If really need to this then you should use a simple align* environment rather than aligned inside equation, and issue \allowdisplaybreaks first, preferably contained in a group.

\documentclass[11 pt,a4paper,oneside,openany, notitlepage]{article}
\usepackage{mathtools, amssymb, graphics}
\newcommand{\mathsym}[1]{{}}
\usepackage{amsthm}
\usepackage{amsfonts}
\marginparwidth 0pt
\oddsidemargin 0pt
\evensidemargin 0pt
\marginparsep 0pt
\linespread{1.5}
\topmargin 0pt
\textwidth 6.5in
\textheight 8.5 in

\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{\, #1 \,\delimsize\vert\, #2 \,}

\begin{document}
{\allowdisplaybreaks
\begin{align*}
\Theta^\diamond_o&amp;:=\Set*{\theta  \in \Theta}{
H_l^{(h)}(x;\theta)\mathbb{P}(X=x) \leq \mathbb{P}(G_{\cdot
j}=g_{\bullet}^{(h)},X=x) \leq H_u^{(h)}(x;\theta)\mathbb{P}(X=x)\ h=1,\dots,2^{n-1},\
\forall x \in \mathcal{X}}\\&amp;=\Set*{\theta \in \Theta}{
\begin{pmatrix}
H_l^{(1)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X=x) \leq 0\\ \mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X=x)-H_u^{(1)}(x,n;\theta)\mathbb{P}(X=x) \leq 0\\
H_l^{(2)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X=x) \leq 0\\ \mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X=x)-H_u^{(2)}(x;\theta)\mathbb{P}(X=x) \leq 0\\
\vdots\\
H_l^{(h)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X=x) \leq 0\\ \mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X=x)-H_u^{(h)}(x;\theta)\mathbb{P}(X=x) \leq 0\\
\vdots\\
H_l^{(2^{n-1})}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X=x) \leq 0\\ \mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X=x)-H_u^{(2^{n-1})}(x;\theta)\mathbb{P}(X=x) \leq 0\\
\end{pmatrix}
\forall x \in \mathcal{X}}\\&amp;\subseteq\Set*{\theta \in \Theta}{
\begin{pmatrix}
\sum_{x \in \mathcal{X}}^{}\left(H_l^{(1)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X=x)\right)\mathbb{P}(X=x) \leq 0\\ 
\sum_{x \in \mathcal{X}}^{}\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X=x)-H_u^{(1)}(x;\theta)\mathbb{P}(X=x)\right)\mathbb{P}(X=x)  \leq 0\\
\sum_{x \in \mathcal{X}}^{}\left(H_l^{(2)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X=x)\right)\mathbb{P}(X=x)  \leq 0\\ 
\sum_{x \in \mathcal{X}}^{}\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X=x)-H_u^{(2)}(x;\theta)\mathbb{P}(X=x)\right)\mathbb{P}(X=x)  \leq 0\\
\vdots\\
\sum_{x \in \mathcal{X}}^{}\left(H_l^{(h)}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X=x)\right)\mathbb{P}(X=x)  \leq 0\\ 
\sum_{x \in \mathcal{X}}^{}\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X=x)-H_u^{(h)}(x;\theta)\mathbb{P}(X=x)\right)\mathbb{P}(X=x)  \leq 0\\
\vdots\\
\sum_{x \in \mathcal{X}}^{}\left(H_l^{(2^{n-1})}(x;\theta) \mathbb{P}(X=x)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X=x)\right)\mathbb{P}(X=x)  \leq 0\\ 
\sum_{x \in \mathcal{X}}^{}\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X=x)-H_u^{(2^{n-1})}(x;\theta)\mathbb{P}(X=x)\right)\mathbb{P}(X=x)  \leq 0
\end{pmatrix}
}\\&amp;=\Set*{\theta \in \Theta}{
\begin{pmatrix}
\mathbb{E}_X\left(H_l^{(1)}(X;\theta) \mathbb{P}(X)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X)\right) \leq 0\\ 
\mathbb{E}_X\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(1)},X)-H_u^{(1)}(X;\theta)\mathbb{P}(X)\right)\leq 0\\
\mathbb{E}_X\left(H_l^{(2)}(X;\theta) \mathbb{P}(X)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X)\right)\leq 0\\ 
\mathbb{E}_X\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2)},X)-H_u^{(2)}(X;\theta)\mathbb{P}(X)\right)\leq 0\\
\vdots\\
\mathbb{E}_X\left(H_l^{(h)}(X;\theta) \mathbb{P}(X)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X)\right)\leq 0\\ 
\mathbb{E}_X\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(h)},X)-H_u^{(h)}(X;\theta)\mathbb{P}(X)\right) \leq 0\\
\vdots\\
\mathbb{E}_X\left(H_l^{(2^{n-1})}(X;\theta) \mathbb{P}(X)-\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X)\right) \leq 0\\ 
\mathbb{E}_X\left(\mathbb{P}(G_{\cdot j}=g_{\bullet}^{(2^{n-1})},X)-H_u^{(2^{n-1})}(X;\theta)\mathbb{P}(X)\right) \leq 0
\end{pmatrix}
\forall n \in \mathbb{N}}
\end{align*}}

\end{document}


However, your side conditions contain many repeated patterns, and could be replaced by expressions such as 

 H_^{(h)}.... for h=1,\dots,2^{n-1}...


obviating the need for such a large display.

How to split very long equation in two pages?
I have used facebook to allow users to log into a site.  The page that manages the login will get the users facebook id and then look on the ASP.NET membership tables for a pre-existing user that has a matching ID.

No match: a new user is created and logged in (no problem).

A match:  I want to log the user on.  The problem is that I have the hashed password and I cannot use MembershipService.ValidateUser(logOn.UserName, logOn.Password) to log the user on.

If I just use FormsAuthentication.SetAuthCookie, the user doesn't appear as logged in.

I could set enablePasswordRetrieval to true in the web config and then just get the actual password, but I don't want to.  I want this to be secure.

Does anyone know a way to log a user onto ASP.NET Membership just using the user name?  I have looked at the MembershipProvider class for alternative methods but I cant find any.

Thanks,

Oliver
