Recent content by vrble

Yes, I meant "Take the longest side of the triangle as its base and extend a perpendicular from the vertex opposite the base." Sorry, I must have been thinking about something else while writing up the proof. Is there any improvements that could be made?

From Apostol's Calculus Volume I, "Area as a Set Function" 1. Homework Statement : Right triangular regions are measurable because they are constructed from the intersection of two rectangles. Prove that all triangular regions are measurable and have an area of the product of one-half, their...

Ahh, I see now. Thank you.

I know that. How is that relevant to the issue at hand? Sorry if I sound stupid.

I'm aware to that now thanks to the wonderful members of this board, but I'm still not sure how your notation would allow me to avoid assuming that A is non-empty. Micromass's suggested rewording ""If y is an arbitrary element of X" seems to fix the problem.

Why would I need to assume A was non-empty? If A is empty then it immediately follows that it is a subset of P(A) and that P(A) is a subset of P(P(A)). I thought that it was implicit that P(A) will never be empty despite the contents of A.

Ah, I see. But what exactly is meant by, "whatever you wrote about ##y## holds vacuously."?

I don't believe that was the issue that verty was addressing.The problem is that I assumed there was an arbitrary element y that is an element of the arbitrary set X. In the case where A = ∅, A is a subset of P(A), and X is an element of P(A), it follows that X is indeed a subset of A but this...

Would adding "Let A be a non-empty set" fix this problem?

1. Suppose that A is a subset of the power set of A. Prove that the power set of A is a subset of the power set of the power set of A. Note: I'm going to use P(A) to mean the power set of A and P(P(A)) to mean the power set of the power set of A. 2. None 3. Let X be an...

Theorem: Suppose that ##A\backslash B## is a subset of ##C\cap D##, and x is an element of \large A. If x is not an element of \large D, then x is an element of \large B Proof: Suppose that ##A\backslash B## is a subset of ##C\cap D##, x is an element of \large A, and x is not an element of...

That is exactly what I did, thank you so much. Will post new proof shortly.

I believe that first sentence is a result of my confusion between free and bound variables while examing the definition of intersections. I thought of it as, "For all x, x is in C and x is in D. x is not in D. Therefore, there is not an x such that x is in C and x is in D." Then that lead me to...

Thank you for your response, I thought that would be the case. The set of exercises this problem comes from are specifically of the form, "If P, then Q. Assume P then prove Q." so I'd prefer to stick to that method of attack. Could you point out some of the things that are incorrect about my...

1. Suppose A \ B\subseteqC\capD and x\inA. Prove that if x \notinD then x\inB 2. None 3. Proof: Suppose A \ B\subseteqC\capD, x\inA, and x\notinD. It follows that our first assumption is equivalent to A due to our third assumption. Thus, B\subseteqC\capD is disjoint and either x\notinB\subseteqC...