- #1
SixNein
Gold Member
- 122
- 20
Now, I was told that the above proof was valid by a professor. But I don't see how it could be valid as it is written. The only proof I can arguably see here is a proof that AUB[itex]\subseteq[/itex]BUA.Prove AUB=BUA
Let xεAUB
xεA or xεB (Definition of union)
case 1: xεA
xεBUA (Def of union)
since x is arbitrary, must be true for all x. (inclusion)
therefore, AUB=BUA
Case 2: xεB
xεBUA (Def of union)
since x is arbitrary, must be true for all x. (inclusion)
therefore, AUB=BUA
From the way its written, case 1 shows that A[itex]\subseteq[/itex]BUA while case 2 shows that B[itex]\subseteq[/itex]BUA; therefore, the conclusion would be AUB[itex]\subseteq[/itex]BUA.
Maybe I'm missing something here..?