ProLT · Lecture 02 · 2 of 37
Sets: a language for possible states
In this lecture
- 0:00 · L02.S01 — Give the groups a name
- 0:36 · L02.S02 — A set and its members
- 1:16 · L02.S03 — A card and a set containing that card
- 1:45 · L02.S04 — Is one whole set inside another?
- 2:23 · L02.S05 — Intersection: who belongs to both?
- 3:08 · L02.S06 — Union: who belongs to either?
- 3:52 · L02.S07 — Empty sets and separate groups
- 4:20 · L02.S08 — Complement: who is outside this set?
- 5:01 · L02.S09 — The universe makes a difference
- 5:34 · L02.S10 — A collection can contain sets
- 6:17 · L02.S11 — Your turn, then the next step
Complete transcript
Show transcript
Last time, we used four cards to explore what observations can tell us. A had a dot, B had both features, C had a stripe, and D had neither.
Now we need a precise language for groups of these states. That language is sets.
Our universe is the collection of all the states we are considering. For now, it contains A, B, C, and D. We will call it omega.
Today we are describing groups. Drawing a group does not mean we have learned which card is hidden.
A set is a collection of distinct elements. Here, the elements are card states.
The dot set contains A and B. We write their names inside curly braces: the set containing A and B.
A belongs to this set. C does not.
The membership symbol is read “belongs to.” It relates an element to a set.
Order does not matter: listing B before A describes the same set. Repeating A would not add another element. Membership is what matters.
Here is a distinction that will matter throughout the paper.
A names one state. Braces around A name a set containing that state.
The set has one element, so we call it a singleton.
Think of the difference between an item and a group containing that item. Even when the group has only one member, the two descriptions play different roles.
Membership asks whether an element belongs to a set. Subset asks whether every element of one set also belongs to another.
The singleton containing A is a subset of the dot set: its only element is in that larger set.
But the dot set is not a subset of the singleton. B would be left out.
Subset can also include equality. Every set is a subset of itself. Two sets are equal when they have exactly the same elements.
The dot set contains A and B. The stripe set contains B and C.
Which states belong to both sets?
Only B. It has both features.
The intersection of two sets contains their shared elements. So the intersection here is the singleton containing B.
We are calculating a group. If we later learn that the hidden card has both a dot and a stripe, this calculation tells us that B is the only remaining possibility.
Now ask a different question. Which states belong to at least one of these sets?
A, B, and C.
Their union contains every element belonging to either set, including elements belonging to both.
B appears just once. We do not count it twice because it occurs in both starting sets.
D belongs to neither, so it is outside this union.
Intersection keeps the shared members. Union gathers the members from either set.
What if two sets share no elements? The singleton containing A and the singleton containing C have an empty intersection.
We call them disjoint. The empty set contains no elements at all.
Be careful with card D. It has neither feature, but it is still a state. A set containing D has one element. It is not the empty set.
Return to our universe of four states. The dot set contains A and B.
Its complement contains everything in the universe that is outside the dot set. Here, that is C and D.
You can read the displayed expression as “omega with the dot set removed.” The backslash means set difference: keep elements of the first set that are not in the second.
We can describe the no-dot region mathematically. That does not automatically give our observer a certificate establishing no dot.
For a separate example, suppose we have learned that the card has a stripe.
Our working universe is now B and C. Within that universe, the dot set contains only B, and its complement contains only C.
D does not reappear when we take the complement. We are considering only states inside the specified universe.
Whenever someone says “everything outside this set,” ask: outside it, but inside which universe?
Return to the original four-state universe.
We can also collect sets themselves. Put the dot set and the stripe set into a collection.
This collection has two members: one set containing A and B, and another containing B and C. Its members are those two sets. They are not the individual states A, B and C.
Taking their union would instead give one set containing three states.
Keep track of the level: a state, a set of states, or a collection of such sets. Later, a topology will be a particular collection of subsets of a universe.
Try two different groups: A and C in one set, C and D in the other.
What is their intersection? And what is their union?
The intersection contains only C. The union contains A, C, and D.
We now have a language for grouping states and comparing those groups.
Next, we will learn how logical statements are evaluated, using the same four cards and a truth table.

