Open subspace of a compact space

[de_brook]

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 P_i for corresponding i. The result foolow 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?

 

[de_brook]

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 foolow 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?

 

[HallsofIvy]

Not quite enough! "What I tell you three times is true" and you only said it twice!

How does that result follow? An open cover of G is not necessarily an open cover of X nor can it always be extended to one. If G were closed in X, then its complement in X would be open add adding it to an open cover for G would give an open cover for X which is why a closed subset of a compact set is closed- but you can't do that for an open set. In particular, suppose X is the compact space [0, 1] with the usual metric on the real numbers (d(x,y)= |x -y|) Let G be (0,1). Finally, let U_n be (1/n, 1-1/n). That is an open cover for G. Does it have a finite subcover?

 

[de_brook]

Not quite enough! "What I tell you three times is true" and you only said it twice!

what do you mean?

 

[HallsofIvy]

Sorry, that was a joke (based on a line by the famous mathematician Rev. Dodgson) in his poem "The hunting of the snark". I meant to continue and accidently posted it. I have editted it.

 

[Edwin]

Probably the easiest way to disprove this theorem, would be to find a counter example.



The interval [0,1] is a topological space that has the Heine-Borrel property.

It follows that every closed and bounded subset of [0,1], is compact.

The interval (0,1) is an open subset of [0,1] that is not a closed subset of [0,1].

Therefore, (0,1) is not compact, and so there exists an open cover of (0,1) that has no finite subcover. This provides, I think, a counter-example to the claim.

Please let me know if I am mistaken.

 

[Edwin]

Nevermind, HallsofIvy constructed one! Even better. Should the open cover be

(1-1/n, 1/n) though?

 

[de_brook]

Not quite enough! "What I tell you three times is true" and you only said it twice!

How does that result follow? An open cover of G is not necessarily an open cover of X nor can it always be extended to one. If G were closed in X, then its complement in X would be open add adding it to an open cover for G would give an open cover for X which is why a closed subset of a compact set is closed- but you can't do that for an open set. In particular, suppose X is the compact space [0, 1] with the usual metric on the real numbers (d(x,y)= |x -y|) Let G be (0,1). Finally, let U_n be (1/n, 1-1/n). That is an open cover for G. Does it have a finite subcover?

thank you. i can understand that not every open cover in G has a finite some subcover