# The Unapologetic Mathematician

## The Image of a Connected Space

One theorem turns out to be very important when we’re dealing with connected spaces, or even just with a connected component of a space. If $f$ is a continuous map from a connected space $X$ to any topological space $Y$, then the image $f(X)\subseteq Y$ is connected. Similarly, if $X$ is path-connected then its image is path-connected.

The path-connected version is actually more straightforward. Let’s say that we pick points $y_0$ and $y_1$ in $f(X)$. Then there must exist $x_0$ and $x_1$ with $f(x_0)=y_0$ and $f(x_1)=y_1$. By path-connectedness there is a function $g:\left[0,1\right]\rightarrow X$ with $g(0)=x_0$ and $g(1)=x_1$, and so $f(g(0))=y_0$ and $f(g(1))=y_1$. Thus the composite function $g\circ f:\left[0,1\right]\rightarrow Y$ is a path from $y_0$ to $y_1$.

Now for the connected version. Let’s say that $f(X)$ is disconnected. Then we can write it as the disjoint union of two nonempty closed sets $B_1$ and $B_2$ by putting some connected components in the one and some in the other. Taking complements we see that both of these sets are also open. Then we can consider their preimages $f^{-1}(B_1)$ and $f^{-1}(B_2)$, whose union is $X$ since every point in $X$ lands in either $B_1$ or $B_2$.

By the continuity of $f$, each of these preimages is open. Seeing as each is the complement of the other, they must also both be closed. And neither one can be empty because some points in $X$ land in each of $B_1$ and $B_2$. Thus we have a nontrivial clopen set in $X$, contradicting the assumption that it’s connected. Thus the image $f(X)$ must have been connected, as was to be shown.

From this theorem we see that the image of any connected component under a continuous map $f$ must land entirely within a connected component of the range of $f$. For example, any map from a connected space to a totally disconnected space (one where each point is a connected component) must be constant.

When we specialize to real-valued functions, this theorem gets simple. Notice that a connected subset of $\mathbb{R}$ is just an interval. It may contain one or both endpoints, and it may stretch off to infinity in one or both directions, but that’s about all the variation we’ve got. So if $X$ is a connected space then the image $f(X)$ of a continuous function $f:X\rightarrow\mathbb{R}$ is an interval.

An immediate corollary to this fact is the intermediate value theorem. Given a connected space $X$, a continuous real-valued function $f$, and points $x_1,x_2\in X$ with $f(x_1)=a_1$ and $f(x_2)=a_2$ (without loss of generality, $a_1), then for any $b\in\left(a_1,a_2\right)$ there is a $y\in X$ so that $f(y)=b$. That is, a continuous function takes all the values between any two values it takes. In particular, if $X$ is itself an interval in $\mathbb{R}$ we get back the old intermediate value theorem from calculus.

January 3, 2008

## Connectedness

Tied in with the fundamental notion of continuity for studying topology is the notion of connectedness. In fact, once two parts of a space are disconnected, there’s almost no topological influence of one on the other, which should be clear from an intuitive idea of what it might mean for a space to be connected or disconnected. This intuitive notion can be illustrated by considering subspaces of the real plane $\mathbb{R}^2$.

First, just so we’re clear, a subset of the plane is closed if it contains its boundary and open if it contains no boundary points. Here there’s a lot more between open and closed than there is for intervals with just two boundary points. Anyhow, you should be able to verify this by a number of methods. Try using the pythagorean distance formula to make this a metric space, or you could work out a subbase of the product topology. In fact, not only should you get the same answer, but it’s interesting to generalize this to find a metric on the product of two arbitrary metric spaces.

Anyhow, back to connectedness. Take a sheet of paper to be your plane, and draw a bunch of blobs on it. Use dotted lines sometimes to say you’re leaving out that section of the blob’s border. Have fun with it.

Now we’ll consider that collection of blobs as a subspace $X\subseteq\mathbb{R}^2$, and thus it inherits the subspace topology. We can take one blob $B$ and throw an open set around it that doesn’t hit any other blobs (draw a dotted curve around the blob that doesn’t touch any other). Thus the blob is an open subset $B\subseteq X$ because it’s the intersection of the open set we drew in the plane and the subspace $X$. But we could also draw a solid curve instead of a dotted one and get a closed set in the plane whose intersection with $X$ is $B$. Thus $B$ is also a closed subset of $X$. Some people like to call such a subset in a topological space “clopen”.

In general, given any topological space we can break it into clopen sets. If the only clopen sets are the whole space and the empty subspace, then we’re done. Otherwise, given a nontrivial clopen subset, its complement must also be clopen (why?), and so we can break it apart into those pieces. We call a space with no nontrivial clopen sets “connected”, and a maximal connected subspace $B$ of a topological space $X$ we call a “connected component”. That is, if we add any other points from $X$ to $B$, it will be disconnected.

An important property of connected spaces is that we cannot divide them into two disjoint nonempty closed subsets. Indeed, if we could then the complement of one closed subset would be the other. It would be open (as the complement of a closed subset) and closed (by assumption) and nontrivial since neither it nor its complement could be empty. Thus we would have a nontrivial clopen subset, contrary to our assumptions.

If we have a bunch of connected spaces, we can take their coproduct — their disjoint union — to get a disconnected space with the original family of spaces as its connected components. Conversely, any space can be broken into its connected components and thus written as a coproduct of connected spaces. In general, morphisms from the coproduct of a collection of objects exactly correspond to collections of morphisms from the objects themselves. Here this tells us that a continuous function from any space is exactly determined by a collection of continuous functions, one for each connected component. So we don’t really lose much at all by just talking about connected spaces and trying to really understand them.

Sometimes we’re just looking near one point or another, like we’ve done for continuity or differentiability of functions on the real line. In this case we don’t really care whether the space is connected in general, but just that it looks like it’s connected near the point we care about. We say that a space is “locally connected” if every point has a connected neighborhood.

Sometimes just being connected isn’t quite strong enough. Take the plane again and mark axes. Then draw the graph of the function defined by $f(x)=\sin\left(\frac{1}{x}\right)$ on the interval $\left(0,1\right]$, and by $f(0)=0$. We call this the “topologist’s sine curve”. It’s connected because any open set we draw containing the wiggly sine bit gets the point $(0,0)$ too. The problem is, we might want to draw paths between points in the space, and we can’t do that here. For two points in the sine part, we just follow the curve, but we can never quite get to or away from the origin. Incidentally, it’s also not locally connected because any small ball around the origin contains a bunch of arcs from the sine part that aren’t connected to each other.

So when we want to draw paths, we ask that a space be “path-connected”. That is, given points $x_0$ and $x_1$ in our space $X$, there is a function $f:\left[0,1\right]\rightarrow X$ with $f(0)=x_0$ and $f(1)=x_1$. Slightly stronger, we might want to require that we can choose this function to be a homeomorphism from the closed interval $\left[0,1\right]$ onto its image in $X$. In this case we say that the space is “arc-connected”.

Arc-connectedness clearly implies path-connectedness, and we’ll see in more detail later that path-connectedness implies connectedness. However, the converses do not hold. The topologist’s sine curve gives a counterexample where connectedness doesn’t imply path-connectedness, and I’ll let you try to find a counterexample for the other converse.

Just like we had local connectedness, we say that a space is locally path- or arc-connected if every point has a neighborhood which is path- or arc-connected. We also have path-components and arc-components defined as for connected components. Unfortunately, we don’t have as nice a characterization as we did before for a space in terms of its path- or arc-components. In the topologist’s sine curve, for example, the wiggly bit and the point at the origin are the two path components, but they aren’t put together with anything so nice as a coproduct.

[UPDATE]: As discussed below in the comments, I made a mistake here, implicitly assuming the same thing edriv said explicitly. As I say below, point-set topology and analysis really live on the edge of validity, and there’s a cottage industry of crafting counterexamples to all sorts of theorems if you weaken the hypotheses just slightly.

January 2, 2008