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For B given sets, let P(A). P(B), and P(AU B) be the power" sets of the sets A_ B. and AU B respectively. Prove O1 disprove that P(AUB) = P(A)U P(B) (For dispr...

Question

For B given sets, let P(A). P(B), and P(AU B) be the power" sets of the sets A_ B. and AU B respectively. Prove O1 disprove that P(AUB) = P(A)U P(B) (For disprove, an example is enough: (b Prove o1" disprove that P(A) U P(B) € P(AUB)_ (For disprove, an example is enough. Do part (a) O1 part (b) only W! I will NOT grade this problem unless you clearly(WW!) state which part is for grading!

For B given sets, let P(A). P(B), and P(AU B) be the power" sets of the sets A_ B. and AU B respectively. Prove O1 disprove that P(AUB) = P(A)U P(B) (For disprove, an example is enough: (b Prove o1" disprove that P(A) U P(B) € P(AUB)_ (For disprove, an example is enough. Do part (a) O1 part (b) only W! I will NOT grade this problem unless you clearly(WW!) state which part is for grading!



Answers

Use mathematical induction in Exercises $38-46$ to prove results about sets.
Prove that if $A_{1}, A_{2}, \ldots, A_{n}$ and $B$ are sets, then
$$
\begin{aligned}\left(A_{1}-B\right) \cup\left(A_{2}-B\right) \cup \cdots \cup\left(A_{n}-B\right) \\=\left(A_{1} \cup A_{2} \cup \cdots \cup A_{n}\right)-B \end{aligned}
$$

In this problem We wanna short of aces. If development is a subtitle A compliment. So we're gonna issue We're gonna start by assuming this to be true so soon Hey, is the substance of the then for every ex that is an 11 to pay x will be in a lento Be pricing, Cesar. Subset of feet. Now we know that this is true. Yeah, for everyone, that is a l ittle b Pullman. This implies that it wise and element will be a compliment. Then why is not an elemental said right? And since be known at, um, for every X that is an elemental fight, it's also the element be You can say that. When will I be that it's not an elemental Hey, right. So this is basically equal saying that it was an element of Republicans. Then why's an elemental eh? And since this is true for each and every wire, since that is true for oh wise, this means that be compliment is a subset of foam. So we just showing that if be compliment is a subset of a compliment than eight is a subset

Okay, so in this question, we want to prove that a is a subset of B if and only if being confident is sub set off a comet. So we prove this one first. So let's a b a subset of B then if X is, it is not be these implies that X is lost in bay. This is pretty obvious because if you have set B and you have a set a, well, just contained inside of it, even element is outside of big, then it must be outside. So using definitions off compliment if exits and be this implies that X is in a compliment. So therefore, being confident is a subset of a compliment. Now in the other direction. Well, let me compliment is a subset off a in. Similarly, if an element is not in a compliment, then it follows. Our eggs must not be in his confidence. So therefore, if X is in a a, then X must be in B Therefore a subset puppy

Hi. In this question, we are asked to show that the said, uh equation here is true for any positive integer in on. We asked to use induction even though I believe proving it directly from set properties. Much simpler, but it's an exercise. So let's do it. You see induction less this that maybe p of in first step the basic step. We would have just a one intersect be on both sides, right? And they are obviously equal now for in that keeps that suppose that it is true for P l n For some, in I Wish order, we will show that is true fall in plus one as well. And here you have many ways to do it. You can try to use the information from P O in to prove P up in plus one. Or you can also use some information about said that we know is true to prove it. And it was still like the proof by induction will still be well, all right, I'll do shoes the latter. So let's go over the statement off P and pass one on how you prove it. So first, since I'm proving in the equation. All set, I would I will prove that both of them being a subset off each other, Right? First we have this. Ah, this direction. So suppose x ys and element in this set on the left hand side, it would mean since this is intersection, it would mean that X is in some a J. We don't know what it is. Doesn't matter. And it is also inside b so inside a J and B therefore, it is in the intersection between this too, right? By just the definition of intersection, that means it must be inside this June. The end off off a I the sick be because a J in the second is in there somewhere. All right, It's nothing. Complicate that. So we have a show. One implication specifically Is this one The union is that B? He's a something off union off intersection, right? And we can do the same fall for the reverse implication. So it will start like let X be in this union off intersection sins they are joined by by union. The meaning of this is that eggs must be in some some of the terms inside here. Right? We don't know which, but it doesn't matter is in some genius iq be somewhere. And so by the meaning of this intersection, it would mean that X is inboard this a j and board. This ain't a and B all right? And so, But with with this, it would mean that X is in the union off. They won, too. Itchy. Sorry. It want to a in pass water, right? Because it is in there somewhere. So it is in this whole union thing and is in be so is inside intersection off. This would be all right. Nothing complicated there. So I have shown DeVry was like being a subset. We started the right hand side. Now we arrive at the left hand side. When I show both board us said is a subset off each other. It would mean the state may everyone. So it means that where should I write this? It means the union. Oh, a 12 a and plus one in the sick Me is equal to a one in the state bee You in the n onto impossible on music Be all right that Yes, nothing public it going on it Just symbol said Properties. And so we have shown that the statement is true for peon plus one and so back to our induction. We just showing the inductive step is true. So this together with basic step, means that this statement is true. Fall all positive integer in buy meth induction and that is it. Thank you.

In this problem Where us to prove April the easing of a private seat When, uh, Sethi and see our ego. So let's say that element A is insects, see, And some reading orbiter Element X is in such B So from this, we know that a s will be in port. Eh? Be now I assume a protectee is equal to a C. Then this means that eggs we'll be in a what I see Hence element X we'll be in said See Right. So if he is, you hate to be a fix. Uh, element here eggs isn't returning elements in your victory element X That is instead be we found that it will also being set seed. So this means that B is a subset of C No, this time, let's use April showers. Let's say that again, eh? A. Some Biggs elements of settlement is an element of Hey, now it's some run random arbitrary Ellman is an Elden offset. See? So from this we know that airs will be a said all by a former C. Again, let's assume a product be to be equal to a vaccine that from this we see that a X we'll be in. Hey, brother. See? Right. Some pens. Thanks. Sorry. We already started with a C. So that should be it for me and x will be inset Be so again X could be any element. So this means that any element that's he has he should also has that. So this means that sea is a subset of the if bees the substance e and it sees a subsequent feet This means that being and see our people So this means that product baby is evil for they see it says B and C are people.


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