the projection onto them is continuous.

**Theorem:** Let be path connected and a quotient space of , then is path connected.

**Proof:** This is blindingly obvious since the quotient map which defines the topology on is automatically a continuous surjection.

**Theorem:** If are two non-empty topological spaces, then is not path connected.

**Proof:** This obviously follows since we proved earlier that they aren’t even *connected*.

That said, there are some fairly neat theorems which are rooted very closely in geometric intuition. If for some fixed point we can connect it to any other point by a path we should be able to connect any two points by connecting the first point to , the second point to , and then “connecting” the paths. We formalize in the following paragraphs, but first we need a very important lemma.

**Theorem (Gluing Lemma):** Let be a topological space and a closed cover of it. Furthermore, suppose that for each there exists a continuous map such that for each . Then, there exists a unique continuous map such that .

**Proof:** First note that if then

and so the map

is at least well-defined, and in fact this map clearly satisfies . To see that it’s continuous we let be closed. Clearly

but each is a closed subset of a closed subspace of and thus closed in . Thus, is the finite union of closed sets and thus closed. Lastly, to see that is unique we merely note that if where another such map then given we have that for some and so

from where the conclusion follows.

**Corollary:** By an infinitesimal modification the above proves the same if is replaced by an arbitrary open cover

**Theorem (*):** Let be a topological space and suppose that for some fixed there is a path from to any other point of , then is path connected.

**Proof: **Let be arbitrary. By assumption there exists a path from to . Define a “reverse path” from to as follows

This is readily verified to be path from to . So, let be the path from to . Clearly the maps and are continuous and since

It follows from the gluing lemma that is continuous and so noting that

and

and so the theorem follows.

With this we have an easy way to prove another nice theorem

**Theorem:** Let be a topological space and a collection of path connected subspaces such that , then is path connectected.

**Proof:** Let , then given any we have that for some and so by assumption there exists some path from to . Clearly, if we extend the codomain we see that is also a path from to in . Considering our previous theorem finishes the argument.

With this we can prove a satisfying theorem, which in a very real sense is just mimicry of a previous theorem.

**Theorem:** Let be any normed vector space over . Then, is path connected.

**Proof:** We merely need to note that the map is continuous. Thus, by previous theorem is path connected. But,

and thus noting that

and appealing to the previous theorem finishes the argument.

We finish our discussion with a very, very nice theorem which in essence will show that every manifold is path connected. But, first we need to define the local analogue of path connectedness.

**Locally Path Connected:** Let be a topological space, then is said to be *locally path connected* if for each there exists a path connected neighborhood of it.

The role of local path connectedness plays a much more important role that it’s plain old connectedness analogue. This can be seen from the following theorem.

**Theorem:** Let be a locally path connected connected space, then is path connected.

**Proof:** Let and let be the set of all points in for which there is a path from to that point. Since () it suffices to prove that it is both open and closed (and the result will then follow from ‘s connectedness). So, let . By assumption there exists some neighborhood of it such that is path connected, we claim that . To see this, let . Using the exact same argument as in theorem (*) we can then construct a path from to and so . It follows that is open. Now, to show that is closed we show it is invariant under closure, and sine it suffices to show the reverse inclusion. So, let . Then, by ‘s local connectedness there exists a neighborhood of which is path connected. But, since there must be some point . So, using the exact same technique described in theorem (*) again we may connect the path from to and the path from to to get a path from to . It follows that .

Thus, is open and closed and since it’s non-empty it must be that . But, that means that every point of may be connected to by a path. Thus, by theorem (*) is path connected.

As stated this will prove that every manifold is path connected, but maybe a little closer to home is that this implies that *every* open connected set in is path connected, we state this in the following theorem.

**Theorem:** Let be a normed vector space and open and connected. Then, is path connected.

**Proof:** By assumption for each point there exists some open ball , but this open ball is convex and thus path connected. So, is locally path connected and connected thus by the previous theorem path connected. .

This confirms everyone’s intuition from back in complex analysis or calculus when it “seemed” that every connected “blob” in could have a “line drawn in it connecting two points”.

This ends our brief discussion of path connectedness.

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