5

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,

enter image description here

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?

11

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}

enter image description here

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 $t\in[t_{1},t_{2}]$.

enter image description here

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*}
\eta(x_{1},t)=\eta(x_{2},t)=\alpha,\qquad \eta(x,t_{1})=\eta(x,t_{2})=\beta,
\end{equation*}
for all $x\in[x_{1},x_{2}]$ and for all $t\in[t_{1},t_{2}]$.

enter image description here

Long description with the set builder notation should be avoided.

5

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.

enter image description here

\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}
  • 1
    The answers are all fantastic. Thank you very much....always to all. – Sebastiano Aug 24 at 21:35
4

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} 

enter image description here

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