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Set theory/topology proof

  • #1
i want to prove that given a set K in G that is closed and a set F in K that is closed, then K is closed in G. K, F, and G are all topological spaces.

so to reword the problem i instead want to show that given G-K is open in G and K-F is open in K show that G-F is open in G. so since G is a topological space and K is a subspace of G, i endow K with the subspace topology. then there exists some open set O in G such that any open subset of K such as K-F = O ∩ K. then F = K - (O∩K) then i drew myself a picture and found that F also equals (G-O)∩K which works because G-F = G-(G-O) ∪ (G-K) = O U (G-K) and since O is open in G and (G-K) is open in G therefore the union of those 2 sets is also open in G and G-F is then open in G which means F is closed in G.

i wish to know that in these types of proofs, is it common to have to draw pictures in order to continue? i was stuck at F = K - (O∩K) for a while until i drew a picture and saw that F = (G-O)∩K which i could take the complement of since it was a subset of G. my question is could F = (G-O)∩K be derived from F = K - (O∩K) or any of the other information i was given, without having to draw pictures to derive it? or is drawing pictures the usual way of going through these set theory types of proofs. thanks.
 

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
i want to prove that given a set K in G that is closed and a set F in K that is closed, then K is closed in G. K, F, and G are all topological spaces.
so haven't you said K is closed in G already? and you want to show K is closed in G?

do you mean:
- F is closed in K
- K is closed in G
show F is closed in G
 
  • #3
oh whoops. i made a typo there. yes you are correct. if K in G is closed and F in K is closed i want to show that F in G is closed.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
The most direct way to show F is closed in G is to show that its complement is open. Let H be the complement of F in G. Of course, The complement of F in K is just H intersect K.

The topology on K is the topology of G restricted to K. That is, every open set in K is equal to a set open in G intersect K.
 
  • #5
thanks! your proof was a lot more concise than mines.
 

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