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Martin Sleziak
Martin Sleziak
6:53 AM
After I saw Kasun Fernando's edits replacing \cl with \operatorname{cl}, this reminded me to check for this macro. math.stackexchange.com/posts/169363/revisions math.stackexchange.com/posts/168866/revisions
One of those posts was mine. Again, this was the case where macro was defined in the question and after the changes to the SE software, the macro is no longer valid in the answers.
In the "naive" search just for the exact string, the problem with \cl will be that it is a prefix of some other macros defined in LaTeX.
data-explorer posts One solution is to explicitly exclude some of other macros: data.stackexchange.com/math/revision/1071496/1323755/…\cl&string2=\clubsuit&string3=\class&string4=\clearpage
data-explorer posts Another possibility is to use a regex to exclude cases where \cl is followed by a letter: data.stackexchange.com/math/revision/1094683/1351462/…
I see that the first link does not work - here it is urlencoded: data.stackexchange.com/math/revision/1071496/1323755/…
Of course, the same thing can be done in comments
Here are some comments where \cl is not rendered:
I don’t understand the choice of the sets $V(B)$ and $W(B)$. I guess, that should be non-empty, but this is not stated. Also, in this case, why $|B\cap U|>1$ implies $|(B\cap U)\setminus(\cl V_n\cup\cl W_n)|>1$? — Alex Ravsky Aug 21 '17 at 14:35
Beautiful answer. One doubt about the last paragraph. Is the condition for hereditarily collectionwise normal really $\cl F\cap\bigcup(\mathscr{F}\setminus\{F\})=\varnothing$? In a T1 space, if $\mathscr{F}$ is a family of disjoint singletons, the condition would always be satisfied, but the conclusion would not be true in general. Maybe $F\cap\cl\bigcup(\mathscr{F}\setminus\{F\})=\varnothing$ would work better, but I am not sure it matches exactly the notion of hereditarily collectionwise normal. — PatrickR Jul 20 '17 at 2:53
@M.Sina: $f$ is continuous on $Y$, so $|f(y)-a|\ge\epsilon$ for all $y\in Y\cap\cl_XV$, and $|f(y)-a|\le\frac{\epsilon}2$ for all $y\in Y\cap\cl_XW$. — Brian M. Scott Sep 11 '13 at 8:27
$p$ is the only point in the $\cl_XV\cap\cl_XW$? if there is another point in $\cl_XV\cap\cl_XW$ how can we show that $Y\cap\cl_XV\cap\cl_XW=\varnothing$? — M.Sina Sep 11 '13 at 5:23
@M.Sina: Because $p\in\cl_XU$ for any open nbhd $U$ of $a$ in $\Bbb R$: see the second paragraph of the proof. — Brian M. Scott Sep 10 '13 at 20:23
@Klara: You’re welcome, as always. $U_x\subseteq N_x$, so $\cl U_x\subseteq\cl N_x$. But $N_x$ is closed, so $\cl N_x=N_x$, and therefore $\cl U_x\subseteq N_x$. For the other part, have you seen the proof that a compact Hausdorff space is regular? — Brian M. Scott Apr 25 '13 at 7:27
@Stefan: All you have to prove is that $K\cap\cl U=\varnothing$, which is clear, since $K$ is relatively open in $Y$. — Brian M. Scott Feb 21 '13 at 20:05
Since $Y$ is normal, there are disjoint open sets containing $\cl_Y A$ and $\cl_Y B$. Does the fact that $Y$ is open give us that these open sets are subsets of $Y$, and so are not only open sets in the subspace topology on $Y$ but open sets in $X$ as well, giving us our result? — Alex Petzke Jan 31 '13 at 2:07
@TXC: Yes. If $y\in Y\cap\cl_XU$, and $V$ is any open nbhd of $y$ in $Y$, let $W$ be open in $X$ with $V=Y\cap W$. Then $W\cap U\ne\varnothing$, and since $U\subseteq Y$, $W\cap U=V\cap U$. Thus, $V\cap U\ne\varnothing$, and $y\in\cl_YU$. — Brian M. Scott Jan 30 '13 at 5:34
@Mathematics: No, it doesn’t. But it’s true that $A\cup U\ne A\setminus U$, simply because $U\ne\varnothing$. I think that you’re making this more complicated than it really is; all I’m saying is that if $A$ is not connected, there must be an open $U$ in $\Bbb R$ such that $A\cap U$ and $A\setminus\cl U$ are non-empty relatively open subsets of $A$ whose union is $A$ $-$ in other words, a disconnection of $A$. — Brian M. Scott Nov 17 '12 at 6:40
@ege: You’re welcome. By the way, $\{X\cap\int_Y\cl_YB:B\in\mathscr{B}\}$ is also a local base at $x$, since $Y$ is regular, and if you use it instead of $\mathscr{B}$, then you automatically get $x$ to be in the $Y$-interior of each member of the local base. But you don’t need this for the argument. — Brian M. Scott Nov 9 '12 at 15:58
@ege: More generally, if $G$ is open in $X$ and $x\in G$, then $x\in\int_Y\cl_YG$. There is an open $U$ in $Y$ such that $G=U\cap X$. $X$ is dense in $Y$, so $\cl_YG=\cl_YU$. Thus, $x\in U\subseteq\int_Y\cl_YU=\int_Y\cl_YG$. — Brian M. Scott Nov 8 '12 at 17:52
How do you know that it can’t happen that $x\in(\cl B)\setminus B$, and the only member of the countable network containing $x$ is $\{x\}$? — Brian M. Scott Feb 23 '13 at 11:22
While it’s true that each open nbhd of $a\in\cl A$ meets $A$, this isn’t quite immediate from the OP’s definition of $\cl A$; at the level of experience implied by the question it requires proof. The assertion that $\cl A\cap\cl(X\setminus A)=\bdry A$ also requires proof. — Brian M. Scott Dec 4 '12 at 1:31
One comment popped up several times in the SEDE queries but it no longer exists on the page: math.stackexchange.com/questions/240656/…
The text shown in SEDE is: "I have added a \cl command to be available on this page." And it was posted by Asaf Karagila.
I suppose that Asaf looked at that post after I have edited his answer and removed the comment. (Clearly, he is very fast.)
 

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