How do I prove that if A, B and C are sets, then A × (B −C) = (A ×B) − (A ×C)?
As usual, prove it using the general heuristic that[math]X=Y[/math] if and only if [math]X\subseteq Y \text{ and }X\supseteq Y.[/math]In other words, take a generic element of [math]A\times(B-C)[/math] and show that it also must be an element of [math](A\times B)-(A\times C)[/math], and vice versa.
Prove that for all sets A, B, and C, (A - B) - C = (A - C) - (B-C)?
Let be [math]x \in (A-B)-C[/math]. Therefore, [math]x\in(A-B)[/math] and [math]x\notin C[/math]. So[math]x\in A \wedge x\notin B \wedge x\notin C[/math].[math]x\in A \wedge x\notin C \Rightarrow x\in(A-C)[/math].[math]x\notin B \Rightarrow x\notin (B-C)[/math].[math]x\in(A-C) \wedge x\notin (B-C)\Rightarrow x\in(A-C)-(B-C)[/math].Therefore, as every [math]x\in(A-B)-C[/math] is in [math]x\in(A-C)-(B-C)[/math], then[math]x\in(A-B)-C \subseteq x\in(A-C)-(B-C)[/math].On the other hand, if [math]x\in(A-C)-(B-C)[/math] then[math]x\in (A-C) \wedge x\notin (B-C)[/math].[math]x\in (A-C) \Rightarrow x\in A \wedge A\notin C[/math].[math]x\notin (B-C) \Rightarrow x\notin B \vee x\in C[/math].As for the first we know that [math]x\notin C[/math], we can deduce that [math]x\notin B[/math].[math]x\in A \wedge x\notin B\Rightarrow x\in(A-B)[/math].[math]x\in(A-B)\wedge x\notin C\Rightarrow x\in(B-C)-A[/math].We deduced that for every [math]x\in(A-C)-(B-C), x\in(A-B)-C[/math]. Hence[math](A-C)-(B-C)\subseteq (A-B)-C[/math].The double inclusion shows that both sets are equal.
How do you prove that for all sets A, B and C, (A∪B) \C = (A\C) ∪ (B \C)?
We have the following definitions:[math]A \cup B = \{x: x \in A \text{ or } x \in B \}[/math][math]A \backslash C = \{x: x \in A \text{ and } x \notin C \}[/math]Combining the two, we have:[math](A \cup B) \backslash C = \{x: x \in A \text{ or } x \in B \text{ and } x \notin C \}[/math][math](A \backslash C) \cup (B \backslash C) = \{x: x \in A \text{ and } x \notin C \} \cup \{x: x \in B \text{ and } x \notin C \} [/math][math]= \{x: x \in A \text{ or } x \in B \text{ and } x \notin C \}[/math]which proves the proposition. I'm hoping you can visualize the result with Venn-diagrams as well.
For the sets A, B and C prove that (A × B) U (A × C) = A × (B U C).?
Let (x,y) be an arbitray element of (A × B) U (A × C) => (x,y) Є (A × B) or (x,y) Є (A × C) => { x Є A and y Є B } or { x Є A and y Є C } => x Є A and { y Є B or y Є C } => x Є A and y Є (B U C) => (x,y) Є A × (B U C) This shows that whenever (x,y) belongs to (A × B) U (A × C) , it also belongs to A × (B U C). So, (A × B) U (A × C) is a subset of A × (B U C). Similary, going in reverse order, it can be shown that A × (B U C) is a subset of (A × B) U (A × C). Hence, (A × B) U (A × C) = A × (B U C)
How would you prove that, for all sets A, B and C, if A ⊆ C and B ⊆ C then A ∪ B ⊆ C?
It becomes very easy if you look up the formal definition of [math]\subseteq[/math]:[math]A \subseteq B \iff \forall x \in A: x \in B[/math]At the same time, [math]\forall x : x \in A \cup B \iff x \in A \lor x \in B[/math].Now you can reduce the problem to simple calculations with logical formulas.
Prove: If A and B are sets, show that A is subset of B if and only if A intersection B = A?
(For notation, let A ^ B = intersection of A and B.) Whenever you try to prove that X=Y as sets, you should show that X is a subset of Y and that Y is a subset of X. To show that X is a subset of Y, show that any element x in X is also in Y. (1) if part: Suppose A ^ B = A. Let a be an element of A. Then by hypothesis, a is also an element of A ^ B, whence a is in B. Thus A is a subset of B. (2) only if part: Suppose A is a subset of B. Let x be an element of A. Then x is also in B. So x is in A ^ B. Thus A is a subset of A ^ B. Now let x be an element of A ^ B. Then, in particular, x is in A. Thus A ^ B is a subset of A. So A = A ^ B.
How do I prove if a>b and b>c then a>c?
My initial answer suggested that this was an axiom - but on reflection :Assuming that a,b,c are all real numbers (i.e. not complex).The definition of A > B is that A-B is a positive real number.Since a>b then a-b is a positive real number (call it x) [math]\rightarrow[/math] a-b = xsince b>c then b-c is a positive real number (call it y) [math]\rightarrow[/math] b-c = ySo :a-c = a-(b-y) = a-b + y = x + ysince x & y are positive real numbers we know that x + y is a positive real number.so we can show that a-c = x + y, and therefore a > c
Let a, b and c be sets. How do I prove that a⊆b and b⊂c, then a⊂c?
Suppose c={1,2,3,4,5,6,...10}b={3,6,8,9}a={6,9}hence prove
Let A, B, and C be sets. Prove or disprove that (a) If A ∪ C = B ∪ C, then A = B (b) If A ∪ C = B ∪ C and B ∩ C = A ∩ C, then A = B (c) If (A − C) ∪ (C − A) = (B − C) ∪ (C − B), then A = B?
b) is true prove [math]A\subset B[/math] and [math]B\subset A[/math]For the first inclusion:Let [math]x\in A[/math] distinguish 2 cases[math]x\in C[/math] or [math]x\notin C[/math]- if [math]x\in C[/math] then [math]x\in A\cap C[/math] as [math] A\cap C= B\cap C[/math], [math]x\in B[/math]- if [math]x\notin C[/math], as [math]x\in A[/math], [math]x\in A\cup C[/math] and according to [math] A\cup C= B\cup C[/math], [math]x\in B\cup C[/math] and as [math]x\notin C[/math], [math]x\in B[/math]In all case [math]x\in B[/math].c) is true. Idem prove [math]A\subset B[/math] and [math]B\subset A[/math]Only the first inclusion:Let [math]x\in A[/math] distinguish 2 cases[math]x\in C[/math] or [math]x\notin C[/math]- Suppose [math]x\in C[/math]. By reduction if [math]x\notin B[/math] then [math]x\in C-B[/math]. According to [math](B-C)\cup (C-B)=(A-C)\cup (C-A)[/math] , [math]x\in (A-C)\cup (C-A)[/math] absurdity, then [math]x\in B[/math].-if [math]x\notin C[/math] then [math]x\in A-C[/math]. According to [math](A-C)\cup (C-A)=(B-C)\cup (C-B)[/math], [math]x\in (B-C)\cup (C-B)[/math]. Impossible for the last, it follows that [math]x\in B-C[/math].In all case [math]x\in B[/math].
Prove the Statement (A U (B n C) = (A U B)n(A U C) for sets A, B, C)?
A U (B ∩ C) = (A U B) ∩ (A U C): Now we'll prove that A U (B ∩ C)⊂ (A U B) ∩ (A U C), and (A U B) ∩ (A U C) ⊂ A U (B ∩ C) A U (B ∩ C) = (A U B) ∩ (A U C) follows. ======================================... A U (B ∩ C)⊂ (A U B) ∩ (A U C), Statament (1.1) proof: Let xϵ A U (B ∩ C), then xϵA or xϵ (B ∩ C). a) Suppose first that xϵA then xϵ A U B and xϵ A U B, so: xϵ(A U B)∩(A U C). b) suppose now that xϵ (B ∩ C), then xϵB and xϵC, hence xϵ(A U B) ∩ (A U C). we obtain that A U (B ∩ C)⊂ (A U B) ∩ (A U C) ======================================... (A U B) ∩ (A U C) ⊂ A U (B ∩ C) Statement (1.2) let xϵ(A U B) ∩ (A U C) , then xϵ(A U B) and xϵ(A U C), if xϵA it's simple to show (1.2) is true. Suppose that xϵB and xϵC then xϵ(B ∩ C) hence xϵA U (B ∩ C). ======================================... Statements (1.1) and (1.2) mean A U (B ∩ C) = (A U B) ∩ (A U C):