KDE 4.0.0!

[Sylviane et Perry White]

On Friday 18 January 2008 11:40, Ignazio Palmisano wrote:
>
> /me been wandering around something similar long time ago...
>
> KDE 4.0.0 is a piece of software and it is a container of pieces of
> software, which is the same kind of KDE4... but then the set of all
> possible sets is contained in itself? :) Can't remember which famous
> mathematician had a hard time defining this problem, I think it was
> about first and second order logic, and in the end if you can put a
> container in itself then the logical results take a while to be
> computed... like forever :)
>
> /me feels off topic and in need of a coffee, got a headache because of
> these containers...
> I.
>
> --
> Ignazio Palmisano

Thanks Ignazio,

We are getting out of topic here but any mathematician will smile at he 
"Impeccable logic".
If a set has only one element the set is the same as the element, right?
What about a classroom with only one student, so class 3B=Eric, and you could 
add other students into class 3B or into Eric. _Ouch!		;O)

About sets not containing themselves look up "Russell's Paradox"
(*) see at bottom

Derek said: 
> Not buying it.  KDE 4 isn't a container.  	At this time, it _is_ KDE 4.0.0.
(that above makes me uncomfortable)		^--------------^
Well, perhaps we should accept the fact that in common language we use words 
that are not understood/defined-yet. The definition of some terms may come or 
change with usage. Yet patterns concerning the "decimal" notation of versions 
and revisions has already been established long ago and we all expect that 
version X.0.0 will be folloved by something; should the definition of the 
version change, from not-container to container, "at this time", when a new 
revision is added?

Billie added:
> Isn't language, and playing with words, fun.
I second that.


(*)
Russel's paradox
It might be assumed that, for any formal criterion, a set exists whose members 
are those objects (and only those objects) that satisfy the criterion; but 
this assumption is disproved by a set containing exactly the sets that are 
not members of themselves. If such a set qualifies as a member of itself, it 
would contradict its own definition as a set containing sets that are not 
members of themselves. On the other hand, if such a set is not a member of 
itself, it would qualify as a member of itself by the same definition. This 
contradiction is Russell's paradox.


Cheers		Perry


-- 
BOFH excuse #192: runaway cat on system