I = {{dog, car, rug}, dog, car, rug}


   I = {dog, car, rug, dog, car, rug}
  Once again, removing the inside braces will have no effect when comparing the dog to the car, or the car to the rug, or . . . . Also, regardless of the relationships being unaltered, there are now duplicate elements. And this negates the set.
  Consider d The set of all abstract concepts is itself an abstract concept, therefore it contains itself. This statement follows the logic of c.
  Consider e The set of all non-integers is itself a non-integer, therefore it contains itself. This statement cannot have a finite setting, because every set, other than the set of integers is being considered. In this scenario, the set of all non-integers is infinite; therefore, it cannot translate to Tspace.
  There is another reason that this set cannot contain itself in Tspace. When it does, there will be a duplication of the elements, which means it is no longer a set.
  Consider f The set of all non-trees is itself a non-tree, therefore it contains itself. This statement follows the logic of e.
   So, all of these examples are forever false. (Any idea that cannot be transferred to reality is false until a way has been found to transfer it. If it is forever false, then it can never be transferred.) In conclusion, a set in Tspace cannot contain itself.
  Since both premises are false, Russell's Paradox does not exist in Tspace and never has, which means it is of no consequence.

In Ispace, the Solution of Russell's Paradox

  All paradoxes can be solved, but when infinity is involved the paradox becomes much more complicated and it is more difficult to find the solution. To help with these complications Venn diagrams will be used in solving Russell's paradox.

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