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Cover of the domain of the characteristic function [closed]

0

$begingroup$

Let $X$ be a locally compact Hausdorff space and $f:Xrightarrow mathbbR$
a continuous function with compact support,$chi_A$ the characteristic function and $A$ a subset of $X$ .

Assume that $chi_A ≤ f$ and let $epsilon in(0,1)$, and define $U_epsilon$ by $U_epsilon = x ∈ X : f (x) > 1 − epsilon$,
how can we proof that that $U_epsilon$ covers $A$ ?

real-analysis general-topology

share|cite|improve this question

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

$endgroup$

closed as off-topic by Kavi Rama Murthy, Thomas Shelby, Javi, José Carlos Santos, mrtaurho Mar 20 at 21:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Thomas Shelby, Javi, José Carlos Santos, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Kavi Rama Murthy
  Mar 20 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $A in U_epsilon $ since $ chi_A ≤ f$ and by the definition of $U_epsilon$
  $endgroup$
  – amilton moreira
  Mar 20 at 10:39
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $X$ be a locally compact Hausdorff space and $f:Xrightarrow mathbbR$
a continuous function with compact support,$chi_A$ the characteristic function and $A$ a subset of $X$ .

Assume that $chi_A ≤ f$ and let $epsilon in(0,1)$, and define $U_epsilon$ by $U_epsilon = x ∈ X : f (x) > 1 − epsilon$,
how can we proof that that $U_epsilon$ covers $A$ ?

real-analysis general-topology

share|cite|improve this question

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

$endgroup$

closed as off-topic by Kavi Rama Murthy, Thomas Shelby, Javi, José Carlos Santos, mrtaurho Mar 20 at 21:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Kavi Rama Murthy, Thomas Shelby, Javi, José Carlos Santos, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What have you tried?
  $endgroup$
  – Kavi Rama Murthy
  Mar 20 at 10:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $A in U_epsilon $ since $ chi_A ≤ f$ and by the definition of $U_epsilon$
  $endgroup$
  – amilton moreira
  Mar 20 at 10:39
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ be a locally compact Hausdorff space and $f:Xrightarrow mathbbR$
a continuous function with compact support,$chi_A$ the characteristic function and $A$ a subset of $X$ .

Assume that $chi_A ≤ f$ and let $epsilon in(0,1)$, and define $U_epsilon$ by $U_epsilon = x ∈ X : f (x) > 1 − epsilon$,
how can we proof that that $U_epsilon$ covers $A$ ?

real-analysis general-topology

share|cite|improve this question

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

$endgroup$

share|cite|improve this question

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

share|cite|improve this question

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

asked Mar 20 at 10:21

amilton moreiraamilton moreira

1179

1 Answer
1

active

oldest

votes

1

$begingroup$

$x in A$ implies $f(x) ge 1 > 1-varepsilon$ (as $chi_A le f$ and $chi_A(x)=1$ for $x in A$) so $x in U_varepsilon$ showing $A subseteq U_varepsilon$.

share|cite|improve this answer

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

$endgroup$

add a comment | 

1

$begingroup$

$x in A$ implies $f(x) ge 1 > 1-varepsilon$ (as $chi_A le f$ and $chi_A(x)=1$ for $x in A$) so $x in U_varepsilon$ showing $A subseteq U_varepsilon$.

share|cite|improve this answer

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

$endgroup$

add a comment | 

1

$begingroup$

$x in A$ implies $f(x) ge 1 > 1-varepsilon$ (as $chi_A le f$ and $chi_A(x)=1$ for $x in A$) so $x in U_varepsilon$ showing $A subseteq U_varepsilon$.

share|cite|improve this answer

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

$endgroup$

add a comment | 

$begingroup$

$x in A$ implies $f(x) ge 1 > 1-varepsilon$ (as $chi_A le f$ and $chi_A(x)=1$ for $x in A$) so $x in U_varepsilon$ showing $A subseteq U_varepsilon$.

share|cite|improve this answer

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

$endgroup$

share|cite|improve this answer

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124

answered Mar 20 at 17:42

Henno BrandsmaHenno Brandsma

114k348124