[SOLVED] Metric Space- Distance Function

**tidus89** · December 15th 2009, 04:36 PM

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.
Thanks in Advance.

**Drexel28** · December 15th 2009, 08:32 PM

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.
Thanks in Advance.

This notation is hard to understand (at least for me), but I would tend to agree. If $D>1$ then $\frac{D}{1+D}=1-\frac{1}{1+D}<1=\min\left(D,1\right)$

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