Topology: An Inquiry-Based Approach

Note that $S_2$ is an open subset of $S\text{,}$ since $S_2=\left(\bigcup _{z=(x,y)\in S_2}B(z,\frac{x}{2})\right)\cap S\text{.}$ So the continuity of $p$ implies that $V$ is an open subset of $[0,1]\text{.}$ Also, the fact that $p(0)\in S_1$ means that $U\ne \mathrm{\varnothing }\text{,}$ and the fact that $p(1)\in S_2$ means that $V\ne \mathrm{\varnothing }\text{.}$ If we demonstrate that $U$ is an open subset of $[0,1]\text{,}$ then Equation (19.2) will imply that $[0,1]$ is not connected, a contradiction. So we proceed to prove that $U$ is open in $[0,1]\text{.}$

Let $x\in U\text{,}$ and so $p(x)$ in $S_1\text{.}$ The set $O=B_S(p(x),\frac{1}{2})\cap S$ is open in $S\text{.}$ The continuity of $p$ then tells us that $p^{-1}(O)$ is open in $[0,1]\text{.}$ So there is a $\delta >0$ such that the open ball $B=B_{[0,1]}(x,\delta )$ is a subset of $p^{-1}(O)\text{.}$ We will prove that $p(B)\subseteq S_1\text{.}$ This will imply that $B\subseteq U$ and so $U$ is a neighborhood of each of its points, and $U$ is therefore an open set.

Every element in $B$ is mapped into $O$ by the path $p\text{.}$ The set $O$ is complicated, consisting of infinitely many sub-curves of the curve $S_2\text{,}$ along with points in $S_1\text{,}$ as illustrated in Figure 19.18. To simplify our analysis, let us consider the projection onto the $x$-axis. The function $P_x:{\mathbb{R}}^2\to \mathbb{R}$ defined by $P_x(x,y)=x$ is a continuous function. Let $I=P_x(p(B))\text{.}$ Since $p(B)\subseteq O\text{,}$ we know that $I\subseteq P_x(O)\text{.}$ Let $Z=P_x(O)\text{.}$ So $I\subseteq Z\text{.}$ Since $B$ is a connected set ($B$ is an interval), we know that $p(B)$ is a connected set. The fact that $P_x$ is continuous means that $I=P_x(p(B))$ is connected as well. Now $I$ is a bounded subset of $\mathbb{R}\text{,}$ so $I$ must be a bounded interval. Recall that $x\in B$ and so $p(x)\in p(B)\text{.}$ The fact that $p(x)\in S_1$ tells us that $0=P_x(p(x))\in P_x(p(B))=I\text{.}$ So $I\ne \mathrm{\varnothing }\text{.}$ There are two possibilities for $I\text{:}$ either $I=\{0\}\text{,}$ or $I$ is an interval of positive length. We consider the cases.

Suppose $I=\{0\}\text{.}$ Then the projection of $p(B)$ onto the $x$-axis is the single point $0$ and $p(B)\subseteq S_1$ as desired. Suppose that $I$ is an interval of the form $[0,d]$ or $[0,d)$ for some positive number $d\text{.}$ The structure of $O$ would indicate that there must be some gaps in the set $Z\text{,}$ the projection of $O$ onto the $x$-axis. This implies that $I$ cannot be a connected interval. We proceed to show this. In other words, we will prove that $I\setminus Z\ne \mathrm{\varnothing }$ (which is impossible since $I\subseteq Z$). Remember that $p(x)\in S_1\text{,}$ so let $p(x)=(0,q)\text{.}$ We consider what happens if $q<\frac{1}{2}$ and when $q\ge \frac{1}{2}\text{.}$

Suppose $q<\frac{1}{2}\text{.}$ Then the ball $B_S(p(x),\frac{1}{2})$ contains only points with $y$ value less than 1. Let $N\in {\mathbb{Z}}^{+}$ so that $t=\frac{1}{\pi /2+2N\pi }<d\text{.}$ Then $t\in I\text{.}$ But $\sin \left(\frac{1}{t}\right)=\sin (\pi /2+2N\pi )=\sin (\pi /2)=1\text{,}$ and so $(t,\sin \left(\frac{1}{t}\right))$ is not in $O\text{.}$ Thus, $t\notin Z\text{.}$ Thus we have found a point in $I\setminus Z\text{.}$

Finally, suppose $q\ge \frac{1}{2}\text{.}$ Then the ball $B_S(p(x),\frac{1}{2})$ contains only points with $y$ value greater than $-1\text{.}$ Let $N\in {\mathbb{Z}}^{+}$ so that $t=\frac{1}{3\pi /2+2N\pi }<d\text{.}$ Then $t\in I\text{.}$ But $\sin \left(\frac{1}{t}\right)=\sin (3\pi /2+2N\pi )=\sin (3\pi /2)=-1\text{,}$ and so $t\notin Z\text{.}$ Thus we have found a point in $I\setminus Z\text{.}$

We conclude that there can be no path in $S$ from $(0,0)$ to any point in $S_2\text{,}$ completing our proof that $S$ is not path connected. (In fact, the argument given shows that there is no path in $S$ from any point in $S_1$ to any point in $S_2\text{.}$