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7) (De Morgan'$ Laws) Suppose X is a set, and for AnV subset S of X let Se = XIS Suppose that Aa $ X for every helonging to some indlex set ^_ Prove that (Uxea...

Question

7) (De Morgan'$ Laws) Suppose X is a set, and for AnV subset S of X let Se = XIS Suppose that Aa $ X for every helonging to some indlex set ^_ Prove that (Uxea An)" Oea Ai; (Oxea 4,)" = UAe^ A;

7) (De Morgan'$ Laws) Suppose X is a set, and for AnV subset S of X let Se = XIS Suppose that Aa $ X for every helonging to some indlex set ^_ Prove that (Uxea An)" Oea Ai; (Oxea 4,)" = UAe^ A;



Answers

Prove each, where $A, B,$ and $C$ are any sets.
$$
(A \cup B \cup C)^{\prime}=A^{\prime} \cap B^{\prime} \cap C^{\prime}
$$

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

Right. We're going to prove this by showing both of these sets are subsets of each other. So proof you can let XP an element in the set and be a compliment. Okay, This means that so then X. Is not in A. And B. Okay by the mark. But the migrants laws, you can say that um X is not an A. Or access not M. P. All right. Um It follows so yeah, it follows that X is in a compliment or X is in the compliments. Ah Then X is in a compliment or the a compliment us um was set is a subset off a compliment. Big compliments. Okay, now we're going to look at the congress. So conversely, we can let X be in a compliment. The compliments. Right then access in a compliment or access in the compliments. Again, we apply the morgans law. So by by the morgans law, you can say that X. Uh is Yeah. So by the morgans law, we can say that X is not and A and B. Thus, so since X is not a name or be right, we can say that. So then access in. So the definition of Exxon A. And B. Is just excess and A. And be compliments. Thus, you can say a compliment or be compliments is a subset of A. And B. Compliment

Okay, so we want to prove the commentated laws. So first prove egging and easy to be union. So we'll prove the four condition. Say, let's X b in a union be then my definition of union you. This implies that X is in a oh, X is in B. Oh, hey, So then this implies if you just which around the ex is in B All ex is in a You really did nothing here's just stripped around within. This implies them by union that's X is in be union eh s so therefore you have a union be is a subset off be unit A that's approved the reverse condition off this you do the same thing where you switch so you do the exact same thing as always, you hear Will you switch be an A B and a Yeah. Yeah, And then you get the reverse condition. So then you should get the union, eh? Is a substance off, eh? Base soon. So therefore because have conditions will definitions, you have a union B is equal to be in a hey now intersection, so it will prove this one again. So if we get that X B A intersect with B. This implies that implies that X is in a and X is in big, so switch gets around again. You have X is in B on DDE X is in a which implies that X is in okay. Intersection is a so therefore a intersect with B the subset o b second Hey a and then similarly this i m i l A. I think that's how it's supposed to similarly, hopefully using the dis reversing. Maybe you have be insect with a is a subset. Oh, a distinct be. So now because you have two conditions both to fourth and reverse, you have a second B is equal to be intersected with a

Theme this problem. We're us to show that a union some 78 union the M percent is these set a intersection? New York's upset is saying, Hey, let's say that for always where X isn't Ellen Toe said A. We can write a union the intercept as X such that X will be an elemental and or eggs will be an elemental empty set. So what do you know? This? We know that every cent doesn't help and the elements on all the elements. So this part is Oh, so it means that a union in percent we'll have elements, I said. Old elements will be in L A. So this means that a union, uh, the cassette movie sets itself so embarked me. We have a section u Universal side. So imagine that we have this universal that complains or the els. And we have said so. This is the universe aside. It contains every element in any open set. Since A is a subset off the university sent, we can say that intersection you will be sent a


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