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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 Maybe someday we can get rid of this outdated crutch.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

17 Answers 17

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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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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 $x/\mathscr E = y/\mathscr E$ "

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 $\equiv z\mathscr E y \Rightarrow z \mathscr E x$}}$


"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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"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


"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


Feel free filling frame for future feasibly fruitful feats.


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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.




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


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 $\displaystyle \prod_{j=1}^B \mathbb Z_{b_j}$



  • $\mathcal G = \mathbb Z_{60} \times \mathbb Z_{84} \times \mathbb Z_{210}$
  • $s_1 = \begin{bmatrix} 30\\ 28 \\ 42 \end{bmatrix},\; s_2 = \begin{bmatrix} 12 \\ 42 \\ 30 \end{bmatrix}$


  • $\operatorname{ord}(s_1) = 30$ 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.