Sharepoint sync sets every document to read-only

I’ve made some adjustments to the library we are using, but now users can’t direct edit the items in the desktop. However, I can.

They see the following:

  • Green locks on every document
  • When they open a document its in read-only mode
  • When they try to upload a new document they are asked for administrator rights.

enter image description here

How do I fix this? So that every user is able again to edit and upload documents on the fly.

Thank you.

Cardinality of permutations of 3 finite sets

I’m trying to understand the definition of the cardinality of permutations, but after looking online I’m still struggling with this:

If for example I have 3 finite sets:

$ A = \{2,2,4\}\$ $ B = \{1,2,3\}\$ $ C = \{4,5,1\}\$

And I want to find the cardinality of permutations, how should I do it? Does it mean to find the number of distinct permutations of each set and then multiply those 3 numbers? Can someone help me clarify the process? Thank you!

Free skew fileds over sets of different cardinal

Let $ K$ be a field and let $ X$ be a set. Denote by $ \mathcal D_K(X)$ the free skew $ K$ -field on $ X$ .

Assume that $ |X|\ne |Y|$ . Is it true that $ \mathcal D_K(X)$ and $ \mathcal D_K(Y)$ are not isomorphic?

There are several constructions of $ \mathcal D_K(X)$ . The standard one is the following. Let $ F_X$ be a free group on $ X$ . Fix a bi-invariant total order $ \le$ on $ F_X$ and let $ K_{\le}((F_X))$ be the Malcev-Neumann ring of formal series (it consists of formal series having well-ordered support). One proves that $ K_{\le}((F_X))$ is a skew field. Then $ \mathcal D_K(X)$ is isomorphic to the division closure of the group algebra $ K[F_X]$ in $ K_{\le}((F_X))$ .

Can closure and complement generate 14 distinct operations on a topological space if no subset generates more than 6 distinct sets?

$ \newcommand{\XT}{(X,\mathcal{T})}$ From Definition 1.2 in Gardner and Jackson (GJ): The $ K$ ‑number $ K(\XT)$ of a topological space $ \XT$ is the cardinality of the Kuratowski monoid of operators on $ \XT$ generated by closure $ b$ and complement $ a$ under composition. For any $ A\subset X,$ the $ k$ ‑number $ k(A)$ of $ A$ is the cardinality of the family of subsets generated by $ A$ under $ \{a,b\}.$ The $ k$ ‑number of the space is $ $ k((X,\mathcal{T}))=\max\{k(A):A\subset X\}.$ $

Question. What values does the ordered pair $ (K(\XT),k(\XT))$ take?

Partial Answer. Lemma 2.8 in GJ states that $ k(\XT)=4$ iff $ \XT$ is a non-discrete door space or $ \mathcal{T}\setminus\varnothing$ is a filter in $ 2^X.$ Each implies $ K(\XT)<14,$ hence $ (14,4)$ is impossible (GJ p. 25).

The case $ (14,6)$ only gets mentioned once in GJ, on p. 28: $ $ \unicode{x201C}\text{We do not know of … any Kuratowski space with }k\text{-number }6.\unicode{x201D}$ $ (A Kuratowski space is one that satisfies $ K(\XT)=14.)$

It is well known that $ k(A)$ is always even (GJ p. 15). It follows from the definitions and the identity $ bababab=bab$ that $ $ 2\leq k(\XT)\leq K(\XT)\leq14.$ $

By Theorem 2.1 in GJ, the only possible values of $ K(\XT)$ are 14, 10, 8, 6, 2. Clearly, $ k(\XT)=2$ iff $ \XT$ is discrete. Thus, besides possibly $ (14,6),$ the pair can only be:

$ (2,2),$
$ (6,4),$ $ (6,6),$
$ (8,4),$ $ (8,6),$ $ (8,8),$
$ (10,4),$ $ (10,6),$ $ (10,8),$ $ (10,10),$
$ (14,8),$ $ (14,10),$ $ (14,12),$ $ (14,14).$

Each occurs in some space $ \XT$ with $ |X|\leq7$ (this holds for both Kuratowski monoids that satisfy $ K(\XT)=10).$ Examples are listed below.

Theorem 2.10 in GJ states that for any $ A\subset X,$ the family of subsets generated by $ A$ under $ \{b,i\}$ where $ i$ denotes interior satisfies exactly one of 30 possible collapses of the Hasse diagram:

$ \hspace{268px}$

The table below lists these collapses in the same order as they appear in Table 2.1 in GJ. Each entry is labeled by what I am calling the $ h$ ‑number of $ A$ , or $ h(A).$ The identity operator is denoted by $ \textsf{id}.$

\begin{array}{|c|c|c|c|} \hline h(A) & h(aA) & \text{collapse} & k(A) \ \hline 1 & 1 & \varnothing & 14 \ \hline 2 & 3 & bi=ibi & 12 \ \hline 3 & 2 & ib=bib & 12 \ \hline 4 & 5 & bib=b & 12 \ \hline 5 & 4 & ibi=i & 12 \ \hline 6 & 6 & ib=ibi,\ bi=bib & 10 \ \hline 7 & 7 & ib=bib,\ bi=ibi & 10 \ \hline 8 & 9 & ib=bib,\ ibi=i & 10 \ \hline 9 & 8 & bi=ibi,\ bib=b & 10 \ \hline 10 & 11 & bi=ibi=i & 10 \ \hline 11 & 10 & ib=bib=b & 10 \ \hline 12 & 12 & bib=b,\ ibi=i & 10 \ \hline 13 & 13 & ibi=bi=ib=bib & 8 \ \hline 14 & 16 & ib=ibi=i,\ bi=bib & 8 \ \hline 15 & 17 & ib=bib,\ bi=ibi=i & 8 \ \hline 16 & 14 & ib=ibi,\ bi=bib=b & 8 \ \hline 17 & 15 & ib=bib=b,\ bi=ibi & 8 \ \hline 18 & 19 & bi=ibi=i,\ bib=b & 8 \ \hline 19 & 18 & ib=bib=b,\ ibi=i & 8 \ \hline 20 & 21 & ibi=bi=ib=bib=i & 6 \ \hline 21 & 20 & ibi=bi=ib=bib=b & 6\ \hline 22 & 22 & ib=ibi=i,\ bi=bib=b & 6 \ \hline 23 & 24 & \textsf{id}=b,\ ib=ibi=i,\ bi=bib & 6 \ \hline 24 & 23 & \textsf{id}=i,\ bi=bib=b,\ ib=ibi & 6 \ \hline 25 & 25 & ib=bib=b,\ bi=ibi=i & 4,6 \ \hline 26 & 27 & \textsf{id}=bi=bib=b,\ ib=ibi=i & 4 \ \hline 27 & 26 & \textsf{id}=ib=ibi=i,\ bi=bib=b & 4 \ \hline 28 & 29 & \textsf{id}=b,\ ibi=bi=ib=bib=i & 4 \ \hline 29 & 28 & \textsf{id}=i,\ ibi=bi=ib=bib=b & 4 \ \hline 30 & 30 & ibi=bi=ib=bib=b=i=\textsf{id} & 4 \ \hline \end{array}

When $ h(A)=25,$ if $ A$ is dense with empty interior (e.g., $ \mathbb{Q}$ in $ \mathbb{R}$ under the usual topology) then $ k(A)=4,$ otherwise $ k(A)=6.$

Conjecture. Let $ A$ and $ B$ be subsets of a topological space. If $ $ \tag1h(A)\in\{22,23,24,26,27\}\text{ and }h(B)=25,$ $ then $ $ \tag2\min\{h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB)\}<20.$ $

If true, the conjecture implies $ (14,6)$ is impossible. For, suppose $ \XT$ is a Kuratowski space with $ k$ ‑number 6. Since $ K(\XT)=14,$ there exist subsets $ A$ and $ B$ of $ X$ such that $ biA\neq ibiA$ and $ ibB\neq ibiB.$ Since $ k(\XT)=6,$ it follows from the table that $ (1)$ holds. The conjecture then contradicts $ k(\XT)=6.$

Computer experiments have verified the conjecture for all 2450 inequivalent $ \text{non-}T_0$ spaces such that $ 1\leq|X|\leq7$ (these were recently posted here) and roughly 40,000 others such that $ 8\leq|X|\leq16.$ We ignore $ T_0$ spaces because finite ones satisfy $ K(\XT)\leq10$ by Theorem 3 in Herda and Metzler. We also ignore sets $ A$ such that $ h(A)\in\{24,27\}$ because our C program scours entire power sets and De Morgan’s laws imply that $ A$ provides a counterexample iff $ aA$ does.

Our list of 136 known (at the time of writing) $ 4$ ‑tuples $ $ (h(A\cup B),h(A\cap B),h(A\cup aB),h(A\cap aB))$ $ satisfying $ h(A)\in\{22,23,26\}$ and $ h(B)=25$ can be found here.

A few weeks ago I posted a question here based on the case $ (26,25,30,12),$ where $ h(A)=26.$ Currently no answer has appeared.

Here is the list promised above. It is surely complete, but we lack proof. The cardinality of each space is minimal. It is assumed that $ X=\{1,\ldots,n\}$ when $ |X|=n.$

\begin{array}{|c|c|c|c|} \hline \text{pair} & |X| & \text{base for }\mathcal{T} & h\text{-numbers that occur} \ \hline (2,2) & 1 & \{\{1\}\} & 30 \ \hline (6,4) & 2 & \{\{1,2\}\} & 25,30 \ \hline (6,6) & 3 & \{\{1\},\{2,3\}\} & 25,30 \ \hline (8,4) & 2 & \{\{1\},\{1,2\}\} & 28\text{-}30 \ \hline (8,6) & 3 & \{\{1\},\{1,2\},\{1,2,3\}\} & 20,21,28\text{-}30 \ \hline (8,8) & 4 & \{\{1\},\{2\},\{1,3\},\{2,4\}\} & 13,28\text{-}30 \ \hline (10,4) & 3 & \{\{1\},\{2\},\{1,2,3\}\} & 26\text{-}30 \ \hline (10,6) & 4 & \{\{1\},\{2\},\{1,2,3\},\{1,2,4\}\} & 22,26\text{-}30 \ \hline (10,8) & 4 & \{\{1\},\{2\},\{1,3\},\{1,2,4\}\} & 14,16,23,24,26\text{-}30 \ \hline (10,10) & 5 & \{\{1\},\{2\},\{1,3\},\{2,4\},\{1,2,5\}\} & 6,14,16,23,24,26\text{-}30 \ \hline (14,8) & 4 & \{\{1\},\{2,3\},\{1,2,3,4\}\} & 18,19,26\text{-}30 \ \hline (14,10) & 5 & \{\{1\},\{2,3\},\{1,4\},\{1,2,3,5\}\} & 10,11,14,16,18,19,23,24,26\text{-}30 \ \hline (14,12) & 6 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{1,2,6\}\} & 4,5,12,14\text{-}17,23\text{-}30 \ \hline (14,14) & 7 & \{\{1\},\{2\},\{3,4\},\{1,5\},\{2,6\},\{1,2,7\}\} & 1,4,5,6,12,14\text{-}17,23\text{-}30 \ \hline \end{array}

Countable union of well ordered sets

Assume I have a sequence $ (A_i)_{i<\omega}$ of well-ordered subsets of an ordered set $ S$ . Assume that $ A:=\underset{i<\omega}{\cup}A_i$ is also well-ordered. Let $ \alpha$ be an ordinal upper bound on the order types of the $ A_i$ ‘s. Does anyone know a condition such that $ A$ has order type at most $ \alpha\otimes\omega$ (natural multiplication of ordinal) ?

Actually, I know that we can say that $ \underset{i=0}{\overset{n-1}{\cup}}A_i$ is well ordered and has order type at most $ \alpha\otimes n$ . This fact is true since if $ U$ and $ V$ are well ordered with ordered type $ \alpha$ and $ \beta$ , then $ U\cup V$ is well ordered with order type $ \alpha\oplus\beta$ . I don’t have any reference to this fact but many article seems to say that it is known. But it is not known by me. Using this fact, the intuition about my above question should be true, at least for reasonnable condition on $ S$ and the $ A_i$ ‘s.

Thanks in advanced for reading and (possibly) answering.

SharePoint Document Sets with template from content type

Document Sets are created and simple but in the default content – I do not want static documents. I want the users to be able to modify the document as and when. I have attempted to add the template in a content type and used that content type as default content, but it does not stay in likewise the document does not arrive in the document set – any ideas ?

Partitioning $\{0,1\}^n$ into $n$ sets

I am working on an answer to the question

Magic trick based on deep mathematics

and came across the following problem: I am trying to partition the cube $ \{0,1\}^n$ into $ n$ sets $ P_1,\dots,P_n$ such that, for any point $ x\in\{0,1\}^n$ and any $ i\in\{1,\dots,n\}$ , there exists a point $ y\in P_i$ such that $ x$ and $ y$ differ in at most one term. Using my computer I know that this is possible for $ n\leq 4$ , impossible for $ n=5$ and $ 6$ , and possible for $ n=7$ and $ 8$ (by brute force integer programming). For example, one solution for $ n=4$ is \begin{align*} P_{1} & =xyyy\text{ or }xxxx\ P_{2} & =xyxx\text{ or }xyyx\ P_{3} & =xxyx\text{ or }xyxy\ P_{4} & =xxxy\text{ or }xxyy \end{align*} where $ x$ and $ y$ are separate values in $ \{0,1\}$ . My question is: can this partition (or something like it) extend to $ n=8$ in a simple way? Alternatively, are there any values of $ n$ for which there is a simple and natural partition?

Ionic 3: How to calculate sum of values in sets of arrays

I have few sets of array which looks like this enter image description here

How can i sum up all the cartTotal?

in my ts file code are like this

with this method i am getting constantly 0 as answer

itemTotal() {     let total = 0;     for (var i = 0; i < this.bag.length; i++) {       if (this.bag.cartTotal[i]) {         total += Number(this.bag.cartTotal[i]);       }     }     return total;   } 

How can i achieve this?

$ \inf_{ \text {open sets covering E} } ( \sum length(I_n) ) \leq \inf_{ \text {closed sets covering E} } ( \sum length(J_n) ) $

Let $ E$ be a subset of $ \mathbb R $ , $ J_n$ be a closed cover of $ E$ , where $ \forall n, J_n = [a_n,b_n]$ and we build the open cover $ I_n$ like this : $ \forall n, I_n = ]a_n – \frac {\epsilon }{2^{n+1}} , b_n + \frac {\epsilon }{2^{n+1}} [ $ .

I’m struggling to prove that the outer measure on open sets is inferior to the one on closed sets. So here I’d like to show that :

$ $ \inf_{ \text {open sets covering E} } ( \sum length(I_n) ) \leq \inf_{ \text {closed sets covering E} } ( \sum length(J_n) ) $ $

Why am I struggling ?

For now I’ve proved this :

$ $ \sum length(I_n) = \sum length(J_n) + \epsilon$ $ here I’m stuck.


Please help me to complete this proof. Moreover, I ask for not giving me a proof using “it works for all closed sets so we take the infinimum on the RHS”. I’m asking for something really formal.