## 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 $U$ 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 $X$ is a connected space, then the following are equivalent.

1. There are no open sets $V, W\subseteq U$ such that $\bar{V}\cap W = \emptyset$, $V\cap\bar{W} =\emptyset$ and $V\cup W = U$.  (These bars above the subspaces denote the closure of the space in $U$.)
2. The only subspaces of $X$ that are both open and closed are $\emptyset$ and $X$.
3. $X$ 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 $U\subseteq {\mathbb R}^{n}$, then $U$ is connected if and only if $U$ 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 $X$ is not connected, so we have two closed sets $A$ and $B$ such that $A\cap B = \emptyset$ and $A\cup B = U$ where neither $A$ nor $B$ 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 $W$ is connected for some set $W$, then if $f$ is a continuous map, then $f(W)$ is also connected.  We usually use this in the following way:

We usually take the interval $[0,1]$ and map it continuously into some space, as I’ve tried to show above in this picture.  Notice that $\gamma:[0,1]\rightarrow X$ where $X$ is that space on the bottom.  We usually think about about $\gamma$ 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 $\gamma$, 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 $[0,1]$ is connected, the squiggle $\gamma([0,1])$ is connected if we assume $\gamma$ is continuous.

This example turns out to be extremely important, because what does being path connected mean?  It means that from any two points $p,q\in U$ we have that there exists a continuous function $\gamma :[0,1]\rightarrow U$ such that $\gamma (0) = p$ and $\gamma (1) = q$ and $\gamma$ is continuous does this look familiar?  It should!

Let’s take our two closed sets $A$ and $B$ from before.  Since they’re both not empty, consider $p\in A$ and $q\in B$.  Since $U$ is path connected, we have that there is a $\gamma : [0,1]\rightarrow U$ such that $\gamma (0) = p$ and $\gamma (1) = q$ with $\gamma$ continuous.  Moreover, the image of $\gamma$ lies completely within $U$, and this is the important part: we now have our contradiction.  Do you see it?

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

Note that, obviously, the partition of $U$ could be much more complex, and, therefore, the pictures for that the intersections of the path with $A$ and $B$ could be much more complex.  The point here is that the intersections $\gamma([0,1])\cap A$ and $\gamma([0,1])\cap B$ 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 $A$ or $B$, but both of those are closed…).  But this means that the path in $U$ 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 $U$ cannot exist.  Thus, $U$ 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 $U$ 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 $U$.  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 $U$ is connected so that the only subspaces which are both open and closed in $U$ are the empty set and $U$ itself.  Then we’re going to prove that $U$ must be path connected by introducing something called the path component of  $U$ from a point  $p$.  This is a really intuitive concept: it’s just the set of all points in $U$ you can start at $p$ and make a path to, as we noted.

In the picture above, we have that if we start at $p$ we can make a path to $r$, as shown in red.  But we can’t start at $p$ and make a path to $q$.  Poor, lonely $q$.  Either way, looking at the picture we see that the path component of $U$ from the point $p$ 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 $p\in U$ and let’s agree to call the path component of $U$ from a point $p$ something nice: let’s just denote it $P$.  The plan of attack is to show that, in fact, $P = U$.  We’ll do this by showing that $P$ is both open and closed in $U$ (why does this suffice, again?).

First, let’s show $P$ is open in $U$.  Given a point $r\in P$ we want to show that there is some neighborhood around $r$ which is also in $P$.  Since $r\in U$ which is open, we know that there exists a neighborhood about $r$ which is inside of $U$; now, here is where we use the Euclidean part.  IN FACT, we have a ball of radius $\epsilon$ about $r$ for some $\epsilon > 0$.  Let’s denote this $B(r,\epsilon)$

Here’s the picture.  Now, because every point in the ball is at most $\epsilon$ away from $r$ we can just make some other path $\rho$ from $r$ to $q$ and then kind of "glue" the path from $p$ to $r$ together with our new path $\rho$.  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 $p$ to $q$.  Hence $q\in P$ for every $r\in B(r,\epsilon)$, which implies that every point $r\in P$ has an open neighborhood about it which is completely contained in $P$.  This makes $P$ an open set.

Now, let’s prove that $P$ 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 $r$ is a limit point of $P$ if for every open set $V$ containing $r$ we have that

$P\cap (V\setminus \{r\}) \neq \emptyset$

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

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

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 $X$, let the map be $f$.  Then we have that $f(X) = A\cup B$ for $A$ and $B$ 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

$f^{-1}(A) \cup f^{-1}(B)$.

First, is this all of $X$?  If $x\in X$ then since $A\cup B$ cover the image, then either $f(x) \in A$ or $f(x) \in B$ or both.  Thus, these pre-images cover $X$.  All we need to show now is that these two pre-images are disjoint and we’ll have a separation of $X$ given by $f^{-1}(A)$ and $f^{-1}(B)$.  Suppose $x\in f^{-1}(A) \cap f^{-1}(B)$; then we have $f(x)\in A\cap B$.  But we’ve assumed that $A$ and $B$ were disjoint, so this cannot happen.

This gives a separation for $X$, but $X$ is connected, so this is a contradiction.  Thus, $f(X)$ is connected.  $\hfill\Box$

### 4 Responses to “In Euclidean Space, Path Connected if and only if Connected.”

1. Anonymous said

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

2. James said

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.

3. daniel ghebresilassie said

send me a brief explanation.

• Matin said

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.