Is a subspace of a k-space is a k-space?

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

My definition of $k$-space is:

Let be $(X,tau)$ a topological subspace and $Asubset X$.

$A$ is a $k$-closed in $X$ if $forall Ksubset X$ compact it happens that $Acap K$ is closed in $K$.

$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.

My question is the following:

I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.

Let be $(X,tau)$ a $k$-space and $Asubset X$ closed, then $(A,tau_A)$ is a $k$-space.

Proof:

Let be $Bsubset A$ a $k$-closed.

Then $forall Ksubset A$ compact it happens that $Bcap K$ is closed in $K$.

As $Ksubset A subset X$ and compactness is an absolute property, then $K$ is compact in $X$

$Rightarrow$ $forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$

$Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.

Finally $overlineB^A=overlineB^Xcap A=Bcap A=B$

$Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space

Obs: I mean absolute compactness for the following:

Theorem: Let be $(X,tau)$ a topological space and $Asubset X$, then $Bsubset A$ is compact in $A$ if and only if $B$ is compact in $X$.

general-topology

share|cite|improve this question

edited Jul 20 at 3:23

asked Jul 19 at 23:34

Israel

589

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
  – Oliver Jones
  Jul 20 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
  – bof
  Jul 20 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
  – Israel
  Jul 20 at 3:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
  – bof
  Jul 20 at 3:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
  – bof
  Jul 20 at 3:36
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

My definition of $k$-space is:

Let be $(X,tau)$ a topological subspace and $Asubset X$.

$A$ is a $k$-closed in $X$ if $forall Ksubset X$ compact it happens that $Acap K$ is closed in $K$.

$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.

My question is the following:

I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.

Let be $(X,tau)$ a $k$-space and $Asubset X$ closed, then $(A,tau_A)$ is a $k$-space.

Proof:

Let be $Bsubset A$ a $k$-closed.

Then $forall Ksubset A$ compact it happens that $Bcap K$ is closed in $K$.

As $Ksubset A subset X$ and compactness is an absolute property, then $K$ is compact in $X$

$Rightarrow$ $forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$

$Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.

Finally $overlineB^A=overlineB^Xcap A=Bcap A=B$

$Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space

Obs: I mean absolute compactness for the following:

Theorem: Let be $(X,tau)$ a topological space and $Asubset X$, then $Bsubset A$ is compact in $A$ if and only if $B$ is compact in $X$.

general-topology

share|cite|improve this question

edited Jul 20 at 3:23

asked Jul 19 at 23:34

Israel

589

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
  – Oliver Jones
  Jul 20 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
  – bof
  Jul 20 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
  – Israel
  Jul 20 at 3:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
  – bof
  Jul 20 at 3:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
  – bof
  Jul 20 at 3:36
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

up vote
0
down vote

favorite

My definition of $k$-space is:

Let be $(X,tau)$ a topological subspace and $Asubset X$.

$A$ is a $k$-closed in $X$ if $forall Ksubset X$ compact it happens that $Acap K$ is closed in $K$.

$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.

My question is the following:

I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.

Let be $(X,tau)$ a $k$-space and $Asubset X$ closed, then $(A,tau_A)$ is a $k$-space.

Proof:

Let be $Bsubset A$ a $k$-closed.

Then $forall Ksubset A$ compact it happens that $Bcap K$ is closed in $K$.

As $Ksubset A subset X$ and compactness is an absolute property, then $K$ is compact in $X$

$Rightarrow$ $forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$

$Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.

Finally $overlineB^A=overlineB^Xcap A=Bcap A=B$

$Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space

Obs: I mean absolute compactness for the following:

Theorem: Let be $(X,tau)$ a topological space and $Asubset X$, then $Bsubset A$ is compact in $A$ if and only if $B$ is compact in $X$.

general-topology

share|cite|improve this question

edited Jul 20 at 3:23

asked Jul 19 at 23:34

Israel

589

My definition of $k$-space is:

Let be $(X,tau)$ a topological subspace and $Asubset X$.

$A$ is a $k$-closed in $X$ if $forall Ksubset X$ compact it happens that $Acap K$ is closed in $K$.

$X$ is a k-space if every $k$-closed of $X$ is closed in $X$.

My question is the following:

I was trying to prove that every closed subspace of a $k$-space is a $k$-space, but in my demonstration I do not use that subspace to be closed. If someone can gave me a clue that is wrong in my demonstration I would appreciate it.

Let be $(X,tau)$ a $k$-space and $Asubset X$ closed, then $(A,tau_A)$ is a $k$-space.

Proof:

Let be $Bsubset A$ a $k$-closed.

Then $forall Ksubset A$ compact it happens that $Bcap K$ is closed in $K$.

As $Ksubset A subset X$ and compactness is an absolute property, then $K$ is compact in $X$

$Rightarrow$ $forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$

$Rightarrow$ $B$ is a $k$-closed of $X$ and $X$ is a $k$-space, then $B$ is closed in $X$.

Finally $overlineB^A=overlineB^Xcap A=Bcap A=B$

$Rightarrow$ $B$ is closed in $A$ and therefore $A$ is a $k$-space

Obs: I mean absolute compactness for the following:

Theorem: Let be $(X,tau)$ a topological space and $Asubset X$, then $Bsubset A$ is compact in $A$ if and only if $B$ is compact in $X$.

general-topology

share|cite|improve this question

edited Jul 20 at 3:23

asked Jul 19 at 23:34

Israel

589

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 3:23

edited Jul 20 at 3:23

edited Jul 20 at 3:23

asked Jul 19 at 23:34

Israel

589

asked Jul 19 at 23:34

Israel

589

asked Jul 19 at 23:34

Israel

589

589

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
  – Oliver Jones
  Jul 20 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
  – bof
  Jul 20 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
  – Israel
  Jul 20 at 3:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
  – bof
  Jul 20 at 3:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
  – bof
  Jul 20 at 3:36
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
  – Oliver Jones
  Jul 20 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
  – bof
  Jul 20 at 0:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
  – Israel
  Jul 20 at 3:27
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
  – bof
  Jul 20 at 3:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
  – bof
  Jul 20 at 3:36
  

  
  
  
  

This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
– Oliver Jones
Jul 20 at 0:10

This question was answered here math.stackexchange.com/q/1416556/55622. Compare proofs and maybe you can find your mistake.
– Oliver Jones
Jul 20 at 0:10

Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
– bof
Jul 20 at 0:25

Your statement "$forall K$ compact of $X$ it happens that $Bcap K$ is closed in $K$" does not seem to follow from anything stated previously.
– bof
Jul 20 at 0:25

what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
– Israel
Jul 20 at 3:27

what I would like to show is that $B$ is a $k$-closed of $X$ to use the hypothesis that $X$ is a $k$-space and conclude that $B$ is closed in $X$ and then use the paragraph that says finally in my demonstration. But to see that $B$ is a $k$-closed in $X$, I need to see that for all compact $K$ of $X$ (not compact in $A$) the intersection $Bcap K$ is closed in $K$ and according to me that's where I use that compactness is an absolute property.
– Israel
Jul 20 at 3:27

1

1

Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
– bof
Jul 20 at 3:32

Suppose $K$ is a compact subset of $X,$ how do you know that $Bcap K$ is closed in $K?$ If $Ksubset A$ then $K$ is a compact subset of $A,$ so you can use the fact that $B$ is $k$-closed in $A;$ but what if $Knotsubset A?$
– bof
Jul 20 at 3:32

Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
– bof
Jul 20 at 3:36

Now, if $A$ happens to be closed, then $Kcap B$ is compact so you can use that; but you say you are not assuming that $A$ is closed.
– bof
Jul 20 at 3:36

 | 
show 1 more comment

active

oldest

votes

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%2f2857141%2fis-a-subspace-of-a-k-space-is-a-k-space%23new-answer', 'question_page');

);

Post as a guest

Name Email

active

oldest

votes

active

oldest

votes

active

oldest

votes

active

oldest

votes

 

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%2f2857141%2fis-a-subspace-of-a-k-space-is-a-k-space%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