3

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?

Community
  • 1
Maryam
  • 69
  • 1
    Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be closed, see meta. – Martin Sleziak Aug 12 '13 at 08:32

2 Answers2

6

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.

Brian M. Scott
  • 631,399
  • Why set difference between the interval and the compact (in said order) is non-empty? What hinders a sufficiently large compact to cover the interval? – Incnis Mrsi Dec 17 '14 at 14:59
  • Generally, why shouldn’t ℚ be locally compact? – Incnis Mrsi Dec 17 '14 at 15:51
  • Got it finally. Any more-than-singleton interval in ℚ may not be compact because it contains closed but not compact subsets such as intervals bound by irrational points. Then, compactification fails T₂ and diagonal is not closed. – Incnis Mrsi Dec 17 '14 at 15:57
  • @IncnisMrsi: Alternatively, $\Bbb Q$ itself is metrizable, so a subset is compact iff it’s sequentially compact, and every interval in $\Bbb Q$ has sequences with no convergent subsequences (because in $\Bbb R$ they converge to irrationals). – Brian M. Scott Dec 17 '14 at 20:35
  • 1
    Maybe easier, $\Delta$ is not closed in $X$ because $\Bbb{Q}^*$ is not Hausdorff (because $\Bbb Q$ is not locally compact). – PatrickR Oct 04 '21 at 00:24
0

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.

Martin Sleziak
  • 56,060
Supriyo
  • 6,329
  • 7
  • 33
  • 64