- #1
ozkan12
- 149
- 0
Let $d$ be a metric on $X$. Fix ${x}_{0}\in X$. Let ${d}_{\lambda}\left(x,y\right)=\frac{1}\lambda{}\left| x-y \right|$ and The two sets
${X}_{w}={X}_{w}\left({x}_{0}\right)=\left\{x\in X:{d}_{\lambda}\left(x,{x}_{0}\right)\to0\left( as \lambda\to\infty\right) \right \}$
and
${X^*}_{w}={X^*}_{w}\left({x}_{0}\right)=\left\{x\in X:\exists\lambda=\lambda\left(x\right)>0 such that {d}_{w}\left(x,{{x}_{0}}\right)<\infty\right\}$
Then, it is clear that ${X}_{w}\subset{X^*}_{w}$...How this happens ? Please can you explain ? Thank you for your attention...Best wishes...
${X}_{w}={X}_{w}\left({x}_{0}\right)=\left\{x\in X:{d}_{\lambda}\left(x,{x}_{0}\right)\to0\left( as \lambda\to\infty\right) \right \}$
and
${X^*}_{w}={X^*}_{w}\left({x}_{0}\right)=\left\{x\in X:\exists\lambda=\lambda\left(x\right)>0 such that {d}_{w}\left(x,{{x}_{0}}\right)<\infty\right\}$
Then, it is clear that ${X}_{w}\subset{X^*}_{w}$...How this happens ? Please can you explain ? Thank you for your attention...Best wishes...