Compactess of the sum closed balls whose centers is a compact set.

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Let $(mathcalX,d) $ be a compact and metric space and let $A subset mathcalX$ be a compact subset and let $alpha >0 $. Consider the set
$$
C(A, alpha) = x in mathcalX ,
$$
How to prove or disprove that $C(A, alpha) $ is compact?

Thanks in advance.

general-topology compactness

share|cite|improve this question

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

asked Aug 2 at 16:39

jaogye

405413

add a comment | 

up vote
0
down vote

favorite

Let $(mathcalX,d) $ be a compact and metric space and let $A subset mathcalX$ be a compact subset and let $alpha >0 $. Consider the set
$$
C(A, alpha) = x in mathcalX ,
$$
How to prove or disprove that $C(A, alpha) $ is compact?

Thanks in advance.

general-topology compactness

share|cite|improve this question

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

asked Aug 2 at 16:39

jaogye

405413

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $(mathcalX,d) $ be a compact and metric space and let $A subset mathcalX$ be a compact subset and let $alpha >0 $. Consider the set
$$
C(A, alpha) = x in mathcalX ,
$$
How to prove or disprove that $C(A, alpha) $ is compact?

Thanks in advance.

general-topology compactness

share|cite|improve this question

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

asked Aug 2 at 16:39

jaogye

405413

Let $(mathcalX,d) $ be a compact and metric space and let $A subset mathcalX$ be a compact subset and let $alpha >0 $. Consider the set
$$
C(A, alpha) = x in mathcalX ,
$$
How to prove or disprove that $C(A, alpha) $ is compact?

Thanks in advance.

general-topology compactness

share|cite|improve this question

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

asked Aug 2 at 16:39

jaogye

405413

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

edited Aug 2 at 16:42

José Carlos Santos

112k1696172

112k1696172

asked Aug 2 at 16:39

jaogye

405413

asked Aug 2 at 16:39

jaogye

405413

asked Aug 2 at 16:39

jaogye

405413

405413

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

The set $C(A,alpha)$ is closed because the function$$beginarrayrcccDeltacolon&mathcalX&longrightarrow&mathbb R\&a&mapsto&d(a,A)endarray$$is continuous and $C(A,alpha)=Delta^-1bigl([0,alpha]bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

share|cite|improve this answer

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
  – jaogye
  Aug 2 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I've edited my answer. I hope that everything is clear now.
  – José Carlos Santos
  Aug 2 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos many thanks for answer.
  – jaogye
  Aug 3 at 6:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I'm glad I could help.
  – José Carlos Santos
  Aug 3 at 7:30
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870262%2fcompactess-of-the-sum-closed-balls-whose-centers-is-a-compact-set%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The set $C(A,alpha)$ is closed because the function$$beginarrayrcccDeltacolon&mathcalX&longrightarrow&mathbb R\&a&mapsto&d(a,A)endarray$$is continuous and $C(A,alpha)=Delta^-1bigl([0,alpha]bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

share|cite|improve this answer

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
  – jaogye
  Aug 2 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I've edited my answer. I hope that everything is clear now.
  – José Carlos Santos
  Aug 2 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos many thanks for answer.
  – jaogye
  Aug 3 at 6:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I'm glad I could help.
  – José Carlos Santos
  Aug 3 at 7:30
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

The set $C(A,alpha)$ is closed because the function$$beginarrayrcccDeltacolon&mathcalX&longrightarrow&mathbb R\&a&mapsto&d(a,A)endarray$$is continuous and $C(A,alpha)=Delta^-1bigl([0,alpha]bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

share|cite|improve this answer

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
  – jaogye
  Aug 2 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I've edited my answer. I hope that everything is clear now.
  – José Carlos Santos
  Aug 2 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos many thanks for answer.
  – jaogye
  Aug 3 at 6:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I'm glad I could help.
  – José Carlos Santos
  Aug 3 at 7:30
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The set $C(A,alpha)$ is closed because the function$$beginarrayrcccDeltacolon&mathcalX&longrightarrow&mathbb R\&a&mapsto&d(a,A)endarray$$is continuous and $C(A,alpha)=Delta^-1bigl([0,alpha]bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

share|cite|improve this answer

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

The set $C(A,alpha)$ is closed because the function$$beginarrayrcccDeltacolon&mathcalX&longrightarrow&mathbb R\&a&mapsto&d(a,A)endarray$$is continuous and $C(A,alpha)=Delta^-1bigl([0,alpha]bigr)$.

Now, use the fact that a closed subset of a compact metric space is always compact.

share|cite|improve this answer

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

share|cite|improve this answer

share|cite|improve this answer

edited Aug 2 at 20:38

edited Aug 2 at 20:38

edited Aug 2 at 20:38

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

answered Aug 2 at 16:41

José Carlos Santos

112k1696172

112k1696172

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
  – jaogye
  Aug 2 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I've edited my answer. I hope that everything is clear now.
  – José Carlos Santos
  Aug 2 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos many thanks for answer.
  – jaogye
  Aug 3 at 6:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I'm glad I could help.
  – José Carlos Santos
  Aug 3 at 7:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
  – jaogye
  Aug 2 at 17:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I've edited my answer. I hope that everything is clear now.
  – José Carlos Santos
  Aug 2 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Dear @José Carlos Santos many thanks for answer.
  – jaogye
  Aug 3 at 6:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @jaogye I'm glad I could help.
  – José Carlos Santos
  Aug 3 at 7:30
  

  
  
  
  

Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
– jaogye
Aug 2 at 17:39

Dear @José Carlos Santos ,Why $C(A,alpha)$ is closed?. So far I know $C(A,alpha)$ is an uncontable sum of closed balls. But any uncontable sum of closed balls not necessary it closed.
– jaogye
Aug 2 at 17:39

@jaogye I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Aug 2 at 20:39

@jaogye I've edited my answer. I hope that everything is clear now.
– José Carlos Santos
Aug 2 at 20:39

Dear @José Carlos Santos many thanks for answer.
– jaogye
Aug 3 at 6:58

Dear @José Carlos Santos many thanks for answer.
– jaogye
Aug 3 at 6:58

@jaogye I'm glad I could help.
– José Carlos Santos
Aug 3 at 7:30

@jaogye I'm glad I could help.
– José Carlos Santos
Aug 3 at 7:30

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870262%2fcompactess-of-the-sum-closed-balls-whose-centers-is-a-compact-set%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email