# Another solution to create a set with two conditions

With this MWE

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\usepackage{newtxtext}
\usepackage{newtxmath}
\begin{document}
Y_{\alpha\beta}=\left\{\eta\coloneq\eta(x,t) \colon \quad \eta\in\mathcal{C}^1(D)\colon \quad \begin{aligned} \phantom{a} & \eta(x_{1},t)=\eta(x_{2},t)=\alpha, \,\forall t\in[t_{1},t_{2}]\\ \phantom{b} & \eta (x,t_{1})=\eta(x,t_{2})=\beta,\,\forall x\in[x_{1},x_{2}] \end{aligned} \right\}
\end{document}


and this output,

I have realized a set with two conditions. Is there another best solution to delete \phantom and to have both \forall t\in[t_{1},t_{2}] and \forall t\in[t_{1},t_{2}], vertically aligned?

Writing \eta:=\eta(x,t) has no mathematical meaning whatsoever. Since apparently D is a subset of the plane, functions over D are two-variable by definition; how you call the variables is completely irrelevant.

I wouldn't align the two final intervals. Around the colon I would add some additional space because of the split line on the right.

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\usepackage{newtxtext}
\usepackage{newtxmath}
\begin{document}

Y_{\alpha\beta}=\left\{ \eta\in\mathcal{C}^1(D)\;:\; \begin{alignedat}{3} \eta(x_{1},t)&=\eta(x_{2},t)=\alpha,&\quad&\forall t &&\in[t_{1},t_{2}]\\ \eta(x,t_{1})&=\eta(x,t_{2})=\beta, &\quad&\forall x &&\in[x_{1},x_{2}] \end{alignedat}\, \right\}

Y_{\alpha\beta}=\left\{ \eta\in\mathcal{C}^1(D)\;:\; \begin{alignedat}{2} \eta(x_{1},t)&=\eta(x_{2},t)=\alpha,&\quad&\forall t \in[t_{1},t_{2}]\\ \eta(x,t_{1})&=\eta(x,t_{2})=\beta, &\quad&\forall x \in[x_{1},x_{2}] \end{alignedat}\, \right\}

\end{document}


My preference would go to a textual description:

We define $Y_{\alpha\beta}$ as the set of all functions $\eta\in C^1(D)$ such that
\begin{align*}
\eta(x_{1},t)&=\eta(x_{2},t)=\alpha,\\
\eta(x,t_{1})&=\eta(x,t_{2})=\beta,
\end{align*}
for all $x\in[x_{1},x_{2}]$ and for all $t\in[t_{1},t_{2}]$.


If not constrained by the line width, a one liner might be even preferable:

We define $Y_{\alpha\beta}$ as the set of all functions $\eta\in C^1(D)$ such that
\begin{equation*}
\end{equation*}
for all $x\in[x_{1},x_{2}]$ and for all $t\in[t_{1},t_{2}]$.


Long description with the set builder notation should be avoided.

Here's a solution that employs an array environment to align the elements of the two rows of conditioning information.

Note that I use a vertical bar to denote "given that" or "conditional on". If you prefer using a colon, you should input it as :, not as \colon.

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,array,newtxtext,newtxmath}
\newcolumntype{C}{>{{}}c<{{}}} % column type for '=' and '\in' symbols
\begin{document}

$Y_{\!\alpha\beta} = \left\{ %\eta\coloneq\eta(x,t) \colon % commented out per egreg's comments \eta\in\mathcal{C}^1(D) \;\middle\vert\; \setlength\arraycolsep{0pt} \begin{array}{rCrClrCl} \eta(x_{1},t)&=&\eta(x_{2},t)&=&\alpha,&\ \forall t&\in& [t_{1},t_{2}] \\[0.5ex] \eta(x,t_{1})&=&\eta(x,t_{2})&=&\beta, &\ \forall x&\in& [x_{1},x_{2}] \end{array} \right\}$
\end{document}

• The answers are all fantastic. Thank you very much....always to all. – Sebastiano Aug 24 '19 at 21:35

To have the \forall and the \in vertically aligned, I used a 3 columns alignedat (due to the difference in width between t and x). Further, as newtx produced error messages on my system, I replaed them with fourier:

\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,amssymb}
\usepackage{fourier}
%\usepackage{newtxtext}
%\usepackage[libertine]{newtxmath}

\begin{document}

Y_{\alpha\beta}=\left\{\eta\coloneqq\eta(x,t) \colon \quad \eta\in\mathcal{C}^1(D)\colon \quad \begin{alignedat}{3} \phantom{a} & \eta(x_{1},t)=\eta(x_{2},t)=\alpha, & \enspace & \forall t & &{}\in[t_{1},t_{2}]\\ \phantom{b} & \eta (x,t_{1})=\eta(x,t_{2})=\beta, & & \forall x& & {}\in[x_{1},x_{2}] \end{alignedat} \right\}

\end{document}