A question from the proof of topologists sine curve being not path connected Let $S = \bigl\{ (x,\sin\frac{1}{x} ) : 0 < x \le 1 \bigr\} \cup \bigl\{ (0,y) : -1 \le y \le 1 \bigr\}$. Show that $S$ is connected but not pathwise connected.
Consider the proof below:

Q 11.8 It is clear that the set $\Gamma=\{(x,\sin(1/x)) : 0<x\leq1\}$ is pathwise connected and therefore connected. So $\overline\Gamma$ is connected. But $\overline\Gamma=S$ (Q 3.4). So $S$ is connected. Now suppose there exists a continuous path $f\colon[0,1]\to S$ that joins $(0,0)$ to some point of $\Gamma$. Let $r=\sup(f^{-1}(S\setminus\Gamma))$. Note that $f^{-1}(S\setminus\Gamma)$ is closed because $S\setminus\Gamma$ is closed, so that $f(r)\in S\setminus\Gamma$. So $r<1$ because $f(1)\in\Gamma$ by hypothesis. Since $f$ is continuous at $r$, there exists $\delta\in(0,1-r]$ such that $f([r,r+\delta])$ is included in the ball $B$ of radius $1/2$ around $f(r)$. It is easy to check that the connected component of $B\cap S$ that contains $f(r)$ is $B\cap(S\setminus\Gamma)$, so that $f(r+\delta)\in S\setminus\Gamma$, contradicting the definition of $r$. So there is no such continuous path $f$.

I couldn't understand why the connected component of $B \cap S$ that contains $f(r)$ is $B \cap (S\setminus\Gamma)$ ?
Could someone please give me some insight.  Thank you for reading through 
 A: A standard method to show that a set $C$ is the connected component of a point $p\in C$ (in some topological space $X$) consists of two parts,


*

*showing that $C$ is connected, and

*showing that every $q \notin C$ can be separated from $p$, i.e. there are disjoint open sets $U_q,V_q$ with $p \in U_q$, $q \in V_q$, and $X = U_q \cup V_q$.


The first part is rather simple here. $L = \{0\} \times [-1,1] = S \setminus \Gamma$ is a convex subset of $\mathbb{R}^2$, the ball $B$ is also a convex subset of $\mathbb{R}^2$, hence their intersection is a convex subset of $\mathbb{R}^2$. Convex sets in $\mathbb{R}^d$ are path-connected, hence connected. So $B \cap L = B \cap (S\setminus \Gamma)$ is a connected subset of $X = B \cap S$ that contains the point $p = f(r)$.
It remains to show that every point of $B \cap \Gamma$ can - in $B \cap S$ - be separated from $p$. The fundamental idea to see this is that the graph $\Gamma$ must leave the ball $B$ between the point $q = (x_0, \sin (1/x_0))$ and the line segment $L$ because $\sin (1/x)$ has infinitely many oscillations of amplitude $2$ in the interval $(0,x_0)$, and the radius of the ball is too small for it to contain points with second component $1$ as well as points with second component $-1$. And if $(t, \sin (1/t)) \in \Gamma \setminus B$ for some $0 < t < x_0$, then the open half-planes $H_1 = \{ (x,y) : x < t\}$ and $H_2 = \{(x,y) : x > t\}$ separate $p$ from $q$. Clearly we have $H_1 \cap H_2 = \varnothing$, $p \in U_q := H_1 \cap X$, and $q \in V_q := H_2 \cap X$. It remains to see that $U_q \cup V_q = X$, that is, that $X = B \cap S$ contains no point with first component $t$. But since $t > 0$ there is only one point with first component $t$ in $S$, namely $(t, \sin (1/t))$. And we specifically demanded that this point lie outside $B$, so we have our separation showing that $q$ does not lie in the connected component (with respect to $X$) of $p$. Since $q \in X \setminus (B \cap L) = B \cap \Gamma$ was arbitrary, it follows that indeed $B \cap L$ is the connected component of $f(r)$ in $B \cap S$.
We have not yet proved our assertion that $\Gamma$ leaves $B$ (infinitely often) between the point $(x_0, \sin (1/x_0))$ and the line segment $L$. For $n \in \mathbb{N}$ define
$$a_n := \frac{1}{\bigl(n + \frac{1}{2}\bigr)\pi}.$$
Then $0 < a_n < 1$ for all $n$, and $a_n \to 0$. Also $\sin (1/a_n) = (-1)^n$ and hence the distance between $(a_n, \sin (1/a_n))$ and $(a_{n+1}, \sin (1/a_{n+1}))$ is larger than $1$, so at most one of these two points can lie in $B$. Pick $t = a_n$ for an $n$ sufficiently large that $a_n < x_0$, and such that $(a_n, \sin (1/a_n))\notin B$.
