# Question about open extensions of sets in metric spaces.

• Jun 14th 2013, 05:28 AM
Question about open extensions of sets in metric spaces.
Let $\displaystyle (X,d)$ be an arbitrary metric space. For every $\displaystyle A\subset X$ and $\displaystyle t>0$, we define the open $\displaystyle t$-extension of $\displaystyle A$ as the set

$\displaystyle A_t=\{x\in X:d(x,A)<t\}$,

where $\displaystyle d(x,A)=\inf\{d(x,y):y\in A\}$.

It seems intuitive that, given $\displaystyle A\subset X$ and $\displaystyle s,t>0$, we should have that $\displaystyle A_{t+s}=(A_t)_s=(A_s)_t$. However, I haven't been able to even begin to construct a rigorous argument to prove it. Any hint or proposed direction would be greatly appreciated (or counterexample in the event that it is false).
• Jun 14th 2013, 06:00 AM
Plato
Re: Question about open extensions of sets in metric spaces.
Quote:

Let $\displaystyle (X,d)$ be an arbitrary metric space. For every $\displaystyle A\subset X$ and $\displaystyle t>0$, we define the open $\displaystyle t$-extension of $\displaystyle A$ as the set
$\displaystyle A_t=\{x\in X:d(x,A)<t\}$,
where $\displaystyle d(x,A)=\inf\{d(x,y):y\in A\}$.
It seems intuitive that, given $\displaystyle A\subset X$ and $\displaystyle s,t>0$, we should have that $\displaystyle A_{t+s}=(A_t)_s=(A_s)_t$. However, I haven't been able to even begin to construct a rigorous argument to prove it. Any hint or proposed direction would be greatly appreciated (or counterexample in the event that it is false).
Can you show that $\displaystyle (A_t)_s\subseteq A_{t+s}~?$