How does ${X}_{w}\subset {X^*}_{w}$ occur in modular metric space?

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SUMMARY

The discussion centers on the inclusion of sets ${X}_{w}$ and ${X^*}_{w}$ in modular metric spaces, specifically demonstrating that ${X}_{w} \subset {X^*}_{w}$. The sets are defined as follows: ${X}_{w} = \{x \in X : d_{\lambda}(x, x_0) \to 0 \text{ as } \lambda \to \infty\}$ and ${X^*}_{w} = \{x \in X : \exists \lambda > 0 \text{ such that } d_{w}(x, x_0) < \infty\}$. The proof approach suggested involves demonstrating that if an element is not in ${X^*}_{w}$, it cannot be in ${X}_{w}$, thus establishing the inclusion. The discussion also touches on the definitions of metrics and the implications of the metric space being a subset of real numbers.

PREREQUISITES
  • Understanding of modular metric spaces
  • Familiarity with the concept of convergence in metric spaces
  • Knowledge of metric definitions, specifically ${d}_{\lambda}$ and ${d}_{w}$
  • Basic principles of set theory and subset relations
NEXT STEPS
  • Study the properties of modular metric spaces and their applications
  • Learn about convergence criteria in metric spaces
  • Examine the implications of different metrics on set inclusions
  • Review relevant literature, such as the paper referenced (arXiv:1112.5561v1) for deeper insights
USEFUL FOR

Mathematicians, researchers in topology, and students studying advanced metric space theory will benefit from this discussion, particularly those interested in modular metric spaces and their properties.

ozkan12
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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...
 
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Well, in fact the inclusion is quite intuitive. To prove it you could do the following. In stead of proving: $x \in X_w \Rightarrow x \in X_w^{*}$ you prove $x \notin X_w^{*} \Rightarrow x \notin X_w$. It will be straightforward then. Do you see?
 
Dear Siron,

First of all, thank you for your attention...But I couldn't prove this...Best wishes..
 
ozkan12 said:
Let $d$ be a metric on $X$. Fix ${x}_{0}\in X$. Let ${d}_{\lambda}\left(x,y\right)=\frac{1}\lambda{}\left| x-y \right|$
What is $|x-y|$? Is it the case that $X\subseteq\Bbb R$?

ozkan12 said:
${X}_{w}={X}_{w}\left({x}_{0}\right)=\left\{x\in X:{d}_{\lambda}\left(x,{x}_{0}\right)\to0\left( as \lambda\to\infty\right) \right \}$
Doesn't this mean that $X_w=X$ since it is always the case that $\frac{1}{\lambda}d(x,y)\to 0$ as $\lambda\to\infty$?

ozkan12 said:
${X^*}_{w}={X^*}_{w}\left({x}_{0}\right)=\left\{x\in X:\exists\lambda=\lambda\left(x\right)>0 such that {d}_{w}\left(x,{{x}_{0}}\right)<\infty\right\}$
What is $d_w$?
 
$X\subseteq R$ and ${d}_{w}$ is wrong, I wrote wrong it, it must be ${d}_{\lambda}$
 
ozkan12 said:
$X\subseteq R$
Then why do you write, "Let $d$ be a metric on $X$" and never use $d$ afterwards?

I believe both $X_w$ and $X_w^*$ are equal to $X$. For $X_w^*$ and $x\in X$, take $\lambda=1$; then $d_\lambda(x,x_0)=|x-x_0|<\infty$, so $x\in X_w^*$.
 
No paper '''arXiv:1112.5561v1.pdf...''' this link, there is something related to modular metric space...And in page 6 you will see definition 2.2...Please can you explain that how happened ${X}_{w}\subset {X^*}_{w}$ ? Thank you for your attention...Best wishes...
 

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