A command to replace \middle| when it's missing?

Question

I'm using this LaTeX plugin for a Wiki (Confluence, blech), and it doesn't support the \middle| command. I'm guessing it doesn't have eTex primitives or some such backwardness. Anyway, I guess I can write things like

\newcommand{\fakemiddlebar}[2]{{\kern-\nulldelimiterspace\left.\vphantom{{#1}{#2}}\right|}}
\newcommand{\condset}[2]{\left\{ #1 \mathrel{\fakemiddlebar{#1}{#2}} #} \right\}}

(improved as per @egreg's suggestion).

Edit: But I want to do better! I want to have an argument-less command with the same effect is \middle|. Also, I can't \include or \usepackage. Is that possible?

Answer 1

You get the same result with \condset{x\in X}{x<\frac{1}{2}} as

\left\{x\in X\middle|x<\frac{1}{2}}

by modifying your macros into

\newcommand{\fakemiddlepipe}[2]{{\kern-\nulldelimiterspace\left.\vphantom{{#1}{#2}}\right|}}
\newcommand{\condset}[2]{\left\{ #1 \fakemiddlepipe{#1}{#2} #2 \right\}}

Note the additional pair of braces around the body of \fakedmiddlepipe and the \kern instruction that takes away the space added by \left.. Of course there would be no space around the pipe, so the final definition might be

\newcommand{\condset}[2]{\left\{ #1 \mathrel{\fakemiddlepipe{#1}{#2}} #2 \right\}}

so that the result of \condset{x\in X}{x<\frac{1}{2}} would be the same as the more correct

\left\{x\in X\;\middle|\;x<\frac{1}{2}}

(and you could even remove the additional pair of braces in the definition of \fakemiddlepipe; but I'd leave them for other possible usages).

Answer 2

If you are only interested in \middle| (and not any other delimiter), just use \vrule.

$$\left\{x\in X\;\vrule\; x< {1 \over 2} \right\}$$
\bye

Answer 3

As @Mico suggests, the braket package has the solution: Change the catcode of |, and use | instead of \middle| . You just have to be careful not to use it out of a group by mistake.