11

I'm denoting the complement of the closure of the set $A$ by $\overline{A}^C$. This looks ugly, though (the C is too high up and not far enough to the right)--might there be any quick fixes?

5

Is this closer?

$\overline{A}\,^C$

It pushes to the right as \, is a thin space and lowers it as it is the space atom that is being superscripted not the capital A.

  • This is great. Somehow, not only does the \, space push the C to the right, but it also lowers it so that its height matches A^C. Thanks! – jamaicanworm Mar 2 '12 at 19:50
8

What about this?

$\smash{\overline{A}}\vphantom{A}^C$

(\smash typesets its argument and "hides" it as if it had zero height; \vphantom puts a zero-width, invisible "something" of height equal to its argument's height (and depth, for that matter).)

  • 2
    Actually, with \vphantom{a} you'd obtain the same output. Only if it gets taller than A, you'd see a difference. – Hendrik Vogt Jun 15 '12 at 17:15
5

How about this:

\documentclass{article}

\renewcommand*\bar[1]{\rlap{$\smash{\overline{#1}}$}\phantom{#1}}

\begin{document}
 $\overline{A}^C \bar{A}^C$.
\end{document}

Basically, I delete all the dimensions of the \overlined symbol and place a box of the base symbol's size where the superscript can find it.

5

In my opinion, the \overline looks ugly for a second reason: it extends way too much to the left:

ugly image

I'd use some appropriate kerns to make the \overline shorter. For lowering the C, you could use {}^C (left version in the image below), but you can also fine-tune it with a rule of zero width (right version):

better image

\documentclass{article}
\begin{document}
$\mkern4.5mu\overline{\mkern-4.5mu A}{}^C$
$\mkern4.5mu\overline{\mkern-4.5mu A}\rule{0pt}{1.6ex}^C$
\end{document}

If you don't want to manually find out the appropriate kerns for the \overline, give the \widebar command from this answer a try (shameless plug).

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