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 $A \cap B$
Best Answer
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?