Let $B = \{(x,y) \in \Bbb{R}^2 \mid -1 \le x \lt 2, 0 \lt y \le 2 \} \cup \{(x,y) \in \Bbb{R}^2 \mid 5 \lt x \le 7, y = 1 \}$. Determine whether a point $(0,1)$ is an interior point of $B$.
I got a little bit confuse here, since a point $(0,1)$ exactly lies on vertical line $y=1$, in which it does mean that $(0,1)$ be an interior point of not.
Here's my attempt:
Fix $0 < r = \frac{1}{2}$. Note that \begin{equation*} B((0,1), \frac{1}{2}) = \{(x,y) \in \Bbb{R} \mid x^2 + (y-1)^2 < \frac{1}{4} \}. \end{equation*} Let $(x,y) \in B((0,1), \frac{1}{4})$. Then, \begin{align*} x^2 + (y-1)^2 < \frac{1}{4} \\ x^2 < \frac{1}{4} \wedge (y-1)^2 < \frac{1}{4} \\ -\frac{1}{2} < x < \frac{1}{2} \wedge -\frac{1}{2} < y-1 < \frac{1}{2} \\ -\frac{1}{2} < x < \frac{1}{2} \wedge \frac{1}{2} < y < \frac{3}{2}. \end{align*}
Hence, $x \in (-\frac{1}{2}, \frac{1}{2}) \subseteq [-1,2)$ and $y \in (\frac{1}{2}, \frac{3}{2}) \subseteq (0,2]$. Thus, forall $(x,y) \in B((0,1), \frac{1}{2})$, we have $(x,y) \in B$. In another words, $B((0,1), \frac{1}{2}) \subseteq B$. Therefore, there exists $r>0$ such that $B((0,1), \frac{1}{2}) \subseteq B$. Hence, $(0,1)$ is an interior point of $B$.
Am I true? If not, any explain how to show it? Thanks in advanced.
