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10:40 AM
Also block \let? — Hagen von Eitzen 8 hours ago
Hagen von Eitzen pointed out that \let can be also problematic in the titles.
I did not find any posts with \let in the title: data.stackexchange.com/math/query/972169/…\let
 
11:03 AM
This link should work: data.stackexchange.com/math/query/1071909/… (to find the posts with \let\ )
To add to Hagen von Eitzen's comment, I did not find posts with \let in the title. There were two questions where \let was in the title, by it was later edited away: chat.stackexchange.com/transcript/88939/2019/6/30Martin Sleziak 19 mins ago
 
 
6 hours later…
5:17 PM
It seems that some people use \sp for span. A minor problem with searching for such posts is that \sp might be a prefix of some other macros, so I might search for something like this: data.stackexchange.com/math/revision/1071496/1323755/…
Anyway, search like this returns mostly posts/comments with \span or \spec rather than \sp: data.stackexchange.com/math/revision/1071496/1323755/… data.stackexchange.com/math/revision/1071502/1323761/…
I found two comments with \spt for support - defined in the post.
@chuyenvien94 Note that the set $\{x; \forall \delta>0: \nu(B(x,\delta))>0\}$ is closed, i.e. $$\text{spt} \nu = \overline{\{x; \forall \delta: \nu(B(x,\delta))>0\}}.$$ Concerning $\int f \, d\mu = 1$: This condition ensures that $\spt \nu \neq \emptyset$. In fact, $\spt \nu \neq \emptyset \Leftrightarrow \int f \, d\mu >0.$$ — saz Jun 15 '14 at 7:07
I think $\spt\nu:=\overline{\{x\in\mathbb{R};\forall\delta>0:\nu(B(x,\delta))>0\}} $, that 's the closure of the set you specified. Because support of $f$ is the closure of the set of all $x$ such that $f(x) \neq 0$ — chuyenvien94 Jun 15 '14 at 1:02
Two comments with macro \sp actually defined in a post and used in a comment: data.stackexchange.com/math/revision/1066862/1318474/…
Well, because $E$ was computed from $S(\{T(b_1),\ldots,T(b_n)\})$ so as to be basis for the subspace they span, which subspace is $S(\operatorname{Im}(T))$. From $\sp(E)=S(\operatorname{Im}(T))$ one gets $S^{-1}(\sp(E))=\operatorname{Im}(T)$, while also $S^{-1}(\sp(E))=\sp(S^{-1}(E))$. — Marc van Leeuwen Feb 21 '15 at 13:27
If $T$ and $K$ are subspaces and $\sp$ is span, then $\sp(T) = T$ and $\sp(K) = K$. The hypotheses $T \subseteq \sp(K)$ and $K \subseteq \sp(T)$ are therefore equivalent to $\sp(T) \subseteq \sp(K)$ and $\sp(K) \subseteq \sp(T)$; it follows immediately that $\sp(K)= \sp(T)$. A more interesting question is whether $T \subseteq \operatorname{K}$ and $K \subseteq \sp(T)$ imply $\sp(K) = \sp(T)$ in the case where $T$ and $K$ are just sets of vectors. — Michael Albanese Dec 20 '14 at 20:19
Hi Saaqib. Take note of the changes I made to your post. One change was in the typesetting (I needed to define \span), and the other was the title (titles should describe the problem). — Omnomnomnom May 6 '15 at 14:42
This one is unusual - a comment on a question, but it relies on a macro defined in one of the answers:
The expression $\span(\{v,Av\})=...$ you write in the question is conceptually wrong. The first term is a set while the second and third terms are a vector. — Jack Jun 21 '17 at 19:49
how do we show that if $x \in \overline{M} = \overline{\span \{e_1, e_2, \ldots\}}$, then the convegeent series $\sum \langle x, e_k \rangle e_k$ necessarily has the sum $x$? — Saaqib Mahmood May 23 '15 at 5:19
 

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