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    • CommentRowNumber1.
    • CommentAuthormatc
    • CommentTimeJul 30th 2015
    if "a partial function is the same as a functional relation seen from a different point of view." then why it is considered that partial functions are generalizations of functions, (at least that says Wikipedia in partial function entry) (if a function is "precisely a relation that is both functional and entire.")?
    • CommentRowNumber2.
    • CommentAuthorDavidRoberts
    • CommentTimeJul 30th 2015

    I don’t see a conflict between those two statements. Functions are special cases of partial functions: they are the ones that are entire as relations.

    • CommentRowNumber3.
    • CommentAuthorRodMcGuire
    • CommentTimeJul 30th 2015
    • (edited Jul 30th 2015)

    “a partial function is the same as a functional relation seen from a different point of view.”

    The understanding problem may involve the phrase “is the same”.

    A partial function is a generalization of an “entire functional relation” in that it is not necessarily entire. However a partial function f:ABf: A \to B is also equivalent to a total function to B+{}B + \{\top\} which is a pointed set where the added point \top is the “trash” point which stands for “undefined” or “unknown”. This added structure is a specialization.

    Thus the relation between partial functions and total functions can be described as generalization, equivalence, and specialization. I expect the equivalence expresses a duality.

    I think one could multi-point BB above with distinct points, 1\top_1, 2...\top_2 ... for each unknown value where the different \tops are isomorphic. Something like this is needed if one wants to capture the ordering of partial functions - if ff is a functional relation and g=f+{ab}g = f + \{a \mapsto b\} augments ff with a new mapping that was undefined then f<gf \lt g.

    • CommentRowNumber4.
    • CommentAuthorTobyBartels
    • CommentTimeAug 27th 2015

    Normally this extra point is called \bot rather than \top because it's at the bottom rather than the top of a partial order. This makes fgf \leq g automatically work out in your example as you want: f(a)=<b=g(a)f(a) = \bot \lt b = g(a). We should say more about this ordering at partial function.

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