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I have wondered for a while what gave rise to the notation $0^\sharp$. According to wikipedia this is due to Solovay in 1967, but (perhaps unsurprisingly) there's no discussion of why that notation was chosen. In contrast, Silver mentions in a paper that he chose the symbol $\Sigma$ to represent the same thing. Nonetheless, I couldn't turn anything up with a bit of searching.

So, why did Solovay choose zero-sharp as the name for this object? (The zero part at least makes sense in the general sense of $a^\sharp$ where we consider $L[a]$ instead of $L$) Especially given that there are plenty of ways of indicating something "a little bit more/different" that are quite normal in mathematical notation.

edited Oct 19 2011 at 13:13

Willie Wong
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asked Oct 19 2011 at 12:57

Chris Le Sueur
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Andreas Blass · Accepted Answer · 2011-10-19 14:44:49Z UTC

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I've heard that the symbol was originally $O^\#$ , with a capital letter O, not zero. This set was viewed as a higher-level analog of Kleene's O (a universal $\Pi^1_1$ set).

answered Oct 19 2011 at 14:44

Andreas Blass
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Yes, that's what I've always heard. Of course, you were probably the one to tell me this, Andreas. ;) – Todd Eisworth Oct 19 2011 at 14:48

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I've also heard that it was in analogy to 0' from recursion theory. – Todd Eisworth Oct 19 2011 at 14:57

My only concrete guess was a relation to zero-jump. Interestingly the jump operator seems to be in part due to Kleene as well. In any case, Solovay's paper does seem to say $O^\#$ (with a hash/pound, rather than a musical sharp, but that could be a limitation of the typesetting...) so that seems the most plausible answer. – Chris Le Sueur Oct 19 2011 at 20:30

What is the etymology of zero-sharp?

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