If $A$ is compact and $B$ is Lindelöf space , will be $A \cup B$ Lindelöf I have 2 different questions:
As we know a space Y is Lindelöf if each open covering contains a countable subcovering.
(1) :If $A$ is  compact and  $B$ is Lindelöf space , will be $A \cup B$ Lindelöf?
If it is right, how can we prove it?
(2) : A topological space is called $KC$ , when   each compact subset is closed.
Is cartesian product of KC spaces also KC space? is infinite or finite number important?
 A: The product of two $KC$ spaces need not be $KC$. Let $\Bbb Q^*$ be the one-point compactification of the rationals; then $\Bbb Q^*$ is $KC$, but $X=\Bbb Q^*\times\Bbb Q^*$ is not. It's well-known that $\Bbb Q^*$ is $KC$. To see that $X$ is not, let $\Delta=\{\langle x,x\rangle:x\in\Bbb Q^*\}$. Then $\Delta$ is homeomorphic to $\Bbb Q^*$, so $\Delta$ is compact; I'll show that $\Delta$ is not closed in $X$ and hence that $X$ is not $KC$.
Let $p=\langle\infty,0\rangle\in X$. For each $\epsilon>0$ let $I_\epsilon=(-\epsilon,\epsilon)\cap\Bbb Q$. For each compact subset $K$ of $\Bbb Q$ and $\epsilon>0$ let $B(K,\epsilon)=(\Bbb Q^*\setminus K)\times I_\epsilon$, and let $\mathscr{B}$ be the family of all such $B(K,\epsilon)$; $\mathscr{B}$ is a local base at $p$. Fix $B(K,\epsilon)\in\mathscr{B}$. Clearly $I_\epsilon\setminus K\ne\varnothing$, so let $y\in I_\epsilon\setminus K$; then $\langle y,y\rangle\in\Delta\cap B(K,\epsilon)$, and since $B(K,\epsilon)$ was arbitrary, it follows that $p\in(\operatorname{cl}_X\Delta)\setminus\Delta$ and hence that $\Delta$ is not closed.
A: I am facing a notational problem. What is a $KC$ space? Answer of your first question is the following.
Lindelof Space: A space $X$ is said to be Lindelof is every open cover of the space has a countable subcover.
Consider an open cover $P = \{P_{\alpha}: \alpha \in J, P_{\alpha}$ is open in $A \cup B\}$
Now $P$ will gives cover for both $A$ and $B$ say $P_1$ and $P_2$ where $P_1 = \{P_{\alpha} \cap A\}$ and $P_{\alpha}\cap B$.
Now $A$ is compact, thus there is a finite cover say $P_1^{'}$. You may write its elements yourself.
$B$ is Lindelof, thus you shall get a countable subcover from $P_2$, say $P_2^{'}$.
Collect all elements of $P_1^{'}$ and $P_2^{'}$ form your original cover $P$, it is countable. So $A \cup B$ is countable.
There are for Latex problems in this answers. Thank you for correction.
