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1. • CommentRowNumber1.
   • CommentAuthormatc
   • CommentTimeJul 30th 2015
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   Author: matc
   Format: Textif "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.")?

   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.")?
2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeJul 30th 2015
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   Author: DavidRoberts
   Format: MarkdownItexI 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.

   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.

3. • CommentRowNumber3.
   • CommentAuthorRodMcGuire
   • CommentTimeJul 30th 2015
   • (edited Jul 30th 2015)
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   Author: RodMcGuire
   Format: MarkdownItex> "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 $f: A \to B$ is also equivalent to a total function to $B + \{\top\}$ 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 $B$ above with distinct points, $\top_1$, $\top_2 ...$ for each unknown value where the different $\top$s are isomorphic. Something like this is needed if one wants to capture the ordering of partial functions - if $f$ is a functional relation and $g = f + \{a \mapsto b\}$ augments $f$ with a new mapping that was undefined then $f \lt g$.

   “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 $f: A \to B$ is also equivalent to a total function to $B + \{\top\}$ 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 $B$ above with distinct points, $\top_1$, $\top_2 ...$ for each unknown value where the different $\top$s are isomorphic. Something like this is needed if one wants to capture the ordering of partial functions - if $f$ is a functional relation and $g = f + \{a \mapsto b\}$ augments $f$ with a new mapping that was undefined then $f \lt g$.

4. • CommentRowNumber4.
   • CommentAuthorTobyBartels
   • CommentTimeAug 27th 2015
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   Author: TobyBartels
   Format: MarkdownItexNormally 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 $f \leq g$ automatically work out in your example as you want: $f(a) = \bot \lt b = g(a)$. We should say more about this ordering at [[partial function]].

   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 $f \leq g$ automatically work out in your example as you want: $f(a) = \bot \lt b = g(a)$. We should say more about this ordering at partial function.