The discussion revolves around proving or disproving a mathematical statement regarding set theory. Specifically, participants are examining the conditions under which the statement "If for any sets A, B, C, it holds that A\C = B\C and A ∩ C = B ∩ C, then A ⊆ B" is valid. Additionally, there is a request for assistance in proving the equality of powersets, P(A∩B) = P(A)∩P(B).
Participants generally engage in clarifying the original statement and exploring proofs, but there is no consensus on the validity of the initial claim or the proofs presented. The discussion remains unresolved regarding the correctness of the statements and proofs.
Participants express uncertainty about the phrasing of the original statement and the implications of the assumptions made. There are also unresolved steps in the proofs proposed, particularly regarding the powerset equality.
Readers interested in set theory, mathematical proofs, and logical reasoning in mathematics may find this discussion relevant.
Do you mean "it is the case" instead of "exists"? That is, is the claim as follows: If $A\setminus C=B\setminus C$ and $A\cap C=B\cap C$ for some sets $A$, $B$ and $C$, then $A\subseteq B$?MathPro17 said:If for any A,B,C sets exist A\C=B\C And A ∩ C =B ∩ C , then A⊆ B.
Evgeny.Makarov said:Do you mean "it is the case" instead of "exists"? That is, is the claim as follows: If $A\setminus C=B\setminus C$ and $A\cap C=B\cap C$ for some sets $A$, $B$ and $C$, then $A\subseteq B$?
Note that $X\setminus Y=X\cap\overline{Y}$, so $X=X\cap(Y\cup\overline{Y})=(X\cap Y)\cup (X\setminus Y)$. Therefore,
\[
A=(A\cap C)\cup (A\setminus C)=\text{ (by assumption) }(B\cap C)\cup (B\setminus C)=B.
\]
Why don't you continue $X\in P(A\cap B)\iff\dots$ yourself? You'll need $X\subseteq A\cap B\iff X\subseteq A\land X\subseteq B$ where $\land$ means "and".MathPro17 said:I would like to your help with another thing - to prove this :
P(A∩B)=P(A)∩P(B)
if you could prove it with:"x ∈ to P()..." ?
Evgeny.Makarov said:Why don't you continue $X\in P(A\cap B)\iff\dots$ yourself? You'll need $X\subseteq A\cap B\iff X\subseteq A\land X\subseteq B$ where $\land$ means "and".
You need to show $X\in P(A\cap B)\iff X\in P(A)\cap P(B)$ for all sets $X$. Why don't you continue the series of equivalences $X\in P(A\cap B)\iff\dots$ using the definition of powerset?MathPro17 said:I Mean How could I prove this : "P(A∩B)=P(A)∩P(B)" with Some X that I take ?