3
$\begingroup$

Let $(X,\tau)$ be a Hausdorff space. $A,B \subseteq (X,\tau)$ which are compact sets. Show that the intersection and the union of A and B are compact sets.

Here is the what i did so far but i think there are some points that is not fit, i can feel but can't figure out :) thanks advance for guidince.

Proof:

Every compact subset of Hausdorff space is closed set. So, $A,B \subseteq X$ which are compact sets are closed sets. Thus $A \cup B$ and $A \cap B$ are closed sets (By the previous topologicial theorem which is union and intersection of two closed sets are closed set.) As a result of this $X \setminus (A \cup B)$ and $X \setminus (A \cap B)$ are open sets (By the definition of closed set.)

Let $\mathcal{U}$ is cover for $A \cup B$. Because of $X \setminus (A \cup B)$ is copen set, we can write a cover for the Hausdorff space which is given as $X \subseteq \mathcal{U} \cup (X \setminus (A \cup B) ) $ .Thus, By being compact set, for every cover of X there exists a finite subcover of X. So difference of the finite subcover and $X \setminus A \cup B$ gives us a finite subcover of $A \cup B$.

Let $\mathcal{V}$ is cover for $A \cap B$. Because of $X \setminus (A \cap B)$ is copen set, we can write a cover for the Hausdorff space which is given as $ X \subseteq \mathcal{V} \cup (X \setminus (A \cap B) ) $.Thus, by being compact set, for every cover of X there exists a finite subcover of X. So difference of the finite subcover and $X \setminus A \cap B$ gives us a finite subcover of $A \cap B$

$\endgroup$
2
$\begingroup$

Hint: (i)For $A \cup B$, when you have an open cover for $A \cup B$, you have an open cover for both $A$ and $B$ and by compactness of $A$ and $B$, you have a finite subcover for $A$ and a finite subcover for $B$ and hence a finite subcover of $A \cup B$.

(ii) For $A \cap B$, $A$ and$B$ are compact in a Hausdroff space. Are they closed? Is $A \cap B$ closed? What can you say about a closed subspace of a compact space? Is it compact again?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.