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The Discrete topology on $mathbb{R}$ is a $T_1$ space and not limit point Compact

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1

$begingroup$

I was hoping someone could review my proofs below. I'm not totally sure both statements are true (or even if one of them is true). Thanks!

Statement 1:

$mathbb{R}$ with the discrete topology is a $T_1$ space

Statement 2:

The discrete topology on $mathbb{R}$ is not limit point compact.

Proof of statement 1:

We wish to show that finite sets are closed. It suffices to show that single point sets are closed, since then the finite union of closed singletons will produce a closed finite set.

Take any set ${a}$. Then $X - {a} = (-infty, a) cup (a,infty)$.
Since each set in the above union is open as it can be written as the infinite union of the open singleton points inside each set, we have that $X - {a}$ is open. Hence ${a}$ is closed.
Hence any finite set is closed.

Proof of statement 2:

Let $X$ be the discrete topology on $mathbb{R}$. Take the interval $[0,1]$. This interval is infinite. The interval also contains no limit points since for any proposed limit point $x$, we can take the open set ${x}$ as an open set that does not intersect $[0,1]$ in any place other than itself. Hence $X$ is not limit point compact.

general-topology

share|cite|improve this question

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just a note, "the discrete topology on R" itself is not a space. One should rather write "$mathbb R$ with the discrete topology is a T1 space".
  $endgroup$
  – Pink Panther
  Mar 10 at 21:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Both statements are true and your proofs are correct! You can simplify them by generalising this to any space with the discrete topology (hint: every subset of a space with the discrete topology is both closed and open)
  $endgroup$
  – nammie
  Mar 10 at 21:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  we could also claim that R with the discrete topology is countably compact right? @nammie
  $endgroup$
  – H_1317
  Mar 10 at 21:45
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @H_1317 any (infinite) space with the discrete topology is not countably compact. You can prove this directly (as every set is open), or use that in $T1$ spaces, countably compact and limit point compact are equivalent.
  $endgroup$
  – nammie
  Mar 10 at 21:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes! The intervals $(n, n+2)$ form an open cover of $mathbb{R}$ with no finite subcover in both the discrete topology and the usual Euclidean one.
  $endgroup$
  – nammie
  Mar 10 at 22:01
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I was hoping someone could review my proofs below. I'm not totally sure both statements are true (or even if one of them is true). Thanks!

Statement 1:

$mathbb{R}$ with the discrete topology is a $T_1$ space

Statement 2:

The discrete topology on $mathbb{R}$ is not limit point compact.

Proof of statement 1:

We wish to show that finite sets are closed. It suffices to show that single point sets are closed, since then the finite union of closed singletons will produce a closed finite set.

Take any set ${a}$. Then $X - {a} = (-infty, a) cup (a,infty)$.
Since each set in the above union is open as it can be written as the infinite union of the open singleton points inside each set, we have that $X - {a}$ is open. Hence ${a}$ is closed.
Hence any finite set is closed.

Proof of statement 2:

Let $X$ be the discrete topology on $mathbb{R}$. Take the interval $[0,1]$. This interval is infinite. The interval also contains no limit points since for any proposed limit point $x$, we can take the open set ${x}$ as an open set that does not intersect $[0,1]$ in any place other than itself. Hence $X$ is not limit point compact.

general-topology

share|cite|improve this question

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  just a note, "the discrete topology on R" itself is not a space. One should rather write "$mathbb R$ with the discrete topology is a T1 space".
  $endgroup$
  – Pink Panther
  Mar 10 at 21:42
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Both statements are true and your proofs are correct! You can simplify them by generalising this to any space with the discrete topology (hint: every subset of a space with the discrete topology is both closed and open)
  $endgroup$
  – nammie
  Mar 10 at 21:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  we could also claim that R with the discrete topology is countably compact right? @nammie
  $endgroup$
  – H_1317
  Mar 10 at 21:45
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @H_1317 any (infinite) space with the discrete topology is not countably compact. You can prove this directly (as every set is open), or use that in $T1$ spaces, countably compact and limit point compact are equivalent.
  $endgroup$
  – nammie
  Mar 10 at 21:50
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes! The intervals $(n, n+2)$ form an open cover of $mathbb{R}$ with no finite subcover in both the discrete topology and the usual Euclidean one.
  $endgroup$
  – nammie
  Mar 10 at 22:01
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I was hoping someone could review my proofs below. I'm not totally sure both statements are true (or even if one of them is true). Thanks!

Statement 1:

$mathbb{R}$ with the discrete topology is a $T_1$ space

Statement 2:

The discrete topology on $mathbb{R}$ is not limit point compact.

Proof of statement 1:

We wish to show that finite sets are closed. It suffices to show that single point sets are closed, since then the finite union of closed singletons will produce a closed finite set.

Take any set ${a}$. Then $X - {a} = (-infty, a) cup (a,infty)$.
Since each set in the above union is open as it can be written as the infinite union of the open singleton points inside each set, we have that $X - {a}$ is open. Hence ${a}$ is closed.
Hence any finite set is closed.

Proof of statement 2:

Let $X$ be the discrete topology on $mathbb{R}$. Take the interval $[0,1]$. This interval is infinite. The interval also contains no limit points since for any proposed limit point $x$, we can take the open set ${x}$ as an open set that does not intersect $[0,1]$ in any place other than itself. Hence $X$ is not limit point compact.

general-topology

share|cite|improve this question

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

$endgroup$

share|cite|improve this question

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

share|cite|improve this question

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

edited Mar 10 at 21:59

Eric Wofsey

189k14216347

asked Mar 10 at 21:37

H_1317H_1317

1519

asked Mar 10 at 21:37

H_1317H_1317

1519