Given that $A=\{(x,y)\mid y=0\}$ and $B=\{(x,y)\mid x>0\text{ and }y\neq0\}$, find the boundary and interior of the following subsets of $\mathbb R^2$
$$ C= A \cup B $$
I was trying to visualize the diagram, here the some of my ideas
From the diagram I can conclude that the boundary of $C$ will be $R^2$ and the interior is also $R^2$.
Is my thinking correct or not? Please verify it and tell me the solutions.
On
The union $C$ is also the (disjoint) union of the half-plane $H$ where $x>0$ together with the the set $K$ consisting of the points $(x,0)$ with $x\le0$.
Clearly $H$ is an open subset of $C$ and no point in $K$ has a neighborhood contained in $C$. Therefore the interior of $C$ is $H$.
Can you prove that the boundary is the union of $K$ and the $y$-axis? Hint: what's the closure of $H$?
You are wrong. The interior of $A\cup B$ is $\{(x,y)\in\mathbb{R}^2\,|\,x>0\}$ and the boundary of $A\cup B$ is$$\{(x,0)\in\mathbb{R}^2\,|\,x<0\}\cup\{(0,y)\in\mathbb{R}^2\,|\,y\in\mathbb{R}\}.\tag1$$
Indeed, if $(x,y)\in\mathbb{R}^2$ and $x>0$, then $D\bigl((x,y),x)\bigr)$ is contained in $\{(x,y)\in\mathbb{R}^2\,|\,x>0\}$, which is a subset of $A\cup B$. Therefore, $(x,y)$ belongs to the interior of $A\cup B$. And if $x\leqslant 0$, there are, for each $r>0$ points in $D\bigl((x,y),x)\bigr)$ with $x<0$ and $y>0$. Therefore, $(x,y)$ is not in the interior of $A\cup B$.
On the other hand, $\overline{A\cup B}=\left\{(x,y)\in\mathbb{R}^2\,\middle|\,x\geqslant0\right\}\cup\{(x,0)\,|\,x\in(-\infty,0)\}$. Since the boundary of $A\cup B$ is its closure minus its interior, the boundary is $(1)$.