I have this problem where I have to find the closure and the interior of the following set $$ A=\{\vec{x} \in \mathbb{R}^3 : x^2+y^2 \leq 1 \} $$
The graph of this is a circle in the space, which is laying on the plane XY, so to find the biggest open set contained in A, the interior, I propose it to be the ball $B_1(0) \subset \mathbb{R}^2 $ but If I take it to be radius one, I won't be able to cover the boundary of A and so to be fully contained in such ball the radious should be something like $1+\epsilon$ or maybe $1+ \frac{1}{n}$. On the other hand, to establish the closure means finding a closed set, the smallest such that is fully contained in A which is the same ball as before $B_1(0)$ but closed $\overline{B_1(0)}$.
The questions here are:
- Is it right to think that $int(A)=B_{1+\epsilon}(0) \subset \mathbb{R}^2$ and $\overline{A}=\overline{B_{1}(0)} \subset \mathbb{R}^2$ ?
- If so, how can I prove this?
I'm still trying to get used to working with topologies on metric spaces, and even though I think is much more visual than the abstract one, having to find closures and interiors of this kind of subsets aren't easy for me. Do you have any book you would recommend which focuses only on metric spaces topology?
Thank you in advance.
As the comments above noted, $A$ is closed is $\mathbb{R}^3$ with the usual Euclidean topology on it.
To see why this is true, note that $f : \mathbb{R}^3 \to \mathbb{R}$ (where $\mathbb{R}$ also has the standard Euclidean topology) defined by $f(x, y, z) = f(\vec{x}) = x^2 + y^2$ is a continuous map. Then observe that $A = f^{-1}\left[[0, 1]\right]$ and certainly $[0, 1]$ is closed in $\mathbb{R}$. By continuity of $f$ we then have $A$ to be closed in $\mathbb{R}^3$ because the preimage of a closed set is closed for any continuous function. Finally since $A$ is closed we can conclude that $\overline{A} = A$.