I'm trying to prove that in a normed vector space the boundary of a set A is a subset of the boundary of the closure of A. I've been using the definition of boundary and concluded that if $x\in boundary(A)$ then the intersection of the ball of radius r>0 centered at $x$ with the closure of A is always non empty, but I'm having trouble proving that the intersection of the ball with the complement of the closure of A is non empty.
Any tips?
Thanks!
[Math] The boundary of a set is subset of the boundary of the closure of the set.
general-topology
Best Answer
Tip: It's not true.
Counterexample: Notice that $\mathbb{R}$ is a normed vector space under the Euclidean norm. Let $A$ be the set $[0,1] \smallsetminus \{1/2\}$, the closed unit interval without its midpoint.
The boundary of $A$ is $\{0,1/2,1\}$. The closure of $A$ is $[0,1]$, the closed unit interval. The boundary of the closure is $\{0,1\}$. Since $1/2 \not \in \{0,1\}$, the boundary of $A$ is not a subset of the boundary of the closure of $A$.