Let
${A_1} = \left\{ {\left( {m,n} \right) \in {\mathbb{R}^2}|{m^2} + {n^2} \leqslant 1} \right\}$
${A_2} = \left\{ {\left( {m,n} \right) \in {\mathbb{R}^2}|0 \leqslant m < 1,n = 0} \right\}$
${A_3} = \left\{ {\left( {m,n} \right) \in {\mathbb{R}^2}|m = 1,n \in \mathbb{R}} \right\}$
$M = \left( {{A_1} \cup {A_3}} \right)\backslash {A_2}$
So my progress so far is:
Closure of M is ${A_1} \cup {A_3}$
Interior of M is $\left\{ {\left( {m,n} \right) \in {\mathbb{R}^2}|{m^2} + {n^2} < 1,} \right\}\backslash {A_2}$.
Boundary of M is $\left\{ {\left( {m,n} \right) \in {\mathbb{R}^2}|{m^2} + {n^2} = 1,} \right\}$.
Interior of closure of M is essentially the interior of M.
Closure of interior of M is ${A_1} \cup {A_3}$.
I am not sure if I got it all correct so please help enlighten me.
Thank you all.
You have a ball. In that ball you remove a line with origin in the center of the ball. To the boundary you add another line as a tangent vector. Therefore your closure is right. Interior also right. For the boundary check your definition. The set $A_3$ is also contained. The interior of the closure is not always the interior again. In your case, your closure is a full ball with straight line attached, therefore the ball without boundary is the interior of the closure. Closure of interior is also wrong. It‘s the full ball.