# [SOLVED] Metric Space- Distance Function

• Dec 15th 2009, 04:36 PM
tidus89
[SOLVED] Metric Space- Distance Function
Hey guys, I have a problem from my Topology course that I can't seem to figure out. Any help would be appreciated. Here it is:

The distance functions D'=D/(1+D) and D''= min{D,1} are each metrics on M. Let x $\in$M and let $\epsilon \in$ R>0.
Let:

S $\epsilon$'(x)={y $\in$M|D'(x,y)< $\epsilon$}
S $\epsilon$''(x)={y $\in$M|D''(x,y)< $\epsilon$}

a) Prove that S $\epsilon$''(x) $\subseteq$ S $\epsilon]$'(x).
b)Prove that S $(\epsilon/(1+\epsilon))$'(x) $\subseteq$ S $\epsilon$''(x).

It almost seems to me like the containment should be switched in part a at least, but I don't know.
• Dec 15th 2009, 08:32 PM
Drexel28
Quote:

Originally Posted by tidus89
Hey guys, I have a problem from my Topology course that I can't seem to figure out. Any help would be appreciated. Here it is:

The distance functions D'=D/(1+D) and D''= min{D,1} are each metrics on M. Let x $\in$M and let $\epsilon \in$ R>0.
Let:

S $\epsilon$'(x)={y $\in$M|D'(x,y)< $\epsilon$}
S $\epsilon$''(x)={y $\in$M|D''(x,y)< $\epsilon$}

a) Prove that S $\epsilon$''(x) $\subseteq$ S $\epsilon]$'(x).
b)Prove that S $(\epsilon/(1+\epsilon))$'(x) $\subseteq$ S $\epsilon$''(x).

It almost seems to me like the containment should be switched in part a at least, but I don't know.
This notation is hard to understand (at least for me), but I would tend to agree. If $D>1$ then $\frac{D}{1+D}=1-\frac{1}{1+D}<1=\min\left(D,1\right)$