I have been having some trouble showing conditions are met before applying green's theorem. For example, showing a set is a regular closed region is pretty hard. Showing that a set is compact is easy enough, but I am not sure how to tackle showing that the closure of it's interior is the set.
For example, how would you show the set $S = x^2 + y^2 <= 1$ is a regular closed region? It's boundary is $x^2 + y^2 = 1$, but I'm not sure how to rigorously prove that the closure of the interior is the set itself.
Thank you.
The interior of $S$ is the union of all open balls contained entirely inside $S$. From this it follows that the interior of $S$ is the set $$\textrm{int }S=\bigl\{(x,y)\in\mathbb R^2\colon x^2+y^2<1\bigr\}.$$ (To prove it rigorously, you would need to show two things each of which is straightforward: 1. for every $(x,y)$ satisfying $x^2+y^2<1$ there exists an $r>0$ such that the ball centered at $(x,y)$ of radius $r$ is entirely contained inside $S$, and 2. for every $(x,y)$ satisfying $x^2+y^2=1$ there does not exist any $r>0$ such that the ball centered at $(x,y)$ of radius $r$ is entirely contained inside $S$.)
Now to find the closure of this set, you can use the fact that the complement of the closure is the interior of the complement, and apply the same argument as I just described but now to the complementary set $$ (\textrm{int } S)^c=\bigl\{(x,y)\in\mathbb R^2\colon x^2+y^2\geq 1\bigr\}. $$ You will find that $$ \textrm{int }\bigl((\textrm{int } S)^c\bigr)=\bigl\{(x,y)\in\mathbb R^2\colon x^2+y^2> 1\bigr\}=S^c, $$ and therefore the closure of $\textrm{int }S$ equals $S$.