Also block \let? — Hagen von Eitzen 8 hours ago

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/88939/2019/6/30 — Martin Sleziak 19 mins ago

@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

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

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