Proof explanation to see that subset of $\mathbb{R}^2$ is not path connected.

Question

Consider the following example from http://www.mathcounterexamples.net/a-connected-not-locally-connected-space/ .

The part that I don't understand it's the bold text. What is $\arg⁡(B)$? And why $\arg⁡(1,1)$ is an isolated point of $\arg⁡(B)$?

I greatly appreciate any assistance you may provide.

$$A = \bigcup_{n \ge 1} [(0,0),(1,\frac{1}{n})] \text{ and } B = A \cup (\frac{1}{2},1]$$

For the proof that $B$ is not path connected, suppose that $\gamma$ is a path joining the point $(3/4,0)$ to the point $(1,1).$ We denote $\arg \gamma(t)$ the angle of $\gamma (t)$ with the $x$-axis. $\arg \gamma(t)$ is a continuous function of $t \in [0,1].$ Hence it range is connected.

This is in contradiction with the fact that $\arg⁡(1,1)$ is an isolated point of $\arg⁡(B).$

Answer 1

Short answer

source code of the figure

$$\arg(B) = \left\{ \mbox{set of all possible angles of oblique lines} \left[ (0,0),\left( 1,\frac{1}{n} \right) \right] \mid n \in \Bbb{N} \right\}$$

One easily observes from the above figure that the angle $\dfrac\pi4$ is an isolated point of $\arg(B)$.

Detailed Explanation

Note that $$\arg(B) = \bigcup_{n \ge 1} \left\{ \arctan\frac1n \right\} \cup \{0\}.$$ Observe that $\arctan$ is a continuous function, so that $$\arctan\frac1n \xrightarrow[n \to \infty]{} \arctan 0 = 0.$$ The first oblique line $[(0,0),(1,1)]$ and the set difference between $B$ and this oblique line \begin{align} &\quad |\arg(1,1) - \arg(B \setminus [(0,0),(1,1)])| \\ &\ge \arctan 1 - \arctan\frac12\\ &= \frac\pi4 - \arctan\frac12 \\ &\approx 0.322 \tag{cor. to 3 sig. fig.}\end{align}

Answer 2

The proof needs a little tweak to avoid the issue of $\operatorname{arg}$ not being defined on all of $B$.

Let $I= ({1 \over 2} , 1] \times \{0\}$. Let $t_0 = \sup \{ t | \gamma(t) \in I \}$. We have $[\gamma(t_0)]_1 \ge {1 \over 2}$ and hence there is some $\delta>0$ such that $[\gamma(t)]_1 \ge {1 \over 4}$ for $t \in [0,t_0+ \delta]$. Also, $\gamma(t) \in A$ for $t > t_0$ and $\gamma(t_0) \in \overline{I}$.

For $ x\in [{1 \over 4} , \infty) \times \mathbb{R}$ define $\alpha(x) = {x_2 \over x_1}$ (a proxy for $\operatorname{arg}$). Note that $\alpha$ is continuous.

Note that $\alpha(\overline{I}) = \{0\}, \alpha(A \cap [{1 \over 4} , \infty)) = \{ {1 \over n}\}_n$.

Hence $\alpha \circ \gamma ([t_0,t_0+\delta]) \subset \{0\} \cup \{{1\over n}\}_n$. However, since $\alpha \circ \gamma$ is continuous, the set $\alpha \circ \gamma ([t_0,t_0+\delta]) $ must be an interval containing the set $[0, \alpha(\gamma(t_0+\delta))]$ which is a contradiction since $\{0\} \cup \{{1\over n}\}_n$ contains no non trivial intervals.