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Why is $emptysetcupnmathbbZ, nin mathbbN$ not a topology on $mathbbZ$?

Is it true that, for accumulation points, $(A cup B)' subset A' cup B'$?Testing for topology on $mathbb R^2$Verifying that a certain collection of intervals of $mathbb R$ forms a topologyProving that $[-n, n] mid n in mathbb N cup varnothing, mathbb R$ is a topology over $mathbb R$Why axiom 3) of topology is redundant?Show that $tau= emptyset, mathbbR cup (-infty,x]:xinmathbbR$ is not a topology on $mathbbR$Homeomorphism of $K$ and $Kcup 0$Show that $tau(frac1n)$ is a topology for $mathbbR$.Proof that the Michael line is a topology.Let $X=BbbN setminus lbrace 1 rbrace$, $A_n=lbrace din X : d|n rbrace$, for $nin BbbN$. Is $tau=lbrace A_n : nin BbbN$ topology?

0

$begingroup$

please why $tau=emptysetcupnmathbbZ, nin mathbbN$ is not a topology on $mathbbZ$?

$emptyset in mathbbZ$ and $mathbbZ=1mathbbZin tau$

let $A=n_1mathbbZ$ and $B=n_2mathbbZ$ then $Acap B= maxn_1,n_2mathbbZin tau$

then $tau$ is stable by finite intersection .

let $(A_i)_iin I$ a family of sets from $tau$ : $A_i=n_imathbbZ$

then $bigcup A_i=min_Iin In_i mathbbZ$

thank you

general-topology

share|cite|improve this question

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

$endgroup$

add a comment | 

0

$begingroup$

please why $tau=emptysetcupnmathbbZ, nin mathbbN$ is not a topology on $mathbbZ$?

$emptyset in mathbbZ$ and $mathbbZ=1mathbbZin tau$

let $A=n_1mathbbZ$ and $B=n_2mathbbZ$ then $Acap B= maxn_1,n_2mathbbZin tau$

then $tau$ is stable by finite intersection .

let $(A_i)_iin I$ a family of sets from $tau$ : $A_i=n_imathbbZ$

then $bigcup A_i=min_Iin In_i mathbbZ$

thank you

general-topology

share|cite|improve this question

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

$endgroup$

add a comment | 

$begingroup$

please why $tau=emptysetcupnmathbbZ, nin mathbbN$ is not a topology on $mathbbZ$?

$emptyset in mathbbZ$ and $mathbbZ=1mathbbZin tau$

let $A=n_1mathbbZ$ and $B=n_2mathbbZ$ then $Acap B= maxn_1,n_2mathbbZin tau$

then $tau$ is stable by finite intersection .

let $(A_i)_iin I$ a family of sets from $tau$ : $A_i=n_imathbbZ$

then $bigcup A_i=min_Iin In_i mathbbZ$

thank you

general-topology

share|cite|improve this question

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

$endgroup$

share|cite|improve this question

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

share|cite|improve this question

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

edited Mar 11 at 20:49

Asaf Karagila♦

306k33438769

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

asked Mar 11 at 20:43

Poline SandraPoline Sandra

997

3 Answers
3

active

oldest

votes

5

$begingroup$

$2mathbbZcup 3mathbbZne nmathbbZ$ for any $n$.

share|cite|improve this answer

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? I don't understand ? and the intersection is right? ?
  $endgroup$
  – Poline Sandra
  Mar 11 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PolineSandra your family is a base for a topology but not a topology. As to this non-equality: $2$ is in the left hand side and can only be in $nmathbbZ$ for $n=1,2$ but the first is too large (as $5$ is not in the left hand side), the second too small as $3$ is in the left hand side and not in $2mathbbZ$.
  $endgroup$
  – Henno Brandsma
  Mar 11 at 22:40
  

  
  
  
  

add a comment | 

1

$begingroup$

It is not true that $cup _i in IA_i = min_i in In_i mathbbZ$. Can you come up with a counter-example where $I$ has size $2$?

share|cite|improve this answer

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

As you said, $emptyset in tau$ and $mathbbZ=1mathbbZ in tau$.

Also, let $A=amathbbZ$ and $B=bmathbbZ$. We have that $A cap B = textrmlcm(a,b)mathbbZ in tau$.

But this set is not a topology, because it doesn't satisfy the union property. Just take any $A=amathbbZ$ and $B=bmathbbZ$ such that $textrmgcd(a,b)=1$. For example: $3mathbbZ cup 5mathbbZ neq nmathbbZ, forall n in mathbbZ$. Note that this union contains all integers that are multiple to $3$ and all integers that are multiple to $5$.

share|cite|improve this answer

answered Mar 11 at 21:05

White CrowWhite Crow

284110

$endgroup$

add a comment | 

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5

$begingroup$

$2mathbbZcup 3mathbbZne nmathbbZ$ for any $n$.

share|cite|improve this answer

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? I don't understand ? and the intersection is right? ?
  $endgroup$
  – Poline Sandra
  Mar 11 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PolineSandra your family is a base for a topology but not a topology. As to this non-equality: $2$ is in the left hand side and can only be in $nmathbbZ$ for $n=1,2$ but the first is too large (as $5$ is not in the left hand side), the second too small as $3$ is in the left hand side and not in $2mathbbZ$.
  $endgroup$
  – Henno Brandsma
  Mar 11 at 22:40
  

  
  
  
  

add a comment | 

5

$begingroup$

$2mathbbZcup 3mathbbZne nmathbbZ$ for any $n$.

share|cite|improve this answer

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  why ? I don't understand ? and the intersection is right? ?
  $endgroup$
  – Poline Sandra
  Mar 11 at 20:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @PolineSandra your family is a base for a topology but not a topology. As to this non-equality: $2$ is in the left hand side and can only be in $nmathbbZ$ for $n=1,2$ but the first is too large (as $5$ is not in the left hand side), the second too small as $3$ is in the left hand side and not in $2mathbbZ$.
  $endgroup$
  – Henno Brandsma
  Mar 11 at 22:40
  

  
  
  
  

add a comment | 

$begingroup$

$2mathbbZcup 3mathbbZne nmathbbZ$ for any $n$.

share|cite|improve this answer

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

answered Mar 11 at 20:46

eyeballfrogeyeballfrog

6,471629

1

$begingroup$

It is not true that $cup _i in IA_i = min_i in In_i mathbbZ$. Can you come up with a counter-example where $I$ has size $2$?

share|cite|improve this answer

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

1

$begingroup$

It is not true that $cup _i in IA_i = min_i in In_i mathbbZ$. Can you come up with a counter-example where $I$ has size $2$?

share|cite|improve this answer

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

It is not true that $cup _i in IA_i = min_i in In_i mathbbZ$. Can you come up with a counter-example where $I$ has size $2$?

share|cite|improve this answer

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 20:46

nammienammie

2399

New contributor

nammie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 11 at 20:46

nammienammie

2399

1

$begingroup$

As you said, $emptyset in tau$ and $mathbbZ=1mathbbZ in tau$.

Also, let $A=amathbbZ$ and $B=bmathbbZ$. We have that $A cap B = textrmlcm(a,b)mathbbZ in tau$.

But this set is not a topology, because it doesn't satisfy the union property. Just take any $A=amathbbZ$ and $B=bmathbbZ$ such that $textrmgcd(a,b)=1$. For example: $3mathbbZ cup 5mathbbZ neq nmathbbZ, forall n in mathbbZ$. Note that this union contains all integers that are multiple to $3$ and all integers that are multiple to $5$.

share|cite|improve this answer

answered Mar 11 at 21:05

White CrowWhite Crow

284110

$endgroup$

add a comment | 

1

$begingroup$

As you said, $emptyset in tau$ and $mathbbZ=1mathbbZ in tau$.

Also, let $A=amathbbZ$ and $B=bmathbbZ$. We have that $A cap B = textrmlcm(a,b)mathbbZ in tau$.

But this set is not a topology, because it doesn't satisfy the union property. Just take any $A=amathbbZ$ and $B=bmathbbZ$ such that $textrmgcd(a,b)=1$. For example: $3mathbbZ cup 5mathbbZ neq nmathbbZ, forall n in mathbbZ$. Note that this union contains all integers that are multiple to $3$ and all integers that are multiple to $5$.

share|cite|improve this answer

answered Mar 11 at 21:05

White CrowWhite Crow

284110

$endgroup$

add a comment | 

$begingroup$

As you said, $emptyset in tau$ and $mathbbZ=1mathbbZ in tau$.

Also, let $A=amathbbZ$ and $B=bmathbbZ$. We have that $A cap B = textrmlcm(a,b)mathbbZ in tau$.

But this set is not a topology, because it doesn't satisfy the union property. Just take any $A=amathbbZ$ and $B=bmathbbZ$ such that $textrmgcd(a,b)=1$. For example: $3mathbbZ cup 5mathbbZ neq nmathbbZ, forall n in mathbbZ$. Note that this union contains all integers that are multiple to $3$ and all integers that are multiple to $5$.

share|cite|improve this answer

answered Mar 11 at 21:05

White CrowWhite Crow

284110

$endgroup$

share|cite|improve this answer

answered Mar 11 at 21:05

White CrowWhite Crow

284110

answered Mar 11 at 21:05

White CrowWhite Crow

284110

answered Mar 11 at 21:05

White CrowWhite Crow

284110