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To show that $R^omega$ is not compact in box topology

2

$begingroup$

I want to show $R^omega$ is not compact under box topology. I am trying to find open cover which won't admit finite subcover but I am certainly lost. Can I take $A(n)$ as $(-infty,n) ×R×R$..... as nth element of open cover, their union seems to be $R^omega$ . Am I correct?

general-topology

share|cite|improve this question

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

$endgroup$

add a comment | 

2

$begingroup$

I want to show $R^omega$ is not compact under box topology. I am trying to find open cover which won't admit finite subcover but I am certainly lost. Can I take $A(n)$ as $(-infty,n) ×R×R$..... as nth element of open cover, their union seems to be $R^omega$ . Am I correct?

general-topology

share|cite|improve this question

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

$endgroup$

add a comment | 

$begingroup$

I want to show $R^omega$ is not compact under box topology. I am trying to find open cover which won't admit finite subcover but I am certainly lost. Can I take $A(n)$ as $(-infty,n) ×R×R$..... as nth element of open cover, their union seems to be $R^omega$ . Am I correct?

general-topology

share|cite|improve this question

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

$endgroup$

share|cite|improve this question

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

share|cite|improve this question

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

edited Mar 25 at 9:14

Henno Brandsma

116k349127

edited Mar 25 at 9:14

Henno Brandsma

116k349127

asked Mar 24 at 23:08

BelieverBeliever

646315

asked Mar 24 at 23:08

BelieverBeliever

646315

1 Answer
1

active

oldest

votes

1

$begingroup$

You are correct, that does the trick. They cover the whole space, are open by definition of the box topology, and taking any finite subcollection fails to cover $mathbbR^omega$. An alternative approach is to note that, since continuous functions sends compact spaces to compact spaces, $mathbbR^omega$ being compact would imply its image via a projection to be compact. However, $mathbbR simeq pi_1[mathbbR^omega]$ is not compact.

As a quick generalization, if $prod_i X_i$ is compact, each $X_i$ is compact. By the contrapositive then, any product of spaces in which one of them is not compact will never be compact itself. The reciprocal is much harder to prove in an arbitrary case, and it is known as Tychonoff's theorem.

share|cite|improve this answer

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got a doubt in my proof as why can't I use the same open cover under product topology to show it is not compact (I think I am missing something silly point) but as we know it is actually compact
  $endgroup$
  – Believer
  Mar 24 at 23:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh sorry R is not compact..shit
  $endgroup$
  – Believer
  Mar 24 at 23:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your cover is comprised of open sets which moreover belong to the product topology, so the argument follows. Also, the box topology is finer than the product topology, which means that projections are still continuous and the argument I gave still holds. In any case, $mathbbR^omega$ is not compact.
  $endgroup$
  – Guido A.
  Mar 24 at 23:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The argument of projections will work in any topology which makes them continuous, i.e. in any topology finer than the product topology. In simpler terms, a cover in a topology $tau$ with no finite subcovers will still be a cover with no finite subcovers in a topology $tau'$, if $tau subset tau'$.
  $endgroup$
  – Guido A.
  Mar 24 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A somewhat harder argument will show that no infinite box product can be compact (unless something trivial occurs, like taking all spaces indiscrete or all but finitely many of them singletons etc. ) likewise for connectedness. You’ll start to see why the standard product topology is a much better choice unless you want counterexamples.
  $endgroup$
  – Henno Brandsma
  Mar 25 at 9:17
  
  

  
  
  
  

 | 
show 2 more comments

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1

$begingroup$

You are correct, that does the trick. They cover the whole space, are open by definition of the box topology, and taking any finite subcollection fails to cover $mathbbR^omega$. An alternative approach is to note that, since continuous functions sends compact spaces to compact spaces, $mathbbR^omega$ being compact would imply its image via a projection to be compact. However, $mathbbR simeq pi_1[mathbbR^omega]$ is not compact.

As a quick generalization, if $prod_i X_i$ is compact, each $X_i$ is compact. By the contrapositive then, any product of spaces in which one of them is not compact will never be compact itself. The reciprocal is much harder to prove in an arbitrary case, and it is known as Tychonoff's theorem.

share|cite|improve this answer

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got a doubt in my proof as why can't I use the same open cover under product topology to show it is not compact (I think I am missing something silly point) but as we know it is actually compact
  $endgroup$
  – Believer
  Mar 24 at 23:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh sorry R is not compact..shit
  $endgroup$
  – Believer
  Mar 24 at 23:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your cover is comprised of open sets which moreover belong to the product topology, so the argument follows. Also, the box topology is finer than the product topology, which means that projections are still continuous and the argument I gave still holds. In any case, $mathbbR^omega$ is not compact.
  $endgroup$
  – Guido A.
  Mar 24 at 23:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The argument of projections will work in any topology which makes them continuous, i.e. in any topology finer than the product topology. In simpler terms, a cover in a topology $tau$ with no finite subcovers will still be a cover with no finite subcovers in a topology $tau'$, if $tau subset tau'$.
  $endgroup$
  – Guido A.
  Mar 24 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A somewhat harder argument will show that no infinite box product can be compact (unless something trivial occurs, like taking all spaces indiscrete or all but finitely many of them singletons etc. ) likewise for connectedness. You’ll start to see why the standard product topology is a much better choice unless you want counterexamples.
  $endgroup$
  – Henno Brandsma
  Mar 25 at 9:17
  
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

You are correct, that does the trick. They cover the whole space, are open by definition of the box topology, and taking any finite subcollection fails to cover $mathbbR^omega$. An alternative approach is to note that, since continuous functions sends compact spaces to compact spaces, $mathbbR^omega$ being compact would imply its image via a projection to be compact. However, $mathbbR simeq pi_1[mathbbR^omega]$ is not compact.

As a quick generalization, if $prod_i X_i$ is compact, each $X_i$ is compact. By the contrapositive then, any product of spaces in which one of them is not compact will never be compact itself. The reciprocal is much harder to prove in an arbitrary case, and it is known as Tychonoff's theorem.

share|cite|improve this answer

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I got a doubt in my proof as why can't I use the same open cover under product topology to show it is not compact (I think I am missing something silly point) but as we know it is actually compact
  $endgroup$
  – Believer
  Mar 24 at 23:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh sorry R is not compact..shit
  $endgroup$
  – Believer
  Mar 24 at 23:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your cover is comprised of open sets which moreover belong to the product topology, so the argument follows. Also, the box topology is finer than the product topology, which means that projections are still continuous and the argument I gave still holds. In any case, $mathbbR^omega$ is not compact.
  $endgroup$
  – Guido A.
  Mar 24 at 23:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The argument of projections will work in any topology which makes them continuous, i.e. in any topology finer than the product topology. In simpler terms, a cover in a topology $tau$ with no finite subcovers will still be a cover with no finite subcovers in a topology $tau'$, if $tau subset tau'$.
  $endgroup$
  – Guido A.
  Mar 24 at 23:43
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  A somewhat harder argument will show that no infinite box product can be compact (unless something trivial occurs, like taking all spaces indiscrete or all but finitely many of them singletons etc. ) likewise for connectedness. You’ll start to see why the standard product topology is a much better choice unless you want counterexamples.
  $endgroup$
  – Henno Brandsma
  Mar 25 at 9:17
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

You are correct, that does the trick. They cover the whole space, are open by definition of the box topology, and taking any finite subcollection fails to cover $mathbbR^omega$. An alternative approach is to note that, since continuous functions sends compact spaces to compact spaces, $mathbbR^omega$ being compact would imply its image via a projection to be compact. However, $mathbbR simeq pi_1[mathbbR^omega]$ is not compact.

As a quick generalization, if $prod_i X_i$ is compact, each $X_i$ is compact. By the contrapositive then, any product of spaces in which one of them is not compact will never be compact itself. The reciprocal is much harder to prove in an arbitrary case, and it is known as Tychonoff's theorem.

share|cite|improve this answer

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

$endgroup$

share|cite|improve this answer

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730

answered Mar 24 at 23:21

Guido A.Guido A.

8,1901730