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Despite the announcement of #2, the changes I made are not being displayed. I’m not seeing any syntax errors.
I’m recording my edit here, to be on the safe side.
+– {: .num_prop}
A countably compact space is a limit compact space. =–
+– {: .proof}
Recall that a space is limit point compact if every closed discrete subspace is finite; equivalently, if every countable closed discrete subspace is finite.
Suppose $X$ is countably compact and $A$ is a countable closed discrete subspace. For each $a \in A$, choose an open neighborhood $U_a$ such that $U_a \cap A = \{a\}$, and let $V_a$ be the open subset $U_a \cup \neg A$. Clearly we still have $V_a \cap A = \{a\}$. Also, $\{V_a: a \in A\}$ is a countable cover of $X$, hence admits a finite subcover $V_{a_1}, \ldots, V_{a_n}$. But then
$A = \left(\bigcup_{i=1}^n V_{a_i} \right) \cap A = \bigcup_{i=1}^n (V_{a_i} \cap A) = \{a_1, \ldots, a_n\}$as was to be shown.
=–
+– {: .num_prop}
A space that is $T_1$ and limit point compact is countably compact. =–
+– {: .proof}
Equivalently (under classical logic), the assertion is that if a $T_1$ space $X$ is not countably compact, then it is not limit point compact. Indeed, under this hypothesis, there is a countable open cover $U_1, U_2, \ldots$ of $X$ that admits no finite subcover. We put $V_n = U_1 \cup \ldots \cup U_n$ so that $V_1 \subseteq V_2 \subseteq \ldots$; since every point belongs to some $V_n$ but no $V_n$ is all of $X$, we may discard repetitions and assume without loss of generality that all the inclusions
$\emptyset = V_0 \subset V_1 \subset V_2 \subset \ldots$are strict. Thus for each $n \geq 1$ we may pick a point $x_n \in V_n \setminus V_{n-1}$. Observe that if $m \lt n$, then $x_n \notin V_m$.
Since $X$ is $T_1$ (points are closed), the set $W_m = V_m \cap \neg \{x_1, \ldots, x_{m-1}\}$ is an open neighborhood of $x_m$ that does not contain $x_n$ whenever $m \lt n$, and does not contain $x_n$ for $n \lt m$. Thus every point $x_n$ is open relative to $A = \{x_1, x_2, \ldots\}$, i.e., $A$ is a discrete subspace.
Finally, any point $x \notin A$ belongs to some $V_n$, and then $V_n \cap \neg \{x_1, \ldots x_n\}$ is an open neighborhood of $x$ that doesn’t intersect $A$. Thus $A$ is an infinite closed discrete subspace, meaning that $X$ is not limit point compact. =–
Apparently I got it to display just now. Not sure what happened earlier.
Hi Todd, probably you have seen here that major changes have been made to the rendering of nLab pages, so there may be a few gremlins. However, in this case I cannot reproduce the problem (just tried a couple of trivial edits on this page, and both were visible immediately). Please let me know if you see the same problem again, I will look into it.
Okay, will do. Yes, I am aware that there may be glitches while you’re working on the system; I’m just reporting them as they arise. Thanks!
Absolutely, it is very helpful that you report them. This issue is a bit mysterious to me for the moment, so I will be very grateful if anyone else who experiences it can also report it. Thanks for your understanding!
Hi Richard. At limit point compact space, the link to metrizable space is broken.
Just reporting. Anything I can do to help?
Thanks Todd! This is the same issue that metrizable space has not been rendered by the new renderer. My plan for the moment is remove the restriction on no duplicate redirects, so that we can generate all pages using the new renderer. Then I can switch it back on for edits. Before that, I need to fix an issue with the table of contents rendering that Urs reported. Will keep updated. Sorry for all inconvenience. We are getting there, slowly.
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