The discussion centers on proving set-theoretic statements involving subsets and powersets. The primary claim is that if for sets A, B, and C, the conditions A\C = B\C and A ∩ C = B ∩ C hold, then A ⊆ B. Participants clarify the notation and provide logical steps to demonstrate this relationship. Additionally, the discussion includes a request to prove that P(A ∩ B) = P(A) ∩ P(B) using the definition of the powerset.
PREREQUISITESMathematics students, educators, and anyone interested in formal proofs in set theory and logic.
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 ?