Fixp inX, r > 0, define A to be the set of all q inX for which
d(p, q) < r, define B similarly, with > in place of <. Prove that A and
B are separated.
Wont the closure of A be { q in X | d(p,q) <=r}
and closure of B be {q in X | d(p,q) >=r }
Fixp inX, r > 0, define A to be the set of all q inX for which
d(p, q) < r, define B similarly, with > in place of <. Prove that A and
B are separated.
Wont the closure of A be { q in X | d(p,q) <=r}
and closure of B be {q in X | d(p,q) >=r }
Not necessarily. If $\displaystyle d$ is the discrete metric, and $\displaystyle r=1$, the statement fails.Wont the closure of A be { q in X | d(p,q) <=r}
and closure of B be {q in X | d(p,q) >=r }
You might be able to show that the closure is contained in there, however.
Note that in a metric space $\displaystyle \left(\mathcal{M},d\right)$ if $\displaystyle E\subseteq\mathcal{M}$ then $\displaystyle \text{cl}_\mathcal{M}\text{ }E=\left\{x\in\mathcal{M}:d(x,E)=0\right\}$...so what?
Also, I'm sure you've proven that if $\displaystyle A,B$ are disjoint open subspaces of a metric space they're separated, right? So, note that $\displaystyle \varphi:\mathcal{M}\to\mathbb{R}:x\mapsto d(x,p)$ is continuous and you're to sets are $\displaystyle \varphi^{-1}((-\infty,r)),\varphi^{-1}((r,\infty))$ so that they are disjoint open.