Determine the boundary $R$ of the area $D_2 $

Question

Determine the boundary $R$ of the area $D_2 =$ {$(x, y) : x^2 + y^2 \leq 1, x > 0$}.

a) $R =$ {$x^2 + y^2 = 1, x \geq 0$} $\cup$ {($x, y) : x = 0, -1 < y < 1$}.

b) $R = ${$x^2 + y^2 = 1, x > 0$} $\cup$ {$(x, y) : x = 0, -1 < y < 1$}.

I picked b because in the definition of the boundary of the area, it says that $x>0$. Not to mention the fact even in the alternative a it says: {($x, y) : x = 0, -1 < y < 1$} and then at the same time it says: ${$x^2 + y^2 = 1, x > 0$}. Anyways can someone explain to me why I am wrong?

Answer 1

These two choices only differ for $x=0$ in the first set.

Note that $$ \{ (x,y) \in \mathbb{R}^2 \mid x^2 + y^2 = 1 \text{ and } x = 0 \} = \{ (0,1) \;,\; (0,-1) \} $$ I think it is sufficiently obvious that these points lie in the boundary of $D_2$. For instance, drawing an open ball about each, the inside of those balls always intersects the semicircle. (You can play with this in this Desmos demo: link.)

You should also see that $(0,1),(0,-1)$ are not in the first set.

Answer 2

The closure of the set is $\{(x,y) \, | \, x^2+y^2 \leq 1, x \geq 0\}=D_2 \cup \{(0,y) \, | \, -1 \leq y \leq 1\}$.

The interior of the set is $\{(x,y) \, | \, x^2+y^2 < 1, x>0\}=D_2-\{(x,y) \, | \, x^2+y^2=1\}$. So, boundary is the set difference (closure - interior).