What are the closed sets of this topology.












0












$begingroup$


I have the topology consisting of these Sets.
$$X = {a,b,c,d,e,f}$$
$$O = {X,∅,{b},{c,d},{b,c,d},{a,c,d,e,f}}$$



The question is what are the closed sets of this topology?



I have found information on the 'closure' and 'interior' of sets, thought that that was not what was asked for and then stumbled across this.



Is that what I am supposed to do?



$$X^c = emptyset$$
$$emptyset^c = X$$
$${b}^c = {a, c, d, e, f}$$
$${c,d}^c= {a, b, e, f}$$
$${b, c, d}^c = {a, e, f}$$
$${a, c, d, e, f}^c = {b}$$






general-topology algebraic-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    Also the infinite intersection finite union of closed sets are closed.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 23:36










  • $begingroup$
    This is correct, assuming $O$ is the set of open sets in the topology.
    $endgroup$
    – platty
    Dec 2 '18 at 23:37












  • $begingroup$
    @JoelPereira So what do I need to add?
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15










  • $begingroup$
    @platty awesome. thank you.
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15
















0












$begingroup$


I have the topology consisting of these Sets.
$$X = {a,b,c,d,e,f}$$
$$O = {X,∅,{b},{c,d},{b,c,d},{a,c,d,e,f}}$$



The question is what are the closed sets of this topology?



I have found information on the 'closure' and 'interior' of sets, thought that that was not what was asked for and then stumbled across this.



Is that what I am supposed to do?



$$X^c = emptyset$$
$$emptyset^c = X$$
$${b}^c = {a, c, d, e, f}$$
$${c,d}^c= {a, b, e, f}$$
$${b, c, d}^c = {a, e, f}$$
$${a, c, d, e, f}^c = {b}$$






general-topology algebraic-topology







share|cite|improve this question









$endgroup$












  • $begingroup$
    Also the infinite intersection finite union of closed sets are closed.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 23:36










  • $begingroup$
    This is correct, assuming $O$ is the set of open sets in the topology.
    $endgroup$
    – platty
    Dec 2 '18 at 23:37












  • $begingroup$
    @JoelPereira So what do I need to add?
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15










  • $begingroup$
    @platty awesome. thank you.
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15














0












0








0





$begingroup$


I have the topology consisting of these Sets.
$$X = {a,b,c,d,e,f}$$
$$O = {X,∅,{b},{c,d},{b,c,d},{a,c,d,e,f}}$$



The question is what are the closed sets of this topology?



I have found information on the 'closure' and 'interior' of sets, thought that that was not what was asked for and then stumbled across this.



Is that what I am supposed to do?



$$X^c = emptyset$$
$$emptyset^c = X$$
$${b}^c = {a, c, d, e, f}$$
$${c,d}^c= {a, b, e, f}$$
$${b, c, d}^c = {a, e, f}$$
$${a, c, d, e, f}^c = {b}$$






general-topology algebraic-topology







share|cite|improve this question









$endgroup$




I have the topology consisting of these Sets.
$$X = {a,b,c,d,e,f}$$
$$O = {X,∅,{b},{c,d},{b,c,d},{a,c,d,e,f}}$$



The question is what are the closed sets of this topology?



I have found information on the 'closure' and 'interior' of sets, thought that that was not what was asked for and then stumbled across this.



Is that what I am supposed to do?



$$X^c = emptyset$$
$$emptyset^c = X$$
$${b}^c = {a, c, d, e, f}$$
$${c,d}^c= {a, b, e, f}$$
$${b, c, d}^c = {a, e, f}$$
$${a, c, d, e, f}^c = {b}$$






general-topology algebraic-topology




general-topology algebraic-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 23:32









thebillythebilly

566




566












  • $begingroup$
    Also the infinite intersection finite union of closed sets are closed.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 23:36










  • $begingroup$
    This is correct, assuming $O$ is the set of open sets in the topology.
    $endgroup$
    – platty
    Dec 2 '18 at 23:37












  • $begingroup$
    @JoelPereira So what do I need to add?
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15










  • $begingroup$
    @platty awesome. thank you.
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15


















  • $begingroup$
    Also the infinite intersection finite union of closed sets are closed.
    $endgroup$
    – Joel Pereira
    Dec 2 '18 at 23:36










  • $begingroup$
    This is correct, assuming $O$ is the set of open sets in the topology.
    $endgroup$
    – platty
    Dec 2 '18 at 23:37












  • $begingroup$
    @JoelPereira So what do I need to add?
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15










  • $begingroup$
    @platty awesome. thank you.
    $endgroup$
    – thebilly
    Dec 8 '18 at 14:15
















$begingroup$
Also the infinite intersection finite union of closed sets are closed.
$endgroup$
– Joel Pereira
Dec 2 '18 at 23:36




$begingroup$
Also the infinite intersection finite union of closed sets are closed.
$endgroup$
– Joel Pereira
Dec 2 '18 at 23:36












$begingroup$
This is correct, assuming $O$ is the set of open sets in the topology.
$endgroup$
– platty
Dec 2 '18 at 23:37






$begingroup$
This is correct, assuming $O$ is the set of open sets in the topology.
$endgroup$
– platty
Dec 2 '18 at 23:37














$begingroup$
@JoelPereira So what do I need to add?
$endgroup$
– thebilly
Dec 8 '18 at 14:15




$begingroup$
@JoelPereira So what do I need to add?
$endgroup$
– thebilly
Dec 8 '18 at 14:15












$begingroup$
@platty awesome. thank you.
$endgroup$
– thebilly
Dec 8 '18 at 14:15




$begingroup$
@platty awesome. thank you.
$endgroup$
– thebilly
Dec 8 '18 at 14:15










1 Answer
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$begingroup$

The closed sets of the topology are given by the complements of the sets in $O$, as well as finite unions and arbitrary intersections of these sets. In fact, in this case, just by taking complements as you did, you have found all closed sets in the topology. If you take a union or intersection of any two sets in $$C={∅,X,{b},{a,e,f},{a,b,e,f},{a,c,d,e,f}}$$ you will obtain another set already in $C$.






share|cite|improve this answer









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    $begingroup$

    The closed sets of the topology are given by the complements of the sets in $O$, as well as finite unions and arbitrary intersections of these sets. In fact, in this case, just by taking complements as you did, you have found all closed sets in the topology. If you take a union or intersection of any two sets in $$C={∅,X,{b},{a,e,f},{a,b,e,f},{a,c,d,e,f}}$$ you will obtain another set already in $C$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The closed sets of the topology are given by the complements of the sets in $O$, as well as finite unions and arbitrary intersections of these sets. In fact, in this case, just by taking complements as you did, you have found all closed sets in the topology. If you take a union or intersection of any two sets in $$C={∅,X,{b},{a,e,f},{a,b,e,f},{a,c,d,e,f}}$$ you will obtain another set already in $C$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The closed sets of the topology are given by the complements of the sets in $O$, as well as finite unions and arbitrary intersections of these sets. In fact, in this case, just by taking complements as you did, you have found all closed sets in the topology. If you take a union or intersection of any two sets in $$C={∅,X,{b},{a,e,f},{a,b,e,f},{a,c,d,e,f}}$$ you will obtain another set already in $C$.






        share|cite|improve this answer









        $endgroup$



        The closed sets of the topology are given by the complements of the sets in $O$, as well as finite unions and arbitrary intersections of these sets. In fact, in this case, just by taking complements as you did, you have found all closed sets in the topology. If you take a union or intersection of any two sets in $$C={∅,X,{b},{a,e,f},{a,b,e,f},{a,c,d,e,f}}$$ you will obtain another set already in $C$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 23:47









        greeliousgreelious

        19410




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