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3. Assume that A, B,C are sets. Prove or disprove: (a) Ax (BUC) = (Ax B) U(Ax C) (b) Ax (BnC) = (Ax B)n(Ax C)...

Question

3. Assume that A, B,C are sets. Prove or disprove: (a) Ax (BUC) = (Ax B) U(Ax C) (b) Ax (BnC) = (Ax B)n(Ax C)

3. Assume that A, B,C are sets. Prove or disprove: (a) Ax (BUC) = (Ax B) U(Ax C) (b) Ax (BnC) = (Ax B)n(Ax C)



Answers

Prove or disprove that if $A, B,$ and $C$ are nonempty sets and $A \times B=A \times C,$ then $B=C$

Hello there. So in this exercise we need to show that if is containing B and B is containing C, then A is contained in C. As well. So this is uh technically this the the the trans activity property of the of the subsets. Okay, so first let's illustrate this with the band diagram drawings always helps to show the things. So let's drove first this part. So there is a set of V. That is going to come, there's going to be the set V. And there is a subset of be, that is the set A. But this part of the statement said that B is a subset of a set. See that means that there exists another set that contains B. And you can see just by this illustration that A. Is also going to be contained in C. In the words in writing here, we know that the meaning that A. Is contained in B. He is equivalent to say that for all the elements A. In this set A then A is also an element of B. Okay, so this is for all the elements in A. But also we have this statement here that say that B is a subset of C. There is a cool way to say that for all the elements B, B, then B is also an element in the set seek. But here we have that all the elements of A are elements of B. And here we say that all the elements will be our elements of C. So therefore we have that all the elements of A are going to be elements. Oh, see mm.

Yeah the on the show. But these two sets are equal and where we can do that is just to show that both are subsets of each other. So well consider the case where there's an element X. And a double compliment. Okay this means that X. Is in access not in any compliments. Okay. Yeah. So X is in hey okay so it's kind of like you're trying to get as simple as possible, right? So X is in a double compliment which just means that it is not in a compliment because a double compliment. In fact it's not a compliment. Then excess and a right? So that's what you have and we actually Done for the 1st case. So you can say thus a compliments compliments the subset of Yeah then we need to show. So conversely really let ex. B. In a right. Yeah. Okay so excess and A. This means that X. Is not in a compliment because excess tonight. Thanks. So X is in hey compliment compliments and you can always go back and forth Lexus and a extraordinary. The next season you're taking the compliment over and over. So this is what this means. And so we say that right? Um A is contained in a compliment, compliment

All right, So we're going to approve this um Equality by showing that both are subsets of each other. The first that X be an element in a minus B. Okay, so this means that excess in A. And X is at and be so access and a. And excess in be compliment. Ah Then access in a end. The compliment us a minus B is a subset of A and big compliment. Now we can look at the converse, you can let ex B n A n B compliments. This just means that accessing A and X is not in the okay, so X is in. Okay and not be. Thus our said is a subset Oh a minus B.

In this problem You ensure that a public be instant dizzy with a product in it Or see anyone else wanna show that a party sexist easy on a fella be And that intersects with a sexy so corporate uh, first part party less saying that a is an elevator. These elements of that B S e t. So that the unity would have elements off B C and a part of the unity. And why a. B C. Now let's get April to be in a palace. It would be where every element A B and a product. See what have the element a. C and protect the union of those stupid even have A and B summer a seat, right? And then she gets you. Those two are equal so that the first given staying in this to now in the second World let's assume that set a as L A again has an element of a B. Be an age Ages is a rush on thespian HR just arbitrary elements. And since he has l. A c so a, um, games a section scene would have the element age and a product being to Section C would have the elements. A eight, right? So now let's look at, uh, a product. Be product. Be would look like a bee in a age and a product, See would be a c in a age. Now, if you look at the intersection of those two, we wouldn't not with the element a repair ate right. And as you can see, this gives the same thing. So this also leave that to give a statement?


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