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Separability of collection of functions

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0

$begingroup$

Let $f: mathcalX times Theta to mathbbR$ be a function that is continuous in both $x$ and $theta$. Suppose that $Theta subset mathbbR^d_theta$ is compact. Consider the class of functions:
$$mathcalG:= g : mathcalX to mathbbR :$$

Suppose I equip $Theta$ with the euclidean norm and $mathcalG$ with the sup-norm.

Since $Theta$ is compact and $f(cdot,theta)$ is continuous in $theta$, I am wondering if we can deduce that $mathcalG$ is separable?

I do not know if this is true, but here is my try:

Compactness of $Theta$ implies that every $varepsilon-$net has a finite subnet. Let $theta_k_k=1^K$ denote the centers of the balls of radius $varepsilon$ that cover $Theta$. By continuity of $f(cdot,theta)$ we know that for every $eta>0$ there exists a $varepsilon>0$ such that $||theta - theta'||< varepsilon$ implies that $||f(cdot,theta) - f(cdot,theta')||<eta$. By adjusting $varepsilon>0$ from the first part, if necessary, we can thus obtain that $||theta - theta_k||< varepsilon$ implies $||f(cdot,theta) - f(cdot,theta_k)||< eta$. Setting $g_k=f(cdot,theta_k)$ we thus have that $g_k_k=1^K$ forms an $eta-$cover of $mathcalG$. Since $eta>0$ was arbitrary, we thus have thus shown that for every $eta>0$ there exists a finite collection $g_k_k=1^K$ such that any $g in mathcalG$ is at most $eta$ away form some $g_k_k=1^K$.

real-analysis

share|cite|improve this question

asked Mar 17 at 14:05

möbiusmöbius

901921

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should the sup norm be finite on $X?$
  $endgroup$
  – zhw.
  Mar 17 at 14:50
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $f: mathcalX times Theta to mathbbR$ be a function that is continuous in both $x$ and $theta$. Suppose that $Theta subset mathbbR^d_theta$ is compact. Consider the class of functions:
$$mathcalG:= g : mathcalX to mathbbR :$$

Suppose I equip $Theta$ with the euclidean norm and $mathcalG$ with the sup-norm.

Since $Theta$ is compact and $f(cdot,theta)$ is continuous in $theta$, I am wondering if we can deduce that $mathcalG$ is separable?

I do not know if this is true, but here is my try:

Compactness of $Theta$ implies that every $varepsilon-$net has a finite subnet. Let $theta_k_k=1^K$ denote the centers of the balls of radius $varepsilon$ that cover $Theta$. By continuity of $f(cdot,theta)$ we know that for every $eta>0$ there exists a $varepsilon>0$ such that $||theta - theta'||< varepsilon$ implies that $||f(cdot,theta) - f(cdot,theta')||<eta$. By adjusting $varepsilon>0$ from the first part, if necessary, we can thus obtain that $||theta - theta_k||< varepsilon$ implies $||f(cdot,theta) - f(cdot,theta_k)||< eta$. Setting $g_k=f(cdot,theta_k)$ we thus have that $g_k_k=1^K$ forms an $eta-$cover of $mathcalG$. Since $eta>0$ was arbitrary, we thus have thus shown that for every $eta>0$ there exists a finite collection $g_k_k=1^K$ such that any $g in mathcalG$ is at most $eta$ away form some $g_k_k=1^K$.

real-analysis

share|cite|improve this question

asked Mar 17 at 14:05

möbiusmöbius

901921

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why should the sup norm be finite on $X?$
  $endgroup$
  – zhw.
  Mar 17 at 14:50
  

  
  
  
  

add a comment | 

$begingroup$

Let $f: mathcalX times Theta to mathbbR$ be a function that is continuous in both $x$ and $theta$. Suppose that $Theta subset mathbbR^d_theta$ is compact. Consider the class of functions:
$$mathcalG:= g : mathcalX to mathbbR :$$

Suppose I equip $Theta$ with the euclidean norm and $mathcalG$ with the sup-norm.

Since $Theta$ is compact and $f(cdot,theta)$ is continuous in $theta$, I am wondering if we can deduce that $mathcalG$ is separable?

I do not know if this is true, but here is my try:

Compactness of $Theta$ implies that every $varepsilon-$net has a finite subnet. Let $theta_k_k=1^K$ denote the centers of the balls of radius $varepsilon$ that cover $Theta$. By continuity of $f(cdot,theta)$ we know that for every $eta>0$ there exists a $varepsilon>0$ such that $||theta - theta'||< varepsilon$ implies that $||f(cdot,theta) - f(cdot,theta')||<eta$. By adjusting $varepsilon>0$ from the first part, if necessary, we can thus obtain that $||theta - theta_k||< varepsilon$ implies $||f(cdot,theta) - f(cdot,theta_k)||< eta$. Setting $g_k=f(cdot,theta_k)$ we thus have that $g_k_k=1^K$ forms an $eta-$cover of $mathcalG$. Since $eta>0$ was arbitrary, we thus have thus shown that for every $eta>0$ there exists a finite collection $g_k_k=1^K$ such that any $g in mathcalG$ is at most $eta$ away form some $g_k_k=1^K$.

real-analysis

share|cite|improve this question

asked Mar 17 at 14:05

möbiusmöbius

901921

$endgroup$

share|cite|improve this question

asked Mar 17 at 14:05

möbiusmöbius

901921

share|cite|improve this question

asked Mar 17 at 14:05

möbiusmöbius

901921

asked Mar 17 at 14:05

möbiusmöbius

901921

asked Mar 17 at 14:05

möbiusmöbius

901921