# Sandbox for drafts of long, complex posts [closed]

This sandbox is intended for saving drafts of long, complex posts, especially posts whose composition takes a long time. It serves to localize to one thread the front-page "bumps" caused by edits to drafts of such posts, so that they may be easily ignored. Also, it helps to guard against losing longly-composed posts due to system crashes.

When you are happy with your draft here, you may simply copy the code and paste it to the desired location.

## Proper Use of the Sandbox

1. Do not post a new answer! We wish all the answers on this page to be owned by the Community user (so that only a non-sentient bot is informed of edits to these answers). Posting a new answer will make you the owner, meaning that you will be notified whenever another user makes an edit to that answer.

The sandbox has been closed to prevent the creation of new answers. There are more than enough existing answers for users to edit over, and this will greatly reduce the frequency at which we request that the answers be disassociated from specific users.

2. Do not delete answers! Deleting seems like a reasonable option, but there are no "hard deletions" on Stack Exchange, and users with sufficient privileges will still see your supposedly deleted postings. Deleted answers will be undeleted and cleared for the use of others.

3. Do look for an answer which indicates that it is free and then edit it to your heart's content. If none appears available, take over the one that has been left unchanged the longest (which will appear at the bottom of the page if you order answers by "activity").

4. Do not expect your draft to remain untouched for days. There are no guarantees that your draft will be the latest revision if you return days later. While users will try not to step over others' toes, it may happen that an unfinished draft is edited out. Your draft will, however, still exist as a revision of the answer it was made in. If your drafting is expected to take place over a longer period of time, either

• take note of the URL of the answer provided by clicking the share button, or
• save a copy of your draft locally (or even "in the cloud").
5. Do clear your draft when you are finished. This includes removing all $\LaTeX$ from your answers. Replacing all code with a simple statement like

This answer is free for anyone to use

is sufficient. Periodically users may go through and free up answer slots that have not been edited in, say, over one month. But you can aid in the smooth running of this sandbox by clearing away your drafts when you are finished with them.

6. Do not "claim" multiple answers concurrently. Since this post is closed, the answers are a limited resource. If you really must compose several long, complex posts at the same time, you can still use a single answer, separating the different drafts using Markup: horizontal rules (---) and/or headings (# Header 1 #) are natural choices.

7. Do not create new such sandboxes. The point of having a unique such sandbox is that it minimizes the noise on the front page when the sandbox is edited. If there were multiple sandboxes they will frequently occupy numerous front page slots, pushing other topics off the front page, and increasing noise.

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## closed as off-topic by inactive... for now♦Sep 18 '14 at 10:37

• This question does not appear to be about Mathematics Stack Exchange or the software that powers the Stack Exchange network within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

I have added a [sandbox] tag to allow people ignore it more easily (via software support of ignoring tags), and since it seems that we have two sandboxes now, a tag may seem a bit more in place here. – Asaf Karagila Jul 18 '12 at 8:35
(+1) For thinking outside the (sand)box. – cardinal Jul 18 '12 at 19:40
At the suggestion of the moderators, I have gone and changed the associated owners of all the answers here to the Community user. This way, the original owners will not receive excess pings for each time another user uses the draft space for their work. Enjoy! – Grace Note Oct 5 '12 at 14:45
To prevent crashes I've found the "Bookmarks to disable/enable MathJax", provided in here, pretty useful. – leo Dec 17 '12 at 18:03
This "sandbox" is being closed to prevent the creation of new answers. To start a draft, simply edit one of the existing free answers. – inactive... for now Sep 18 '14 at 10:37
PSA: Between the creation of this sandbox (in July 2012) and today (December 2015), technology has advanced. Something like StackEdit (or others, it's simply the only one I know) essentially solves all the limitations of this sandbox. You can have multiple concurrent drafts, you don't have to worry about polluting meta's front page, you can leave your draft untouched for days and expect it to still be there, you don't have to explicitly clear up your draft when you're done... Maybe someday we can get rid of this outdated crutch. – Najib Idrissi Dec 2 '15 at 14:07
@GraceNote Could you please also change to Community owned all of the posts in our Formatting Sandbox., including deleted posts too (I just got many pings when someone started using a deleted post). Thanks. – Bill Dubuque Dec 24 '15 at 19:50

This answer is free for anyone to use.

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The four vertices of the described 4-hedron are (6,0,0); (0,3,0); (0,0,2); and (0,0,0). The volume is $\frac16\abs\left[\begin{vmatrix}6&0&0&1\\0&3&0&1\\0&0&2&1\\0&0&0&1\end{vmatrix}\right]$

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This answer is free for anyone to use.

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Does [zEx⇒(xEy⇒zEy)]∧[zEy ⇒(xEy⇒yEx⇒zEx)] imply xEy⇒(zEx=zEy)?

The author seems to consider [zEx⇒(xEy⇒zEy)]∧[zEy ⇒(xEy⇒yEx⇒zEx)] imply xEy⇒(zEx=zEy) and as shown in the text in red.

When it's developed, it's true that it leads to (zEx=zEy), but the hypothesis is not just xEy.
[zEx⇒(xEy⇒zEy)]∧[zEy ⇒(xEy⇒yEx⇒zEx)] ≡(zEx⇒zEy)∧(zEy ⇒zEx) ≡ (zEx=zEy)

If I develop the hypothesis to get xEy [zEx⇒(xEy⇒zEy)]∧[zEy ⇒(xEy⇒yEx⇒zEx)] ≡ zEx⇒(xEy⇒zEy) Simp. ⇒ (xEy⇒zEy)

"Proof of Theorem 3

(c) It follows immediately from (a) and (b) above that x/$\mathscr E$ = y/$\mathscr E \Rightarrow x \mathscr E$ y

We need to prove that x$\mathscr E$ $y\Rightarrow x/\mathscr E = y/\mathscr E$ Let x$\mathscr E$y. Then
$z\in x/\mathscr E \Rightarrow z\mathscr E x$ Def. 6

$\space\space\color{gray} {\text{The author repeats the conjunction of above two hypothesises in symbols as the following}}$ Author's intention is proving x$\mathscr E$ $y\Rightarrow (x/\mathscr E \Rightarrow y/\mathscr E$), and then x$\mathscr E$ $y\Rightarrow (x/\mathscr E \Leftarrow y/\mathscr E)$ to prove that x$\mathscr E$ $y\Rightarrow x/\mathscr E = y/\mathscr E$ with a conjunction of two by Definition 1.

$\underline {z\mathscr Ex \land x\mathscr E y \Rightarrow z\mathscr Ey}$ $\space\space\space\space\space\mathscr E$ is transitive.
$\underline {\Rightarrow z\in y/\mathscr E}$ $\space\space\space\space\space Def. 6$

$\color{gray}{\text {The author regards the above$z\mathscr E x \land x\mathscr E y \Rightarrow z\mathscr E y$is equivaelnt to$z\mathscr E x \Rightarrow z\mathscr E y$}}$
$\color{gray}{\text {I'll check if it's true:}}$
$\color{red}{\text {$z\mathscr E x \land x\mathscr E y \Rightarrow z\mathscr E y\equiv z\mathscr E x \Rightarrow (x\mathscr Ey\Rightarrow z\mathscr E y)$by Exportation law in Example 7}}$
$\color{red}{\text {$\equiv z\mathscr E x \Rightarrow z\mathscr E y$}}$

But what's proved is $\color{red}{z\mathscr E x \Rightarrow (x\mathscr Ey\Rightarrow z\mathscr E y)}$, not $\color{red}{x\mathscr E y\Rightarrow (x/\mathscr E \Rightarrow y/\mathscr E)}$.

Since $z$ is arbitrary, it follows that $x/\mathscr E \subseteq y/\mathscr E$. $\underline {\text {A similar argument gives}\space y/\mathscr E \subseteq x/\mathscr E}$; hence $x/\mathscr E = y/\mathscr E$ "
Q.E.D.

If I develop the logical step of the similar argument, it becomes $\color{red}{\text {$z\mathscr E y \Rightarrow (x\mathscr E y \Rightarrow y \mathscr E x \Rightarrow z\mathscr E x$)$\mathscr E$is symmetric, exportation law$\equiv z\mathscr E y \Rightarrow z \mathscr E x$}}$

FYI

"Theorem 3. Let $\mathscr E$ be an equivalence relation on a nonempty set $X$. Then
(a) Each $x/\mathscr E$ is a nonempty subset of $X$.
(b) $x/\mathscr E \cap y/\mathscr E \neq \emptyset$ if and only if $x\mathscr Ey$.
(c) $x\mathscr Ey$ if and only if $x/\mathscr E = y/\mathscr E$"
...
Definition 6. Let $\mathscr E$ be an equivalence relation on a nonempty set $X$. For each $x\in X$, we define

​$X/\mathscr E=\{\,y∈X\mid y\mathscr E x\,\}$

which is called the equivalence class determined by the element $x$.

The set of all such equivalence classes on $X$ is denoted by $X/\mathscr E$; that is, $X/\mathscr E=\{\,x/\mathscr E\mid x∈X\,\}$.

The symbol $X/\mathscr E$ is read "$X$ modulo $\mathscr E$," or simply "$X$ mod $\mathscr E$".

Example 7 Prove the following Exportation Law: $\space\space\space\space\space p \land q \to r \equiv p \to (q\to r)$

[Solution] $p\to (q \to r) \equiv p \to$ ~($q\land$ ~r) Def. 4
$\space\space\space\space\equiv$ ~[$p\land(q\land$ ~r)]$\space\space Def. 4, D.N.$
$\space\space\space\space\equiv$ ~[($p\land q)\land$~r]$\space\space$ Assoc.
$\space\space\space\space\equiv p\land q\to r\space\space$ Def. 4

Hence, p$\land q \to r \equiv p \to (q \to r)$

Definition 4 The connective $\rightarrow$ is called the conditional and may be placed between any two statement p and q to form the compound statement p→q(reaad: "if p, then q". By definition the statement p→q is equivalent to the statement ~(p∧~q).

Definition 1. Two sets A and B are said to be equal or identical, in symbols: A=B, provided that they contain the same elements.
That is, A= B means that
$\forall x(x\in A\Leftrightarrow x\in B)$ Source: Set Theory by You-Feng Lin, Shwu-Yeng T. Lin

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This answer is currently free to edit.

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This answer is free to for anyone to use

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This answer is free for anyone to use.

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This answer is available to anyone.

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This answer is available to anyone.

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"Theorem 4 Let $\mathscr E$ be an equivalence relation on a nonempty set X. Then X/$\mathscr E$ is a partition of X.

[Proof] By Theorem 3(a) and Definition 6, X/​$\mathscr E$ ={x/$\mathscr E$ | $x \in$X} is a family of nonempty subsets of X. We next show that

x/$\mathscr E \neq$ y/$\mathscr E$ ⇒ x/$\mathscr E \cap$ y/$\mathscr E$ = $\emptyset$

by showing its contrapositive : $\color{red}{x/\mathscr E \cap y/\mathscr E \neq \emptyset \Rightarrow x/\mathscr E=y/\mathscr E}$.

The last assertion is a direct consequence of Theorem 3(b) and (c). Finally, we have to show that $\bigcup\limits_{x\in X} x$/ $\mathscr E$ = $X$. This is also trivial, since each x in X belongs to x/$\mathscr E$."

The partition X/$\mathscr E$ of X satisfies two conditions:

(a) (x/$\mathscr E, y/\mathscr E \in X/\mathscr E) \land$ (x/$\mathscr E \neq$ y/$\mathscr E) \Rightarrow$ x/$\mathscr E \cap$ y/$\mathscr E$ = $\emptyset$

(b) $\bigcup \limits_{x \in X}x/\mathscr E$ = X

FYI

"Definition 5 Let X be a nonempty set. By a partician P of X we mean a set of nonempty subsets of X such that:

(a) If A, B$\in$P and A$\neq$B, then A$\cap$B=$\emptyset$ (b) $\bigcup \limits_{C \in P}C$ = X"

"Definition 6. Let $\mathscr E$ be an equivalence relation on a nonempty set X . For each x∈X, we define
​$\space\space\space\space\space\space\space\space\space$ $x/\mathscr E$={$y\in X∣y\mathscr Ex$}
which is called the equivalence class determined by the element x.
The set of all such equivalence classes on X is denoted by X/$\mathscr E$; that is, X/$\mathscr E$={x/$\mathscr E$∣$x\in X$}.
The symbol $X/\mathscr E$ is read "X modulo $\mathscr E$," or simply "X mod $\mathscr E$".

Source: Set Theory by You-Feng. Lin and Shwu-Yeng T. Lin

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Feel free filling frame for future feasibly fruitful feats.

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This box is free to use for an answer.

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Let $C = \{ (x,t,s) \mid t\le u(x)+s \}$ and let $q:X\times I\times I \to C$ be the map $(x,t,s) \mapsto (x,t(u(x)+s),s)$. This map is easily seen to be surjective. Let us show that $q$ is a quotient map: The map $$\Gamma: X×I×I \to X×I×I×I,\ (x,t,s) \mapsto (x,t,u(x),s)$$ is an embedding onto a closed subspace $D$. Since the map $$m: I×I×I \to [0,2]×I, (a,b,c) \mapsto (a(b+c),c)$$ is perfect, $M = \text{id}_X×m$ is closed. The restriction of $M$ to $D$ is closed, hence $M\Gamma = q$ is closed and therefore a quotient map.

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$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^{T}}-1}$
$====================$
$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c-r}{r}+\frac{1-c}{(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r^2(1+r)^{T+1}}}{\frac{(c-r)(1+r)^{T}+(1-c)r}{r(1+r)^{T}}}$
$====================$
$X=\frac{\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)}}{\frac{(c-r)(1+r)^{T}+(1-c)r}1}$
$====================$
$X=\frac{(c(1+r)^{T+1}+(1-c)r^2)1}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$
$====================$
$X=\frac{c(1+r)^{T+1}+(1-c)r^2}{r(1+r)((c-r)(1+r)^{T}+(1-c)r)}$

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$\begin{array}{c} Donald\\ @\$\$#*\€\\ Trump \end{array}$

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A general Delta complex $X$ can be described combinatorially as a sequence of sets $X_0,X_1,\dots$ together with maps $d_i:X_n\to X_{n-1}$ for any $i\in[n]=\{0<\dots<n\}$ such that $d_j d_i = d_i d_{j+1}$ whenever $j\ge i$. The elements of $X_n$ are called $n$-simplices of $X$, and the maps $d_i$ are the face maps of $X$.
Note that any monotone injection $f:[k]\to[n]$ can be written uniquely as a composite $$f=\delta_{i_{n-k}} \dots \delta_{i_1}, \qquad i_1<\dots<i_{n-k}$$ where $\delta_i:[m-1]\to[m]$ denotes the monotone injection which omits the value $i$, for any $m$. The indices in the decomposition of $f$ are then simply those elements of $[n]$ which are not assumed by $f$. Such an $f$ can be thought of as indicating a certain face of $\Delta^n$, namely the one spanned by all vertices in $\mathrm{Im}(f)$. In the Delta complex $X$, the restriction of $\sigma$ to this face is a simplex $B_f(\sigma)$ in $X_k$.

A delta complex $X$ has a realization $|X|$ which is constructed as follows: For any $n$-simplex $\sigma$ take a standard-$n$-simplex $\Delta^n_\sigma$. Now identify its $i$-th subsimplex with the standard-$(n-1)$-simplex $\Delta^{n-1}$ corresponding to the face $d_i x$ using the linear inclusion $\Delta^{n-1} \to \Delta^n$ which preserves the ordering of the vertices. For example, you may have the delta complex $X_0=\{v\} \leftleftarrows X_1=\{e\}$, where the two arrows represent $d_0$ and $d_1$. Then you have a line $\Delta^1$ and a point $\Delta^0$, and both endpoints of the line are glued to that point, so we get a circle.

The barycentric subdivision $\mathrm{Bd}(X)$ has as $n$-simplices all sequences $\ - I have a conjecture. First, let's define the following objects. •$\displaystyle \mathcal G = \prod_{I=1}^A \mathbb Z_{a_i}$is a finite abelian group. We will identify the elements of$\mathcal G$with "column vectors" of the form$ \overline{x} = \begin{bmatrix} \overline{x_1} \\ \overline{x_2} \\ \vdots \\ \overline{x_A} \end{bmatrix} \in \mathcal G $where$ x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_A \end{bmatrix} \in \mathbb Z^{A \times 1}. $•$\displaystyle \{ \overline{s_j} \}_{j=1}^B$is a linearly independent subset of$\mathcal G$, that is to say, for all integer coefficients$\{ \alpha_j \}_{j=1}^B$,$\displaystyle \sum_{j=1}^B \alpha_j \overline{s_j} = \bar {\mathbf 0}$if and only if$\alpha_j \overline{s_j} = \bar{\mathbf 0}$for all$j$equals$1$to$B$. • For$j$equals$1$to$B$,$\operatorname{ord}(\overline{s_j}) = b_j$. •$D_a = \operatorname{Diag}\{a_1,\, a_2,\, \dots,\, a_A\} \in \mathbb Z^{A \times A}$•$D_b = \operatorname{Diag}\{b_1,\, b_2,\, \dots,\, b_B\} \in \mathbb Z^{B \times B}$•$S = [s_1 \mid s_2 \mid \dots \mid s_A] \in \mathbb Z^{A \times B}$is the integer matrix whos columns are$s_1, s_2, \dots, s_A. Now \begin{align} \forall j, (j=1\dots A),\operatorname{ord}(\overline{s_j}) = b_j &\implies \forall j, (j=1\dots A), b_j \overline{s_j} = \overline{\mathbf 0}\\ & \implies \forall j, (j=1\dots A),\forall k,(k=1\dots B), b_j s_{j,k} \equiv 0 \pmod{a_k} \\ & \implies \exists \tilde{s}_k, k=1\dots A, b_j s_{j,k} = a_k \tilde s_k \end{align} This can be expressed quite elegantly as $$\exists \widetilde S \in \mathbb Z^{A \times B}, S D_b = D_a \widetilde S$$ My conjecture is that the rows of\widetilde S$span$\displaystyle \prod_{j=1}^B \mathbb Z_{b_j}$## Example Let •$\mathcal G = \mathbb Z_{60} \times \mathbb Z_{84} \times \mathbb Z_{210}$•$s_1 = \begin{bmatrix} 30\\ 28 \\ 42 \end{bmatrix},\; s_2 = \begin{bmatrix} 12 \\ 42 \\ 30 \end{bmatrix}$Then •$\operatorname{ord}(s_1) = 30$and$\operatorname{ord}(s_2) = 70$•$S = \begin{bmatrix} 30 & 12\\ 28 & 42\\ 42 & 30 \end{bmatrix}$•$D_a = \begin{bmatrix} 60 & 0 & 0\\ 0 & 84 & 0\\ 0 & 0 & 210 \end{bmatrix}$•$D_b = \begin{bmatrix} 30 & 0\\ 0 & 70 \end{bmatrix}$•$S D_b = \begin{bmatrix} 900 & 840\\ 840 & 2940\\ 1260 & 2100 \end{bmatrix} = \begin{bmatrix} 60 & 0 & 0\\ 0 & 84 & 0\\ 0 & 0 & 210 \end{bmatrix} \cdot \begin{bmatrix} 15 & 14\\ 10 & 35\\ 6 & 10 \end{bmatrix}$•$\widetilde S = \begin{bmatrix} 15 & 14\\ 10 & 35\\ 6 & 10 \end{bmatrix}$My conjecture is that the rows of$\widetilde S$span$\mathbb Z_{30} \times \mathbb Z_{70}\$

Note that \begin{align} 5[15 \; 14] + 4[10 \; 35] + 21[6 \; 10] &\equiv [1 \; 0] \pmod{[30 \; 70]}\\ 4[15 \; 14] + 3[10 \; 35] + 5[6 \; 10] &\equiv [0 \; 1] \pmod{[30 \; 70]} \end{align}

It follows that, in this case, the conjecture is true.

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