## In Euclidean Space, Path Connected if and only if Connected.

### December 3, 2010

There was a question on my topology final that asked: *in a Euclidean space prove that an open subset is connected if and only if it is path connected*. Since this is not a difficult problem and it is available in a number of textbooks, I’m going to answer it with lots and lots of details and draw a few pictures, since I like drawing pictures.

Before we do this theorem let’s just state something about connected spaces.

**Theorem**: If is a connected space, then the following are equivalent.

- There are no open sets such that , and . (These bars above the subspaces denote the closure of the space in .)
- The only subspaces of that are both open and closed are and .
- cannot be the union of two disjoint nonempty closed sets.

All of these are good to remember, and it’s nice to prove each from the other. Usually, one is taken as the definition, and the others are proved from this one. It’s all a matter of taste. In the following proof, we will be using the second and the third characterization, so this is actually a nice proof to work on connected spaces with!

**Theorem**: Given an open subset , then is connected if and only if is path connected.

*Proof.*

### The Easy Direction.

One direction is easy, namely, if it is path connected then it is connected. This proof is available almost everywhere, and it is true more generally. Let’s just outline why it’s true.

What does it mean for a space to be connected? Well, we look above. Let’s use the third characterization: suppose that is not connected, so we have two closed sets and such that and where neither nor is the empty set. Now, let’s state something here without proof for now:

*The continuous image of a connected space is connected. *

What does this mean? It means that if is connected for some set , then if is a continuous map, then is also connected. We usually use this in the following way:

We usually take the interval and map it continuously into some space, as I’ve tried to show above in this picture. Notice that where is that space on the bottom. We usually think about about stretching and bending the interval, then laying it on the new space. Take note here that the green squiggle that you see on the space at the bottom is the *image *of , and *this* is the part that we were talking about above when we say the continuous image of a connected space is connected. Since the interval is connected, the squiggle is connected if we assume is continuous.

This example turns out to be extremely important, because what does being path connected mean? It means that from any two points we have that there exists a continuous function such that and and is continuous does this look familiar? It should!

Let’s take our two closed sets and from before. Since they’re both not empty, consider and . Since is path connected, we have that there is a such that and with continuous. Moreover, the image of lies completely within , and this is the important part: we now have our contradiction. Do you see it?

We have our path crossing from into . So what happens when we intersect these things? Intersecting the path with gives us something that looks like:And similarly, intersecting the path with gives us something like:

Note that, obviously, the partition of could be much more complex, and, therefore, the pictures for that the intersections of the path with and could be much more complex. The point here is that the intersections and are disjoint. Showing these two sets is also closed is not difficult (suppose one had a limit point: then it’s also a limit point of or , but both of those are closed…). But this means that the path in is not connected (by the third criteria above)! This contradicts the theorem we just proved, that the continuous image of a connected space is connected. Thus, such a partition of cannot exist. Thus, is connected.

### The Other Direction.

Normally, there is an "easy" and a "hard" direction for this type of proof. This direction isn’t so much the "hard" direction, as the direction that you have to be a bit more clever for. I’m not sure if this is the standard proof (if there is an easier one, please comment) but this one is kind of nice. The idea behind the proof, and *how to get to the idea *(which is just as important!) is relatively simple: if isn’t path connected, then we would like to know *approximately how path connected is it? *In other words, starting at some point, how far can we get by making paths from our starting point out into the space . Formalizing this idea forces us to introduces the idea of "path components" which we’ll talk about below.

We’re going to use the second criteria for connectedness and assume that is connected so that the only subspaces which are both open and closed in are the empty set and itself. Then we’re going to prove that must be path connected by introducing something called the *path component of ** from a point *. This is a really intuitive concept: it’s just the set of all points in you can start at and make a path to, as we noted.

In the picture above, we have that if we start at we can make a path to , as shown in red. But we can’t start at and make a path to . Poor, lonely . Either way, looking at the picture we see that the path component of from the point is going to be the blob on the left, and it won’t include the little blob on the right.

Now let’s get back to our proof! Let’s pick a point and let’s agree to call the path component of from a point something nice: let’s just denote it . The plan of attack is to show that, in fact, . We’ll do this by showing that is both open and closed in (why does this suffice, again?).

First, let’s show is open in . Given a point we want to show that there is some neighborhood around which is also in . Since which is open, we know that there exists a neighborhood about which is inside of ; now, here is where we use the Euclidean part. IN FACT, we have a ball of radius about for some . Let’s denote this .

Here’s the picture. Now, because every point in the ball is at most away from we can just make some other path from to and then kind of "glue" the path from to together with our new path . Looking at the picture, we’d glue the solid red line with the dotted red line. This is, in fact, continuous (by the pasting lemma) and of the form of a path (when we rearrange the parameters a bit) from to . Hence for every , which implies that every point has an open neighborhood about it which is completely contained in . This makes an open set.

Now, let’s prove that is closed. I’m not sure if there’s an easier way to do this, but I’m a big fan of the limit point way, so let’s do that. Recall that is a* limit point of* if for every open set containing we have that

Recall also that **a set is closed if it contains all of its limit points**. Now suppose that is a limit point of ; we’d like to show that is actually in . Consider to be a ball of sufficiently small radius so that is completely contained in . Let’s say that , the ball centered at with some radius of . Since there is some . This means that and with . But, wait, if then there is a path from to . If then we can make a straight-line path (as we did in the preceding part) from to . Concatenating these paths gives us a path from to , and so .

Thus, since was arbitrary, contains all of its limit points. But then is closed. But then is both open and closed, so must either be empty or all of . Since the trivial path from to is in , trivially, this means that is non-empty. But this means that , which means that there is a path from to any point in . Since was arbitrary, this means that there is a path from any point in to any other point in ; ie, that is path connected.

So this is a pretty cool proof. The last thing we really have to do is to prove one of the major things we used in the beginning of the proof. Luckily, this proof is much, much shorter.

**Theorem**: The continuous image of a connected space is connected.

*Proof.* Suppose not. Let the connected space in the domain be , let the map be . Then we have that for and disjoint non-empty closed sets. By definition of continuity, we have that the inverse image of a closed set is closed, and so let’s consider

.

First, is this all of ? If then since cover the image, then either or or both. Thus, these pre-images cover . All we need to show now is that these two pre-images are disjoint and we’ll have a separation of given by and . Suppose ; then we have . But we’ve assumed that and were disjoint, so this cannot happen.

This gives a separation for , but is connected, so this is a contradiction. Thus, is connected.

When you consider sign(1/x) connected does not imply path-connected.

The topologist’s sine curve is the prototypical example of a space which is connected but not path-connected. I think what you mean here is that this example should contradict what I’ve stated above, but think about the image of Sin(1/x) union the vertical interval [-1,1] at x = 0. Is this open in R^2? No. We can prove this a number of ways, but a good way to do it is to take the compliment and show that there are limit points not in the set: specifically, those on that vertical interval.

send me a brief explanation.

Actually your proof applies to all locally path connected spaces (this one I believe is a classic result you can find in Topology books). Top. Sine Curve is not locally path connected.