Re: Relation Schemata vs. Relation Variables
From: paul c <toledobythesea_at_oohay.ac>
Date: Fri, 25 Aug 2006 00:43:34 GMT
Message-ID: <WIrHg.453246$iF6.390802_at_pd7tw2no>
>
> And one cannot infer anything from a subset of the attributes when one
> is talking about a tuple. The only thing that identifies a tuple is the
> tuple's value. Just as the only thing that identifies the number 5 is
> the number 5.
Date: Fri, 25 Aug 2006 00:43:34 GMT
Message-ID: <WIrHg.453246$iF6.390802_at_pd7tw2no>
Bob Badour wrote:
> paul c wrote:
>
>> Bob Badour wrote: >> >>> paul c wrote: >>> >>>> paul c wrote: >>>> ... >>>> >>>>> PMFJI, I would say that the VALUE of a candidate key identifies one >>>>> and only one tuple FOREVER! >>>> >>>> Stupid me, I have to take part of that back - the value of a >>>> candidate key obviously could identify several tuples but I still >>>> think that would hold forever. Might have been better to say the >>>> value of a candidate key identifies a tuple regardless of time. >>> >>> A candidate key does not identify a tuple. A candidate key is a >>> constraint on a relvar and not on a tuple. >> >> No argument about a candidate key being a constraint. I`m talking >> about the value of a candidate key. If you can infer the values of >> the other attributes from that value, I`d say you have achieved >> identification.
>
> And one cannot infer anything from a subset of the attributes when one
> is talking about a tuple. The only thing that identifies a tuple is the
> tuple's value. Just as the only thing that identifies the number 5 is
> the number 5.
There must be a subtlety here that eludes me. If a candidate key, k, of a relation has a value of 1 in some tuple and a tuple in a relation that has that candidate key has a value of {k 1, x 2} then I would say that the value 1 certainly identifies that tuple.
p Received on Fri Aug 25 2006 - 02:43:34 CEST