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Let $(X,\tau)$ be a KC non-compact space. Then there is a discrete subset $D \subseteq X$ such that $\overline D$ is not compact. Furthermore there is an ultrafilter $F$ in $X$ such that $ D \in F $ and for every $C \in F$, $\overline C$ is not compact in $(X,\tau)$.

Proof. Let $ U = \{U_{i} : i < k \} $ be a strictly increasing open cover of $ X $, where $k$ is an infinite regular cardinal. We’ll construct sets $ D_{\beta } = \{ x_{i }: i < \beta \} $ by transfinite induction. First, let $ D_{0} = \{x_{0} \} $ for some $ x_{0 } ∈ U_{0} $. Let $ \beta $ is ordinal successor. If $ \overline{D}_{\beta - 1 } $ is compact, then there is $ \alpha_{\beta}$ such that $ \overline{D}_{\beta - 1 } \subseteq U_{\alpha_\beta}$. Let $x_\beta \in U_{\alpha+1} - U_{\alpha} $ and $ D_{\beta } = D_{\beta - 1 } \cup {x_{ \beta}}$. For limit ordinals $ \beta$, let $ D_{ \beta} = \bigcup_{i < \beta }D_{i} $. This process stops when $ \overline{D}_{\beta} $ is not compact, which holds at least for $ \beta = k $, because then the open cover $\mathcal{ U } $ witnesses that $ \overline{D}_{k} $ is not compact. It is easy to see that $ D_{ \beta} $ is discrete. The open set, which contains exactly one point $ x_{i+1 } $ is $ U_{ \alpha_{i+1}} - \overline{D_{i}}$. Let’s construct the ultrafilter. Let $ F $ a filter-base, such that $ D \in F $ and for any $ C \in F $, $ \overline {C } $ is not compact. Furthermore let $ F $ be maximal with these properties. If $ C \in F $ then for any $ C' \supseteq C $, holds $C'\in F $, because if $C'$ were compact, then it would be closed from $ KC $ property and $ \overline{C} \subseteq C' $, but $ \overline{C}$ is non-compact closed subset of a compact set $C'$, which is a contradiction. If $ D = D_{0 } \cup D_{1} $ and $ D_{0} \cap D_{1} = \varnothing $ then either $D_{0}$ or $ D_{1} $ is in $ F$. Otherwise both $ \overline{D_{0 }} $ and $ \overline{D_{1}} $ would be compact, hence $ \overline{D} = \overline{D_{0 }} ∪ \overline{D_{1 }} $ would be compact. Finally this gives that $ F $ is an ultrafilter.

(1): why we can say in the first paragraph" If $ \overline{D}_{\beta - 1 } $ is compact, then there is $ \alpha_{\beta}$ such that $ \overline{D}_{\beta - 1 } $ ⊆ $ U_{\alpha_{\beta}}$. Let $ x_{ \beta } ∈ U_{\alpha+1} - U_{\alpha} $ and $ D_{\beta } = D_{\beta - 1 } ∪ {x_{ \beta}}$"?

(2): why we can say " This process stops when $ \overline{D}_{\beta} $ is not compact, which holds at least for $ \beta = k $, because then the open cover $\mathcal{ U } $ witnesses that $ \overline{D}_{k} $ is not compact. It is easy to see that $ D_{ \beta} $ is discrete. The open set, which contains exactly one point $ x_{i+1 } $ is $ U_{ \alpha_{i+1}} -\overline{D_{i}}$."?

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  • $\begingroup$ "$k$ is an" what? $\endgroup$
    – dfeuer
    Aug 11, 2013 at 6:24
  • $\begingroup$ @Habib I do appreciate that you took time to typeset your question. However, it would be nice to add a link to the source where it is from. Google suggests that it is taken from this thesis. $\endgroup$ Aug 11, 2013 at 9:24
  • $\begingroup$ BTW you can typeset $\kappa$ as $\kappa$. See the turorial on meta. $\endgroup$ Aug 11, 2013 at 9:32

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This seems to be from this thesis (BTW, it would be nice to mention such sources in the original question!). There the proof is slightly different in notation and this causes some of the confusion here:

We have a non-compact KC space $(X,\tau)$. Then there is some infinite regular cardinal $\kappa$ such that there is a strictly increasing open cover (of non-empty open subsets) $U_i$, $i < \kappa$ of $X$. (this can be found from a minimal size open cover without a finite subcover). Then the proof goes on to construct an increasing family of discrete subsets $D_i$ for some $i < \lambda \le \kappa$ by transfinite recursion: take any $x_0 \in U_0$, and define $D_0 = \{x_0\}$. For a successor ordinal $\alpha+1$, having defined $D_\alpha$, we have two cases: either $\overline{D_\alpha}$ is non-compact (and then we stop: we have found a discrete set $D_\alpha$ with non-compact closure, and $\lambda = \alpha+1$)) or $\overline{D_\alpha}$ is compact, and then we let $i_\alpha < \kappa$ be minimal such that $\overline{D_\alpha} \subset U_{i_\alpha}$. We then pick $x_{\alpha+1} \in U_{i_{\alpha}+1} \setminus U_{i_{\alpha}}$ and we define $D_{\alpha + 1} = D_\alpha \cup \{ x_{\alpha+1} \}$. Note that $D_0$ is discrete and if $D_\alpha$ is discrete, so is $D_{\alpha+1}$, as $(U_{i_\alpha + 1}\setminus \overline{D_\alpha}) \cap D_{\alpha + 1} = \{ x_{\alpha + 1}\}$, making the added point isolated in $D_{\alpha+1}$ as well. For limit ordinals $\beta$ we just "do nothing": $D_\beta = \bigcup_{\alpha < \beta} D_\alpha$.

Added: $D_\beta$ is still discrete. To see this, let $\alpha<\beta$, and let $V=U_{i_\alpha+1}\setminus\overline{D_\alpha}$. Clearly $V$ is an open nbhd of $x_{\alpha+1}$, and $V\cap D_{\alpha+1}=\{x_{\alpha+1}\}$. Suppose that $\alpha<\gamma<\beta$. Then $x_{\alpha+1}\in D_\gamma$, so $U_{i_\gamma}\supseteq\overline{D_\gamma}\supseteq D_\alpha$, and therefore $i_\gamma\ge i_\alpha$ by the minimality of $i_\alpha$. But $x_{\alpha+1}\in U_{i_\gamma}\setminus U_{i_\alpha}$, so $U_{i_\gamma}\supsetneqq U_{i_\alpha}$, and the open cover is strictly increasing, so $i_\gamma>i_\alpha$, and therefore $x_{\gamma+1}\notin U_{i_\gamma}\supseteq U_{i_\alpha+1}\supseteq V$. Thus, $V\cap D_\beta=\{x_{\alpha+1}\}$, and $D_\beta$ is discrete.

Now $\overline{D_\kappa}$ is non-compact, so for some stage we get a $D_{\alpha}$ with non-compact closure; this is basically a cardinal exhaustion argument.

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  • $\begingroup$ I don't quite see how can $U_n$, $n<\omega$, be chosen in such way that $D_\omega$ would be $\mathbb Q$. We want $x_n \in U_{\alpha_n}$, but any open set containing $x_n$ must contain infinitely many other rational numbers, and those numbers cannot be chosen as $x_m$ for $m>n$. (The construction of $D_\omega$ requires $x_m\notin U_{\alpha_n}$.) $\endgroup$ Aug 11, 2013 at 11:34
  • $\begingroup$ BTW the same fact is given as Lemma 3 in Angelo Bella, Camillo Costantini: Minimal KC spaces are compact; DOI: j.topol.2008.04.005 $\endgroup$ Aug 11, 2013 at 11:41
  • $\begingroup$ @MartinSleziak a) I cannot access the other paper (is the proof any different?) and b) how do you suggest we prove the limit case to be discrete if the previous stages are, given the construction? $\endgroup$ Aug 11, 2013 at 13:15
  • $\begingroup$ Feel free too email me (my email can be found two clicks away from my profile page) and I can send you the paper. However, the proof doesn't seem different to me. This is the part which seems relevant to what we are discussing: Any $x_\beta\in D_\delta$ has the open neighbourhood $V=U_{\alpha_\beta+1}\setminus \overline{D_\beta}$ satisfying $V\cap D_\delta=\{x_\beta\}$ (this depends on the fact, which is easy to check, that $\beta\mapsto\alpha_\beta$ is strictly increasing), and we conclude that $D_\delta$ is discrete. $\endgroup$ Aug 11, 2013 at 13:37
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    $\begingroup$ @BrianM.Scott You're free to edit my post, if you like. Make the argument more complete. $\endgroup$ Aug 11, 2013 at 15:25
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(1): why we can say in the first paragraph" If $ \overline{D}_{\beta - 1 } $ is compact, then there is $ \alpha_{\beta}$ such that $ \overline{D}_{\beta - 1 } \subset U_{\alpha_{\beta}}$. Let $ x_{ \beta } \in U_{\alpha+1} - U_{\alpha} $ and $ D_{\beta } = D_{\beta - 1 } \cup \{x_{ \beta}\}$"?

If $\overline{D}_{\beta-1}$ is compact then the cover $\{U_i; i<k\}$ has a finite subcover $\{U_{i_1},\dots,U_{i_n}\}$. Simply take $\alpha_\beta=\max\{i_1,\dots,i_n\}$.

(2): why we can say " This process stops when $ \overline{D}_{\beta} $ is not compact, which holds at least for $ \beta = k $, because then the open cover $\mathcal{ U } $ witnesses that $ \overline{D}_{k} $ is not compact.

For $\overline{D_k}$ the cover induced by $\mathcal U$ on the subspace $\overline{D_k}$ is an open cover without a finite subcover. Just notice that the transfinite sequence $\alpha_\lambda$ is strictly increasing, which implies $\alpha_\lambda>\lambda$. Any subcover must somehow cover all points $x_{\alpha_i}$ for $i<k$, and this is only possible if supremum of $\alpha_i$'s corresponding to sets from subcover is $k$. Since $k$ is an infinite regular cardinal, this cannot happen for any finite subcover.

It is easy to see that $ D_{ \beta} $ is discrete. The open set, which contains exactly one point $ x_{i+1 } $ is $ U_{ \alpha_{i+1}} -\overline{D_{i}}$."?

Since $x_i\notin U_{\alpha_i}$ and $U_{\alpha_i}\supseteq \overline{D_i}$, you have $x_i\notin\overline{D_i}$. You also have $x_i\in U_{\alpha_i+1}$. Together you get $x_i \in U_{\alpha_i+1} \setminus \overline{D_i}$.

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  • $\begingroup$ @Habib: If some further clarification is needed, feel free to ping me here or (probably better) in general topology chatroom. You need 20 rep points to talk in chat, but if you merge all your accounts, you will have more than sufficient rep for this. $\endgroup$ Aug 11, 2013 at 9:56

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