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Originally Posted by <<>> The explanation f why your solution is wrong is not only 1 sentence, but is short, here I give step to step. Please don't carry on arguing that it reduces to your solution, because it doesn't:
The logical idea of a set that contains something is refeared to the fact that the set, for example S is equal to the following elements {a,b,c,d} such that S contains {a}, {b}, {c}, {d}. In some cases, the requierements to enter a set include that set itself, and then S={a,b,c,d,S} which means that S contains itself. But there are many sets that dont' containt themeselves. For example, the "set of all dogs" isn't a dog itself (the set), so it doesn't contain itself. But the "set of all sets that can be described in less than 30 words" includes itself as it is described in 12 words and a number.
Therefore the set of all sets that do not contain themselves is not an empty set or irrelevant. |
I congratulate you very highly for coming closer to disproving my solution then anyone else was able to do. This is surely a testament of your intelligent and inquisitive nature so you should feel proud of yourself because I am happy that you've tried to logically disprove my solution.
However, despite your notable effort, I deduce that your disproof is still incorrect, and here's why. If a set is a set it automatically contains the set which is itself, and this is what we mean when we say the set contains itself. So to use your example, the set of all dogs is not a dog itself, you're correct, but that doesn't pertain to the fact that the set of all dogs IS the SET of all dogs, therefore the set of all dogs contains itself because itself is the set of all dogs. So for the set of all dogs to contain itself doesn't mean that the set of all dogs has to be a dog itself, just that it has to be the set of all dogs itself. So you're incorrect assumption that you made was that the set itself is a dog, whereas the set itself is a set, therefore the set contains itself.
Therefore I again state that any set automatically contains itself as it's own set not necessarily as a member of it's own set but as the set itself, but this is not NECESSARILY true of the empty set because the empty set is not a set, because it has no value i.e. contains nothing, so whether or not the empty set necessarily contains itself is a non-applicable question. Therefore any set which does not necessarily contain itself must automatically be the empty set because it CANNOT be a set otherwise. So my solution still stands but thank you so much for trying to disprove it because I enjoyed the challenge and I feel my proof is more valid now that I have effectively been able to defend it. Please carry on arguing if you are still not satisfied until we both reach an agreement so that we can consider progress is made.
honestly, SubVersion