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?

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

  • $\begingroup$ Also the infinite intersection finite union of closed sets are closed. $\endgroup$ Dec 2, 2018 at 23:36
  • $\begingroup$ This is correct, assuming $O$ is the set of open sets in the topology. $\endgroup$
    – platty
    Dec 2, 2018 at 23:37
  • $\begingroup$ @JoelPereira So what do I need to add? $\endgroup$
    – thebilly
    Dec 8, 2018 at 14:15
  • $\begingroup$ @platty awesome. thank you. $\endgroup$
    – thebilly
    Dec 8, 2018 at 14:15

1 Answer 1


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 $C$.


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