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Let $x in X$. Show that the subset ${x} subset X$ is closed with respect to $tau(d)$

Show that the infinite intersection of nested non-empty closed subsets of a compact space is not emptyIntersection of topologiesIt is the power set of $Bbb R$ a topology of $Bbb R$?Let $X$ be an infinite set. Show that $tau={{emptyset}}cup{A:A^ctext{ is finite}}$ is a topology on $X$.Prove that an uncountable space with the countable-closed topology satisfies the countable chain condition.Show that $tau$ is a TopologyLet X a set not empty and (X,d) a metric space, prove the intersection of a family of closed sets are closed.$Ainoverline{tau}iffforall ain A-E,exists Bintau,ain Bsubset A$ defines a topologyWhat is the standard topology of real line? Why is $(0,1)$ called open but $[0,1]$ not open on this topology?show that the finite intersection of open sets is non empty. for the cocountable Topology.

0

$begingroup$

Let $x in X$. Show that the subset ${x} subset X$ is closed with respect to $tau(d)$

To solve this problem I have looked at properties of closed sets:

$a)$ The empty set $emptyset$ and the whole space $X$ are closed.

$b)$ If $A subset X$ and $B subset X$ are closed, so is their union $A
cup B$.

$c)$ Let $I$ be a set and let $(A_i)_{i in I}$ be a family of closed subsets of $X$. Then their intersection $bigcap_{i in I} A_i$ is closed as well.

To show that the empty set is closed, we have to see that $X$ $emptyset$ is open, and $X = X$ $emptyset$ is open in any topology by definition. Likewise, $X$ $X = emptyset$ is open, hence $X$ is closed.

Not sure how to answer $a)$ and $b)$

Thanks in advance.

general-topology

share|cite|improve this question

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is $tau(d)$ ?
  $endgroup$
  – TheSilverDoe
  Mar 19 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You'll need additional restrictions on the topology. For instance, if we give $mathbb{R}$ the topology ${ varnothing, mathbb{R}}$, then singletons ${x}$ will not be closed. Is $X$ supposed to be either $T_1$ or Hausdorff?
  $endgroup$
  – rolandcyp
  Mar 19 at 15:57
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $x in X$. Show that the subset ${x} subset X$ is closed with respect to $tau(d)$

To solve this problem I have looked at properties of closed sets:

$a)$ The empty set $emptyset$ and the whole space $X$ are closed.

$b)$ If $A subset X$ and $B subset X$ are closed, so is their union $A
cup B$.

$c)$ Let $I$ be a set and let $(A_i)_{i in I}$ be a family of closed subsets of $X$. Then their intersection $bigcap_{i in I} A_i$ is closed as well.

To show that the empty set is closed, we have to see that $X$ $emptyset$ is open, and $X = X$ $emptyset$ is open in any topology by definition. Likewise, $X$ $X = emptyset$ is open, hence $X$ is closed.

Not sure how to answer $a)$ and $b)$

Thanks in advance.

general-topology

share|cite|improve this question

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What is $tau(d)$ ?
  $endgroup$
  – TheSilverDoe
  Mar 19 at 15:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You'll need additional restrictions on the topology. For instance, if we give $mathbb{R}$ the topology ${ varnothing, mathbb{R}}$, then singletons ${x}$ will not be closed. Is $X$ supposed to be either $T_1$ or Hausdorff?
  $endgroup$
  – rolandcyp
  Mar 19 at 15:57
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $x in X$. Show that the subset ${x} subset X$ is closed with respect to $tau(d)$

To solve this problem I have looked at properties of closed sets:

$a)$ The empty set $emptyset$ and the whole space $X$ are closed.

$b)$ If $A subset X$ and $B subset X$ are closed, so is their union $A
cup B$.

$c)$ Let $I$ be a set and let $(A_i)_{i in I}$ be a family of closed subsets of $X$. Then their intersection $bigcap_{i in I} A_i$ is closed as well.

To show that the empty set is closed, we have to see that $X$ $emptyset$ is open, and $X = X$ $emptyset$ is open in any topology by definition. Likewise, $X$ $X = emptyset$ is open, hence $X$ is closed.

Not sure how to answer $a)$ and $b)$

Thanks in advance.

general-topology

share|cite|improve this question

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

$endgroup$

share|cite|improve this question

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

share|cite|improve this question

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

edited Mar 21 at 11:21

Andrews

1,2812422

edited Mar 21 at 11:21

Andrews

1,2812422

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

asked Mar 19 at 15:51

Ben JonesBen Jones

19511

1 Answer
1

active

oldest

votes

0

$begingroup$

For now let $(X,tau)$ be a general topological space. You have already shown that $X$ and $varnothing$ are closed. For (b), let $A$ and $B$ be closed subsets of $X$. To prove that $A cup B$ is closed, we need only check that $(A cup B)^complement = A^complement cap B^complement$ is open. But this is immediate since $A^complement$ and $B^complement$ are open and finite intersections of open sets are open.

Treating (c), let ${A_i}_{i in I}$ be a family of closed subsets of $X$. Then,
begin{align*}
left( bigcap_{i in I} A_i right)^complement = bigcup_{i in I} A_i^complement
end{align*}
is open because arbitrary unions of open sets are open (and every $A_i^complement$ is open by definition). This proves (c).

Now, given an arbitrary topological space $(X,tau)$, it is not true in general that every singleton ${x}$ is closed in $X$. As I pointed out in the comments, if we give $mathbb{R}$ the indiscrete topology ${mathbb{R}, varnothing}$, then every singleton will not be closed in $mathbb{R}$.

Let us also recall the following definition:

Definition. A space $(X,tau)$ is said to be $T_1$ if for any $x,y in X$ with $x neq y$, there exists an open set $U ni x$ that does not contain $y$.

Without assuming that $(X,tau)$ is either Hausdorff or $T_1$, it will be impossible to show that singletons are closed. In fact, the following equivalence holds:

Theorem. Let $(X,tau)$ be a topological space. Then, $(X,tau)$ is $T_1$ if and only if every singleton ${x}$ is closed in $X$.

In light of this theorem, it's clear that one cannot hope to prove what you want without any additional assumptions on the space $(X,tau)$.

share|cite|improve this answer

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

$endgroup$

add a comment | 

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0

$begingroup$

For now let $(X,tau)$ be a general topological space. You have already shown that $X$ and $varnothing$ are closed. For (b), let $A$ and $B$ be closed subsets of $X$. To prove that $A cup B$ is closed, we need only check that $(A cup B)^complement = A^complement cap B^complement$ is open. But this is immediate since $A^complement$ and $B^complement$ are open and finite intersections of open sets are open.

Treating (c), let ${A_i}_{i in I}$ be a family of closed subsets of $X$. Then,
begin{align*}
left( bigcap_{i in I} A_i right)^complement = bigcup_{i in I} A_i^complement
end{align*}
is open because arbitrary unions of open sets are open (and every $A_i^complement$ is open by definition). This proves (c).

Now, given an arbitrary topological space $(X,tau)$, it is not true in general that every singleton ${x}$ is closed in $X$. As I pointed out in the comments, if we give $mathbb{R}$ the indiscrete topology ${mathbb{R}, varnothing}$, then every singleton will not be closed in $mathbb{R}$.

Let us also recall the following definition:

Definition. A space $(X,tau)$ is said to be $T_1$ if for any $x,y in X$ with $x neq y$, there exists an open set $U ni x$ that does not contain $y$.

Without assuming that $(X,tau)$ is either Hausdorff or $T_1$, it will be impossible to show that singletons are closed. In fact, the following equivalence holds:

Theorem. Let $(X,tau)$ be a topological space. Then, $(X,tau)$ is $T_1$ if and only if every singleton ${x}$ is closed in $X$.

In light of this theorem, it's clear that one cannot hope to prove what you want without any additional assumptions on the space $(X,tau)$.

share|cite|improve this answer

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

$endgroup$

add a comment | 

0

$begingroup$

For now let $(X,tau)$ be a general topological space. You have already shown that $X$ and $varnothing$ are closed. For (b), let $A$ and $B$ be closed subsets of $X$. To prove that $A cup B$ is closed, we need only check that $(A cup B)^complement = A^complement cap B^complement$ is open. But this is immediate since $A^complement$ and $B^complement$ are open and finite intersections of open sets are open.

Treating (c), let ${A_i}_{i in I}$ be a family of closed subsets of $X$. Then,
begin{align*}
left( bigcap_{i in I} A_i right)^complement = bigcup_{i in I} A_i^complement
end{align*}
is open because arbitrary unions of open sets are open (and every $A_i^complement$ is open by definition). This proves (c).

Now, given an arbitrary topological space $(X,tau)$, it is not true in general that every singleton ${x}$ is closed in $X$. As I pointed out in the comments, if we give $mathbb{R}$ the indiscrete topology ${mathbb{R}, varnothing}$, then every singleton will not be closed in $mathbb{R}$.

Let us also recall the following definition:

Definition. A space $(X,tau)$ is said to be $T_1$ if for any $x,y in X$ with $x neq y$, there exists an open set $U ni x$ that does not contain $y$.

Without assuming that $(X,tau)$ is either Hausdorff or $T_1$, it will be impossible to show that singletons are closed. In fact, the following equivalence holds:

Theorem. Let $(X,tau)$ be a topological space. Then, $(X,tau)$ is $T_1$ if and only if every singleton ${x}$ is closed in $X$.

In light of this theorem, it's clear that one cannot hope to prove what you want without any additional assumptions on the space $(X,tau)$.

share|cite|improve this answer

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

$endgroup$

add a comment | 

$begingroup$

For now let $(X,tau)$ be a general topological space. You have already shown that $X$ and $varnothing$ are closed. For (b), let $A$ and $B$ be closed subsets of $X$. To prove that $A cup B$ is closed, we need only check that $(A cup B)^complement = A^complement cap B^complement$ is open. But this is immediate since $A^complement$ and $B^complement$ are open and finite intersections of open sets are open.

Treating (c), let ${A_i}_{i in I}$ be a family of closed subsets of $X$. Then,
begin{align*}
left( bigcap_{i in I} A_i right)^complement = bigcup_{i in I} A_i^complement
end{align*}
is open because arbitrary unions of open sets are open (and every $A_i^complement$ is open by definition). This proves (c).

Now, given an arbitrary topological space $(X,tau)$, it is not true in general that every singleton ${x}$ is closed in $X$. As I pointed out in the comments, if we give $mathbb{R}$ the indiscrete topology ${mathbb{R}, varnothing}$, then every singleton will not be closed in $mathbb{R}$.

Let us also recall the following definition:

Definition. A space $(X,tau)$ is said to be $T_1$ if for any $x,y in X$ with $x neq y$, there exists an open set $U ni x$ that does not contain $y$.

Without assuming that $(X,tau)$ is either Hausdorff or $T_1$, it will be impossible to show that singletons are closed. In fact, the following equivalence holds:

Theorem. Let $(X,tau)$ be a topological space. Then, $(X,tau)$ is $T_1$ if and only if every singleton ${x}$ is closed in $X$.

In light of this theorem, it's clear that one cannot hope to prove what you want without any additional assumptions on the space $(X,tau)$.

share|cite|improve this answer

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421

answered Mar 19 at 16:10

rolandcyprolandcyp

2,234421