PrepShorts · Study sheet · Class 11 Mathematics · Chapter 1, Sets
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Outside {2, 3} and outside {3, 4, 5}, together, name exactly {1, 6} — same as the complement of their union. Complementing each set apart, the tempting shortcut, misses by three.
The idea
Failing to be in either of two sets is the same thing as being outside both, and failing to be in both is the same thing as being outside at least one — so De Morgan's two laws are less facts about sets than the behaviour of not applied to or and to and. The reason the operations must swap rather than stay put is structural: complementing reverses containment, and — because complementing twice returns the original set, which the chapter states on p. 20 — it reverses it reversibly, matching each set to exactly one other. Only a correspondence that strict is forced to carry the smallest set holding both A and B to the largest set sitting inside both complements; reversal on its own would give one half of the equality and leave the other open. The chapter verifies the first law on one six-member universe and then states the general result without proof; supplying that argument is the job of this topic.
What you should be able to do
- State both of De Morgan's laws for two subsets of a stated universe
- Verify either law on given finite sets by computing both sides in full
- Prove either law by taking an arbitrary object and following its membership verdict through both sides
- Explain why the union turns into an intersection and not into another union
- Draw the Venn diagrams for the four expressions involved and use the matching shadings to check the laws
- Use the laws to rewrite an expression so that no complement is applied to a compound
- Place De Morgan's laws inside the chapter's full list of complement properties
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| De Morgan's laws | the pair of rules exchanging union and intersection under complement | printed in this chapter (§1.10, p. 19) |
| complement | the members of the universe outside the given set | printed in this chapter (§1.10, p. 18) |
| union | the set of objects in one set or the other or both | printed in this chapter (§1.9.1, p. 14) |
| intersection | the set of objects in both sets | printed in this chapter (§1.9.2, p. 15) |
| universal set | the set the complements are taken inside | printed in this chapter (§1.7, p. 12) |
| order-reversing | the property by which complement turns a containment round | an added term; the chapter prints the relative version of the fact as an exercise and names it nowhere |
Where people slip up
- "Complementing a union gives the union of the complements." It gives the intersection. Test it on Example 22: the union of the two complements would hold 1, 2, 4, 5 and 6, while the complement of the union holds only 1 and 6.
- "Checking the law on a set of six proves it." It confirms one case. The chapter itself moves from the example to a general statement without a printed proof.
- "The laws need the sets to overlap." They do not. Run the argument on two disjoint sets and both sides still agree — nothing in the walk assumes a shared member.
- "The laws hold without a universe." Every complement in them is taken inside U. Without a fixed universe none of the four expressions names a set.
- "Not both means neither." This is the error the second law exists to correct. Outside the intersection means at least one of the two failed, not that both did.
- "There is one De Morgan law and the other is a rearrangement." They are two statements, and the chapter prints both. Each is proved by its own walk, though the walks are mirror images.
- "The swap is a convention to be memorised." It is forced. Complement reverses containment, and the smallest set containing both must go to the largest set inside both.
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Worked answers: Exercise 1.1 · Exercise 1.2 · Exercise 1.3 · Exercise 1.4 · Exercise 1.5 · Miscellaneous Exercise · this video explains Exercise 1.5 Q4, Miscellaneous Exercise Q5
Transcript1,921 words
Two sentences about one object and two sets. It is not in either of them. It is outside the first, and outside the second. The first asks one question, about the set you get by putting the two together. The second asks two, one about each set on its own. And yet the verdict is always the same. In a universe of five objects there are thirty-two sets you can make, and one thousand and twenty-four ordered pairs of them.
Ask both sentences of every object of every pair. That is five thousand one hundred and twenty verdicts. They disagree on none of them. That equality has a twin, and the pair is what this video is about. Start small. The universe is one to six. The first set holds two and three. The second holds three, four and five. Four of the six numbers are in at least one of them.
Left-hand side. Put the two sets together: two, three, four, five. Outside that union sit one and six. Right-hand side, and it never mentions the union. Outside the first set: one, four, five, six. Four numbers. Outside the second: one, two, six. Three numbers. What do those two outsides share? One and six. The same answer by a completely different route. One law, checked on one pair of sets. First, kill the wrong answer.
Almost everyone's guess is that complementing a union gives the union of the complements. Leave the operation alone, and complement the two pieces. Outside the first: one, four, five, six. Outside the second: one, two, six. Together that is one, two, four, five and six. Five numbers, where the right answer had two. It wrongly admits three of them: two, four and five. Watch why two gets in. Two is inside the first set, so it is not outside the union.
But it is outside the second, and or only needs one yes. Across all one thousand and twenty-four pairs, the wrong version lands on the right answer thirty-two times. Those thirty-two are precisely the pairs where the two sets are the same set. Not one pair of different sets survives it. So it is wrong on nine hundred and ninety-two. What did that six-object check actually establish? Six objects give sixty-four subsets, so four thousand and ninety-six ordered pairs.
The check settled one. It left four thousand and ninety-five untouched. Run the machine over all of them and both laws do hold, every time. So the check was not wrong. It was narrow. And widening it does not help, because the claim is about every universe, not this one. Six objects, nine, a hundred, endlessly many. You cannot finish an endless job by working faster. You have to stop counting and start arguing.
Here is the argument. Take any one object of the universe. Suppose it is outside the union. The union holds everything in either set, so if the object were in either, it would be in the union. It is not, so it is in neither. In neither means outside the first and outside the second, which is what being in both complements means. That is one direction. Now run the sentence backwards.
Suppose it is in both complements. Then it is outside the first and outside the second, so nothing put it in the union. Both directions, so the two sides hold exactly the same objects. Notice what the argument never asked. Never how many members anything has. Never which universe we were in. It took one object and followed its verdict, and one object is all any set is made of.
That is why this is a proof and the counting was not. The twin law swaps the operations the other way. Outside the overlap becomes outside one or outside the other. Not in both does not mean in neither. It means at least one of the two failed. Possibly only one. Back to the small universe. The two sets share only three. So outside their overlap sit one, two, four, five and six.
Five numbers, but only one and six are in neither set. Two, four and five are each in one of them. They just are not in both. Across every pair in the five-object universe there are two thousand five hundred and sixty cases of an object failing to be in both while sitting comfortably in one. Not-both is genuinely weaker than neither, and this law is what keeps them straight.
The walk is the mirror image. An object outside the overlap failed at least one test, so it is in at least one complement. Reverse the sentence for the other direction. One more thing the walk never asked. It never assumed the two sets share anything. These laws get read as facts about overlapping circles. In the five-object universe, two hundred and forty-three of the ordered pairs share nothing at all.
That is three to the fifth, and here is why. Each object has three choices: the first set, the second, or neither. It cannot be in both, or they would share it. Both laws hold on every one of those two hundred and forty-three. With nothing shared, the overlap is empty, so outside it is the whole universe. And the two outsides together are also the whole universe. The law does not merely survive the empty case. It is boring there, which is what a law should be.
The proofs are done. But they do not answer why the operations had to swap. Why not complement a union and get another union? Look at what complementing does to containment. Suppose the first set sits inside the second. Anything outside the second is outside the first, because it was never in the first to begin with. So the outsides sit the other way round. Complementing turns a containment around.
In the five-object universe, two hundred and forty-three ordered pairs have one set inside the other. Three to the fifth again. Every one comes back reversed, and no pair that was not nested comes back looking nested. There is a second fact, and it is the quieter one. Complementing twice returns the set you started with, for all thirty-two. So complementing is a reversal that can be undone. How much of the law does reversal alone give you?
The first set sits inside the union. So does the second. Complement all three, and both containments turn around. Outside the union sits inside the first outside, and inside the second outside too. Anything inside both of two sets is inside what they share. So outside the union sits inside what the two outsides share. Checked on all one thousand and twenty-four pairs, and it never fails. But look at what we have: one side sitting inside the other.
We wanted them equal, and sitting inside is half of that. Reversal handed us that half for free, and it has nothing more to give. Is that half really all reversal buys? Take a universe of three objects. Eight subsets. Now consider every way of assigning a subset to each of those eight, with one rule: whenever one set sits inside another, the assigned sets sit the other way round.
A search finds eight thousand of them. All eight thousand give you the easy half, so that half really is free. Now, how many give the whole equality? Seven hundred and twenty-nine. Seven thousand two hundred and seventy-one reverse containment perfectly well and still get the law wrong. Reversal is not enough. That is a measurement, not a feeling. Now the second fact: undoing itself. Of the eight thousand, exactly four undo themselves.
All four give the whole law. Reversal plus undoing itself forces the equality. Reversal alone does not. Those four are worth naming. Shuffle the three objects however you like, then complement. Six shuffles, six one-to-one reversals, and four of those shuffles undo themselves. Complement, with nothing shuffled, is one of the four. And a reversal that gets the law wrong gets at least two of the sixty-four pairs wrong. Never just one.
The reversal is not really about the universe. Any set will do as the thing you take away from. Suppose the first set sits inside the second. Take a third set and remove the second from it. Then remove the first instead. Removing the bigger one leaves less. An object that survived losing the second was never in the second, so never in the first, so it survives losing the first too.
The same walk, with a third set standing where the universe stood. There are seven thousand seven hundred and seventy-six ways to choose a nested pair and a third set, and it holds for all of them. The nesting is doing real work, not decorating the sentence. Take any of the seven hundred and eighty-one pairs that are not nested, and some third set breaks it. So complement is just this, with the universe in the third place.
Now the picture, and it comes last on purpose. Two sets cut a universe into four cases. In the first only. In both. In the second only. In neither. Every object is in exactly one of the four. Shade the outside of the union. One case: in neither. Shade what the two outsides share. Also just in neither. Same picture. That is the first law, drawn. Now shade the outside of the overlap. Three cases: first only, second only, neither.
Shade outside one or outside the other. The same three. Same picture again, and that is the second law. The two pictures are not each other. One shaded region against three. Drawings are a good check and a bad proof. They convince you about two circles on a page. The walk convinced you about every universe there is. One more check, on both laws at once. The universe is one to nine.
The first set is even: two, four, six, eight. The second is two, three, five, seven. Their union holds seven numbers, everything from two to eight. So outside the union sit one and nine. Outside the first: one, three, five, seven, nine. Outside the second: one, four, six, eight, nine. What they share: one and nine. First law confirmed. Now the other law on the same pair. The two sets share only the number two.
So outside their overlap is everything but two. Eight numbers. And the two outsides together are also everything but two. And the second law left out exactly one object, the only one both sets held. So where do these two belong? Complement comes with four groups of facts. First: a set and its outside make up the whole universe between them, and share nothing. Second: complementing twice returns the set.
Third: the empty set and the universe are each other's outsides. Fourth: this pair. Outside a union is the shared outside. Outside an overlap is the combined outside. Every one of those was checked on all thirty-two sets of a five-object universe. But the second group is not just sitting on that list. It is the ingredient that made the fourth come out the way it did. Reversal on its own gave one half. Being undoable gave the other.
So the swap was never a convention anyone chose. Not is a reversal, and a reversal you can undo has to carry the smallest set holding both to the largest set inside both. Or becomes and, and and becomes or. It could not have come out any other way.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Difference and complement: the same idea with and without a universal setClass 11 · Ch 1, Sets
- Union and intersection, and what it means for two sets to miss each other entirelyClass 11 · Ch 1, Sets
- Turning a claim about sets into a picture you can read offClass 11 · Ch 1, Sets
- Why order and repetition cannot make two sets differentClass 11 · Ch 1, Sets
- Containment, proper containment, and why the empty set is inside everythingClass 11 · Ch 1, Sets
Either side of this one
- Why writing a pair in order carries information a set cannotClass 11 · Ch 2, Relations and Functions