Open subspace of a compact space topological space

  • Thread starter de_brook
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It is a fact that if X is a compact topoloical space then a closed subspace of X is compact.
Is an open subspace G of X also compact?
please consider the following and note if i am wrong;

proof: Since G is open then the relative topology on G is class {H_i}of open subset of X such that the union of all sets in this class is G. but X is compact and each H_i is the intersection of G with an open subset P_i of X for corresponding i. The result follows from the fact {p_i} has a finite subclass which contains X.
hence every open subspace of a compact space is compact.

pls, am i right?
 

Answers and Replies

  • #2
Office_Shredder
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Your proof is wrong, as you can have an open cover of G in the topology of X that doesn't cover X, and then when you reduce it to the induced topology on G it doesn't need to have a finite subcover.

Conclusion: All you did is prove that an open cover of a compact space is also an open cover of its subsets (not a very impressive result :p )

As a quick counterexample, see (0,1) contained in [0,1]
 

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