Hello all,
I am currently working on proving these statements and , by means of subsets (I've heard that is is also referred to as, "proof by double inclusion."). I am having trouble with these sort of problems, so I'd certainly appreciate the help.
Here is my work:
Suppose that ; for to be in the set, can't be in . This means that ; however, by this fact, can be in , , or . Thus, when , is also in , meaning that
I have not commenced the second part, yet, because I feel as though there is something wrong with this, and I feel like I might have left out steps and explanations.
Okay, Plato, is this better:
Suppose that , then can't be an element of ; in order for to not be in , then it must be the case that , that is, it must be the case that x is not in at least one of the three sets. If is not in at least in one of those sets, then , that is, it is at least in of the three sets , , or . By the definition of the union, the set will possess the element , because at least one of the three sets has .
Would this show that the statement is a subset of the other?
And to prove that , would this be sufficient:
Suppose that . For to be in the set, has to be in at least one of the three sets, meaning can't be in , because if this were so, would have to then be in all three sets, falsifying our original assumption. Moreover, if is not in , we can say that .
Yes, though the style may be somewhat improved.
This is correct, though since the fact is so simple, it is difficult to say whether this explanation is not enough, adequate, or too much. I would write, "... has to be in at least one of the three sets. If , then and therefore , and similarly for and . In all three cases, we have , i.e., ."