## Counter-examples: The Topologist’s Sine Curve.

### October 8, 2010

After I learned about the topologist’s sine curve, I started using it almost immediately; it’s a really sweet example of a graph that is connected, but is not path connected or even locally path connected!  Let’s just jump right in and define it.

The equation for the topologist’s sine curve is

$f(x) = \sin(\frac{1}{x})$

for every $x\in (0,1]$.  We also include the vertical line at $x = 0$ from $-1\leq y \leq 1$.  The reason for this is that the closure of the image of $f(x)$ includes it.  This is easy to see if we notice that the curve goes up-and-down very quickly near 0.  It does take a bit of proving, but not much (it suffices to show every point on the vertical line we added is a limit point).

Now, let’s show a few properties that the topologist’s sine curve has.

The topologist’s sine curve is connected.

The proof for this is approximately the following: suppose that we had two disjoint open sets.  Every open set in the reals is made of countably many open intervals, so let’s suppose that one open interval has to contain that vertical line — but then it must contain some of the sine curve as well.  This means the problem reduces to finding a disjoint open set that contains the "rest" of the sine curve.  Of course, this reduces to separating something homeomorphic to the closed interval, which is connected.  Contradiction, so the topologist’s sine curve is connected.

The topologist’s sine curve is NOT path connected.

This is kind of weird, since it kind of looks like it should be: the path being, you know, "take the graph, and the graph itself contains a path from any point to any point."  The problem here is that the topologist’s sine curve contains that vertical line.  In particular, we cannot find any path (remember, a path is the image of the compact closed interval $[0,1]$) from any point on the sine curve to any point on that vertical line.  It takes a bit of thinking to see why, but essentially the sine curve gets "too long" near the line $x = 0$.

The topologist’s sine curve is NOT locally path connected.

This is also pretty clear once we think about what it means.  If we take any open ball of radius, say, $\epsilon > 0$ about some point on some point at the peak of the sine curve sufficiently close to the origin, we get things which look like a bunch of upside-down parabolas which are not connected.  The point here is that as the sine graph gets closer to zero, it alternates up and down too much.

Let’s end with an example of a way to use this curve.  This is a question in Munkres.

Question: For some space $X$, we are given $A\subseteq X$ is path connected.  Does this imply that $\bar{A}$, the closure of $A$, is path-connected?

You may be tempted to say, "sure", on first read: you take some limit points, and connected them to the part that is already path connected.  But this is incorrect, and the topolgist’s sine curve is a counterexample.  Let $A$ be the image of $\sin(\frac{1}{x})$ from $(0,1]$.  This is path connected (yes, it is!  without the vertical line, the distance along the curve from any point to any other point is finite, and so we can inject a path between any two points.  we don’t have that pesky vertical line at this point.), but we’ve proved above that the closure, which is the proper Topologist’s sine curve, is not path connected.  This shows that the closure of a path connected space is not necessarily path connected.