Show that the open ball $B(0,1) = \{(x,y): x^2 +y^2 < 1\}$ has the boundary $x^2+y^2=1$. I understand that the boundary is the closure of the ball minus the interior. So, if i can show that the closure of the ball is less than or equal to one, then i would essentially be done, but im not sure how to do this. I know the closure contains the limit points of the set, but im not sure how to use this to help me prove what the closure of the ball is.
[Math] Boundary of a ball
general-topologymetric-spacesreal-analysis
Best Answer
If you show that some point $z\notin A$ is a limit point, then it must be a boundary point since each neighborhood contains a point in $A$ and also $z\notin A$ itself. On the other hand, if $z$ is a boundary point then each neighborhood intersects $A$, thus $A-\{z\}$, so it is a limit point. That means that outside of $A$ the limit points and the boundary point coincide.
Take $z=(x,y)$ such that $x^2+y^2=1$ and $\epsilon>0$. For a $\lambda>1-\epsilon$, show that $d(z,\lambda z)=||z-\lambda z||<\epsilon$ and $d(0,\lambda z)=||\lambda z||<1$. Then you know that $\lambda z\in B_\epsilon(z)\cap B_1(0)$, and since $\epsilon$ was arbitrary, $z$ is a limit point of the open ball.
What remains is to prove that each $z$ with $||z||>0$ is not a limit point, i.e. there is an $\epsilon$ such that $B_\epsilon(z)$ is disjoint from the ball.