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\begin{document}
\title{\textbf{Meaningful Products}}
\date{Wang-Shiuan Pahngerei}
\maketitle

Let $G$ be a semigroup and $A$ a set. Denote $\bar{G}=G\times \wp A$. Then the operation
\begin{align}
  \begin{pmatrix}g_1\\ A_1\end{pmatrix}\ast \begin{pmatrix}g_2\\A_2\end{pmatrix}= \begin{pmatrix}g_1g_2\\A_1\cup A_2\end{pmatrix}\notag
\end{align} 
gives a structure of a semigroup. Let $\alpha: |1,N|\to G$, written multiplicatively, and $\beta: |1,N-1|\to\N$ be such that $\beta(n)\leq N-n$ for all $n$. Set $A=|1,N|$. Define $S: |0,N-1|\to \{\text{functions from a subset of $|1,N|$ to $\bar{G}$ }\}$ be a sequence such that 

\noindent $S(0)=\left(  \pl\alpha(1)\\\{1\}\pr, \pl\alpha(2)\\\{2\}\pr\cdots, \pl\alpha(N)\\\{N\}\pr\right)$.

%, \pl \alpha(2)\\\{2\}\pr,\cdots,\pl \alpha(\beta_1)+\alpha(\beta_1+1)\\\{\beta_1,\beta_1+1\}\pr,  \pl\alpha(\beta_1+2)\\\{\beta_1+2\}\pr,\cdots, \pl\alpha(N)\\\{N\}\pr \right)$, 

\noindent $S(1)=\left( S_0(1),S_0(2),\cdots, S_0(\beta_1)\ast S_0(\beta_1 +1),\cdots,S_0(N)\right)$ , with $N-1$ terms. 

\noindent $S(2)=\left( S_1(1),\cdots, S_1(\beta_2)\ast S_1(\beta_2 +1),\cdots,S_1(N-1)\right)$ , with $N-2$ terms. 

$\vdots$

\noindent $S(k+1)=\left( S_k(1),\cdots, S_k(\beta_{k+1})\ast S_k(\beta_{k+1} +1),\cdots,S_k(N-k)\right)$ .

Note that \,\, [1] each $S(j)(k)(2)$ is a consecutive set, i.e. the set  $|p,q|$. \,\, [2] Then $S(N-1):\{1\}\to \bar{G}$. At this time we define $\prod_\beta\alpha=S(N-1)(1)(1)$, i.e. $S(N-1)=\left(\pl\prod_\beta\alpha\\\ast\pr\right)$. 

We want to show for different $\beta$ and $\beta'$, $\prod_\beta\alpha=\prod_{\beta'}\alpha$. Intuitively, 
\begin{align}
((((a_1+a_2)+(&a_3+a_4))+a_5)+(a_6+a_7))+(a_8+(a_9+a_{10})) \notag\\
&= ((a_1+a_2)+(a_3+(a_4+a_5)))+((a_6+a_7)+((a_8+a_9)+a_{10}))\notag
\end{align}

\noindent\textsc{Sketch of proof}.  By induction on $N\geq 3$. In case it holds at $n$, then for any two different orders of summations, 
\begin{align}
&(a_1+\cdots+a_k)+(a_{k+1}+\cdots+a_{n+1}) \notag\\
&= 
((a_1+\cdots + a_\ell)+(a_{\ell+1}+\cdots+a_k))+(a_{k+1}+\cdots+a_{n+1})\notag\\
&=(a_1+\cdots + a_\ell)+((a_{\ell+1}+\cdots+a_k)+(a_{k+1}+\cdots+a_{n+1}))\notag\\
&=(a_1+\cdots + a_\ell)+((a_{\ell+1}+\cdots+a_{n+1}))\notag
\end{align}
The first and third equalities follow from induction hypothesis while the second from associativity. \,\,\,\,\,\,\,\,\,\,\,\,$\square$

At this time, we define for a sequence $\alpha=(\alpha_1,\cdots,\alpha_N)$ its product $\prod \alpha$ by $\prod_{(1,1,\cdots,1)}\alpha$. 


\end{document}