Let $(X, d)$ be a metric space and $A\subset X$. The interior of $A$ is defined as
$$\textrm{int}(A):=\{a\in X: \exists r>0; B(a, r)\subset A\},$$
where $B(a, r):=\{x\in X: d(x, a)<r\}$.
Is it true that $$\textrm{int}(B[a, r])=B(a, r)$$ where $B[a, r]:=\{x\in X: d(x, a)\leq r\}$?
The inclusion $B(a, r)\subset \textrm{int}(B[a, r])$ is straightfoward using the fact that $B(a, r)$ is open and $B(a, r)\subset B[a, r]$.
Does the reverse inclusion hold?
I believe we must suppose $x\not\in B(a, r)$ and show that $x\not\in \textrm{int}(B[a, r])$. If $x\not\in B(a, r)$ then $d(x, a)\geq r$. If $d(x, a)>r$ then it is easy to show $x\not\in \textrm{int}(B[a, r])$. However, what about the case $d(x, a)=r$? It is pretty much evident every open ball centered at $x$ will also contain points of $X\setminus B[a, r]$ because $x$ lies on the boundary of $B[a, r]$ but how to show this fact?
Remark: I'm not allowed to use sequences.
Best Answer
The equality need not hold.
Let $X=[-1,1]$ and endow $X$ with the discrete metric, $d(x,y) = 0$ if $x=y$, and $d(x,y)=1$ if $x\neq y$.
Then $B[0,1] = X$, $B(0,1)=\{0\}$. But the interior of $B[0,1]$ is $X\neq B(0,1)$.
(Or take $X=\mathbb{Z}$ with its usual distance, and look at $B[0,1]$ vs $B(0,1)$; essentially the same example, but countable)