# Topologist’s Sine Curve

The topologist’s sine curve is usually defined as the graph of the curve $y=\text{sin}(x^{-1})$ plus a vertical line segment on the y-axis (see [2] and [3]). It is a handy example of a connected space that is not pathwise connected. We illustrate an alternative construction of the topologist’s sine curve and make some observations about this example. The post is concluded with some general observations about connectedness and pathwise connectedness.

Figure 1 below is the boundary of the unit square (just the edges and not the interior), the starting point of the construction.

We draw a vertical line segment at the middle of the square in Figure 1. The middle line segment splits the square into two halves. Next we draw a vertical line segment in the middle of left half of the square, which is then split into two halves. Next we draw a vertical line segment in the middle of the resulting left half. The same process is repeated successively, each time a vertical line segment is drawn in the left half created as a result of the previous vertical line segment. Figure 2 below shows the resulting vertical line segments of the first several iterations.

We start with the corner $(1,1)$ and erase the upper horizontal edge to the next vertical line (from the point $(1,1)$ to the point $(\frac{1}{2},1)$). Next, we start with the point $(\frac{1}{2},0)$ and erase the lower horizontal edge to the next vertical line (from the point $(\frac{1}{2},0)$ to the point $(\frac{1}{4},0)$). The same process is repeated successively by alternating between the upper edge and the lower edge. Figure 3 below shows the resulting topologist’s sine curve.

The left vertical line segment in Figure 3 consists of the points $(0,y)$ where $0 \le y \le 1$. We call the vetical line segment $V$. We call the “sine curve” $C$, which converges to $V$. The topologist’s sine curve is the set $T=V \cup C$, which has the topology inherited from the Euclidean plane. The space $T$ is compacted, connected. On the other hand, $T$ is both not pathwise connected and locally connected.

To see that it is connected, note that the curve $C$ is connected. Also note that the closure of $C$ is the entire topologist’s sine curve ($T=\overline{C}$). Since the closure of a connected set is connected, $T$ is also connected.

However, $T$ is not pathwise connected. For example, there is no connected path linking the point $(0,\frac{1}{2})$ and $(1,\frac{1}{2})$. In general, there is no connected path linking any point in $V$ to any point in $C$.

A space is locally connected at a point if there is a neighborhood base consisting of open connected sets at that point. The topologist’s sine curve $T$ is not locally connected at any point on the vertical line secgment $V$.

An interesting subspace of the topologist’s sine curve is $T_0=\left\{(0,\frac{1}{2})\right\} \cup C$. This is the subset with only one point of the vertical line segment $V$. The space $T_0$ is not locally compact at the point $(0,\frac{1}{2})$. However, $T_0$ is the continuous image of $\left\{ -1 \right\} \cup (0,1]$, demonstrating that continuous image of a locally compact space needs not be locally compact.

Figure 4 below is the extended topologist’s sine curve, which is obtained by the same process as indicated above except that no lower horizonatal edges are removed. What is interesting about the extended sine curve is that it is pathwise connected but not locally connected.

Some general observations. In general, a space $X$ is disconnected if there are two disjoint nonempty open sets $H$ and $K$ such that $X=H \cup K$. We say $X$ is disconnected by $H$ and $K$. A space $X$ is connected if no disconnection exists. A space $X$ is pathwise connected if for any two points $x,y \in X$ with $x \ne y$, there exists a continuous function $f:[0,1] \rightarrow X$ such that $f(0)=x$ and $f(1)=y$. The function $f$ is called a path from $x$ to $y$.

It is well known that the Cartesian product of connected spaces is also a connected space (see [1] or [3]).

Pathwise connectedness is also preserved by taking Cartesian product. Suppose that $X_\alpha$ is a pathwise connected space for each $\alpha \in A$. Then $Y=\prod \limits_{\alpha \in A} X_\alpha$ is pathwise connected. To see this, suppose that $g,h \in Y$ with $g \ne h$.

For each $\alpha \in A$, if $g(\alpha) \ne h(\alpha)$, choose a continuous $w_\alpha: [0,1] \rightarrow X_\alpha$ such that $w_\alpha(0)=g(\alpha)$ and $w_\alpha(1)=h(\alpha)$. For each $\alpha \in A$, if $g(\alpha) = h(\alpha)$, let $w_\alpha: [0,1] \rightarrow X_\alpha$ be defined by $w_\alpha(t)=g(\alpha)$ for all $t \in [0,1]$.

We now define $f: [0,1] \rightarrow \prod \limits_{\alpha \in A} X_\alpha$. For each $t \in [0,1]$, let $f(t)$ be the element of the product space $\prod \limits_{\alpha \in A} X_\alpha$ such that $f(t)(\alpha)=w_\alpha(t)$ for each $\alpha \in A$. It is clear that $f$ is continuous and $f(0)=g$ and $f(1)=h$. In other words, the function $f$ is a path from $g$ to $h$.

Reference

1. Engelking, R., General Topology, Revised and Completed edition, 1989, Heldermann Verlag, Berlin.
2. Steen, L. A., Seebach, J. A.,Counterexamples in Topology, 1995, Dover Edition, Dover Publications, New York.
3. Willard, S., General Topology, 1970, Addison-Wesley Publishing Company.