# removing rationals in topolosist's sine curve

Munkres pg. 160 says, where $S$ is the topologist's sine curve

If one forms a space from $\bar S$ by deleting all points of $V$ having rational second coordinate, one obtains a space that has only one component but uncountably many path components.

then, how can removed $\bar S$ still be connected? Plus, can you describe uncountably many path components?

• Presumably "$V$" is the vertical interval...? Are you happy with the fact that $S$ is connected? If so, how (using the definition of "separating open sets", forgetting "uncountably many path components") does $\bar{S}$ seem not connected? – Andrew D. Hwang Jun 26 '15 at 15:11

As suggested, let \begin{align}S&=\{\,(x,\sin\tfrac1x)\mid 0<x<\infty\,\}\\T&=\{\,(0,y)\mid y\notin \mathbb Q, |y|\le 1\,\}\\X&=S\cup T\end{align} with the induced metric. Clearly, $S$ is (path) connected. Assume $X=U\cup V$ with disjoint open subsets $U,V$. As $S$ is connected, one of $U\cap S$, $V\cap S$ must be empty, say $U\cap S=\emptyset$. If $(0,y)\in U$, then we find a sequence $u_n\to\infty$ with $\sin u_n=y$. Then $(\tfrac 1{u_n},\sin u_n)\to (0,y)$ shows that $U$ intersects $S$, contrary to assumption. Hence also $U\cap T=\emptyset$ and finally $U=\emptyset$. We conclude that $X$ is connected.
On the other hand, for distinct irrational $y_1,y_2\in[0,1]\setminus \mathbb Q$, there is no path from $(0,y_1)$ to $(0,y_2)$ as that would have to pass through some point $(x_0,\sin\tfrac1{x_0})\in S$ and we already know from the topologist's sine curve that no such path exists.