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).

Re: Question about open extensions of sets in metric spaces.

Quote:

Originally Posted by

**RaisinBread** 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).

When you say that you are unable "to construct a rigorous argument", does that mean that you cannot even start?

Can you show that $\displaystyle (A_t)_s\subseteq A_{t+s}~?$

Before I work on this, please post what you have been able to do.