Santa’s singleton

Here’s a question which I’m sure bugs all my logical readers. Modern mathematics standardly recognises partial functions which can take something as input but deliver nothing as output (like the reciprocal function which isn’t defined for zero). Do we also need to allow for co-partial functions which can take nothing as input but deliver something as output? Exciting eh?

Well, perhaps not so very thrilling. But for what it is worth, here’s a version of a paper in dialogue form given to the Serious Metaphysics Group here in Cambridge this evening. Not exactly the talk as delivered (with help from Rob Trueman!): the discussion has led me to cut out a long passage which was getting too involved by half and to tinker elsewhere a small amount.

One comment: if/when I get round to further rewriting this a bit, I’d drop the ‘Santa’ example, as I don’t want any noise coming in from issues about fiction. It would be better to use e.g. ‘Vulcan’ instead.

[Added later] Hmmm. The argument about partial-functions-by-stipulation certainly won’t do as it stands …

3 thoughts on “Santa’s singleton”

  1. This reminds me of the analysis of (especially computable) functions using domain theory. As you probably well know, there we allow functions to act on partially informative input and produce partially informative output, provide they do so in a monotone way (more informative input must give more informative output). (Domain functions must also preserve directed joins.)

    In this setting, it seems co-partial functions correspond to non-strict functions. However, due to the monotonicity requirement, if a function produces a maximal (fully informative) output for the bottom element input (about which nothing is known), it must be constant in that argument.

    I don’t know if this analysis is of any help philosophically, but I was reminded of it while reading your dialogue.

  2. That was a fun read (I had to skim quickly as I have lectures to prepare). One minor point: further unraveling the syntactic sugar, the variable in {x : P(x)} is bound so the reading on which {Santa} turns out empty does not depend on what you call a strong reading of identity, as (Ex)(x=Santa) is always false. On the other hand, what you say on the next page might be a bit quick, for some would like to count Vulcan=Vulcan as true.

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