- #1
WisheDeom
- 12
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Hello,
I am teaching myself Set Theory, and in doing some exercises I came across the problem:
Given sets A and B, prove that [itex]A \subseteq B[/itex] if and only if [itex]A \cap B = A[/itex].
My proof, in natural language, is in two parts:
1) Prove that if [itex]A \subseteq B[/itex], [itex]A \cap B = A[/itex].
By the definition of subsets, all elements [itex]x \in A[/itex] are also [itex]x \in B[/itex]. By definition, [itex]A \cap B = \left\{x: x \in A \wedge x \in B \right\}[/itex]. Since all elements of A are also elements of B, but not all elements of B are necessarily elements of A, the intersection fo the two is A.
2) Prove that if [itex]A \cap B \neq A[/itex], it is not true that [itex]A \subseteq B[/itex].
If the intersection of A and B is not A, then there are necessarily elements of A that are not elements of B, therefore A is not a subset of B.
Therefore, [itex]A \subseteq B[/itex] if and only if [itex]A \cap B = A[/itex]. Q.E.D.
Is this a sufficient proof?
If it is, could someone help me translate it into the language of formal language?
Edit: Fixed wrong logical connective, set notation.
I am teaching myself Set Theory, and in doing some exercises I came across the problem:
Given sets A and B, prove that [itex]A \subseteq B[/itex] if and only if [itex]A \cap B = A[/itex].
My proof, in natural language, is in two parts:
1) Prove that if [itex]A \subseteq B[/itex], [itex]A \cap B = A[/itex].
By the definition of subsets, all elements [itex]x \in A[/itex] are also [itex]x \in B[/itex]. By definition, [itex]A \cap B = \left\{x: x \in A \wedge x \in B \right\}[/itex]. Since all elements of A are also elements of B, but not all elements of B are necessarily elements of A, the intersection fo the two is A.
2) Prove that if [itex]A \cap B \neq A[/itex], it is not true that [itex]A \subseteq B[/itex].
If the intersection of A and B is not A, then there are necessarily elements of A that are not elements of B, therefore A is not a subset of B.
Therefore, [itex]A \subseteq B[/itex] if and only if [itex]A \cap B = A[/itex]. Q.E.D.
Is this a sufficient proof?
If it is, could someone help me translate it into the language of formal language?
Edit: Fixed wrong logical connective, set notation.
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