\rightarrow has math symbol class 3, that is, it is a math relation.
How can I define a macro, let's call it \rightinf, to get the same right arrow character, but as an ordinary math symbol (like \infty, say) rather than as a math relation?
Reason for the question:
I want to use \rightinf to designate intervals in an ordered set that are not bounded above, as in (0, \rightinf), instead of (0, \infty).
How about
\newcommand{\rightinf}{\mathord{\rightarrow}}
or, more concisely,
\newcommand{\rightinf}{{\to}}
(\to is an alias for \rightarrow.)
Creating a macro called \rightinf in this manner may be unnecessary, though, at least for the use case you laid out: $(0,\to)$ and $(0,{\to})$ produce the exact same output.
Here's a full MWE. It illutrates my point about $(0,\to)$ and $(0,{\to})$ producing the exact same output.
\documentclass{article}
\newcommand{\rightinf}{{\to}}
\begin{document}
$(0,\to)$
$(0,\rightinf)$
\end{document}
\to is an alias for \rightarrow and \rightarrow is a math relation (class 3), doesn't the concise definition \newcommand{\rightinf}{{\to}} keep \rightinf as a math relation? Or does the extra grouping around \to in that definition convert it into an ordinary math symbol (class 0)?
\to (or any other math-mode object, really) in curly braces changes its math status to math-ord.
\to is of class 3 (math rel), why does the spacing stay the same for the \mathord version (invoked in the form \mathord{\rightarrow}) as for the original math rel form?
Since \rightarrow is a \mathchardef token, you can define
\mathchardef\rightinf=\numexpr\rightarrow-"3000\relax
Example:
\documentclass{article}
\mathchardef\rightinf=\numexpr\rightarrow-"3000\relax
\begin{document}
$a\rightarrow b$
$a\rightinf b$
\end{document}
On the other hand,
\newcommand{\rightinf}{\mathord\rightarrow}`
is simpler.
