Reid's brave officer remembers the boy he was and is remembered by the general he became, while the general no longer remembers the boy. Memory between stages is therefore not transitive, and identity is. The relation lab shows the gap on five stages, and its closure control shows the chain that fills it.
Core ideas
The previous lesson asked what makes a ship the same ship through change of parts. A person changes too, and the most natural answer is that a person is held together by the mind, not the planks. This lesson tests that answer on its first objection.
Two stages of a life are connected when one directly carries something of the other: a memory, an intention carried out, a belief or a trait. It is a relation between two stages, it comes in degrees, and the lab draws it as a single pair in a grid.
Two stages are continuous when some chain of connected stages leads from one to the other. In the vocabulary of the relation lab it is the transitive closure of connectedness: every pair joined by a path of one or more steps.
Because the tie is direct, it need not pass through. A stage can be connected to its neighbour on each side and to nothing further away, and the stages at the ends are still continuous. That gap is the whole lesson.
The material
Locke's proposal, Reid's objection to it, and the relation that repairs it.
Locke proposed that a person is the same person as far back as the person's consciousness extends, and the plainest evidence of that reach is memory. If I remember doing something, I am the one who did it. The proposal is a criterion for when two stages of a life, a person at one time and a person at another, belong to one person.
Thomas Reid answered with a case. A brave officer, in his first campaign, took a standard from the enemy. As a boy he had been flogged for robbing an orchard. When he took the standard he remembered the flogging. Late in life, as a general, he remembers taking the standard and has quite lost the flogging. So the officer is the boy, since he remembers being flogged. The general is the officer, since he remembers the standard. And by the same criterion the general is not the boy, since he remembers nothing of the orchard. Identity is transitive, so a criterion that says all three things has contradicted itself.
Two stages are psychologically connected when there is a direct tie of the kind memory gives. They are psychologically continuous when there is a sequence of stages from the one to the other in which each is connected to the next. Continuity is the transitive closure of connectedness.
The repair is the one the definition makes. Let the tie be as weak as Reid likes. The general and the boy are not connected, but they are continuous, since the officer stands between them and each neighbouring pair is tied. The criterion for identity is then continuity, not connectedness, and Reid's contradiction is gone: the three stages are one person, and the general does not need to remember the boy for that to be so.
The relation lab draws this on five stages. Read the pair (a, b) as “stage b directly remembers stage a”, and take the pattern in which each stage remembers only the one just before it. The grid holds four pairs, the first with a = 1 and b = 2, and the property table finds the failing triple at once: the pairs (1, 2) and (2, 3) are in the relation and (1, 3) is not. The relation is not transitive, and that is Reid's finding on five stages instead of three.
Set the closure control to transitive. The lab marks six more cells, the pairs (1, 3), (1, 4), (1, 5), (2, 4), (2, 5) and (3, 5), and with the original four they fill the cells above the diagonal. The relation the closure draws is “stage a comes earlier than stage b in one chain”, and every stage is continuous with every other. The closure leaves the original four pairs where they were. It does not invent a memory; it adds the pairs that a chain of memories supplies.
Two limits should be stated. The first is that the chain needs enough strength at each link. A link so faint that nothing remains of the earlier stage is no tie at all, and the lab has only two answers, a pair or no pair, where a real tie comes in degrees. Where to put the threshold is an input, as the cutoff was in the previous lesson.
The second is that continuity as defined does not say that the chain is unique. If a single stage can be continuous with two later stages that are not continuous with each other, a person has been divided, and identity, which is one-one, cannot follow the chain in both directions. The next lesson takes that case as its subject. Here the chain runs in a line, and a line is what the closure of the next-stage relation is.
Five stages, each remembering the one before. The relation has 4 pairs and fails transitivity at (1, 2) and (2, 3). Its transitive closure has 10 pairs, every pair (a, b) with a less than b. Stage 5 is continuous with stage 1 and directly connected to stage 4 only.
Interactive
Decided by checking, not by name
Method
Five moves for any criterion that names a tie between stages.
Say what must hold directly between two stages for them to count as linked: remembering, carrying out an intention, keeping a trait. Write it as a set of pairs.
For each pair of pairs (a, b) and (b, c), look for (a, c). Any missing pair is a stage that the criterion cannot link to a stage it ought to.
Identity is transitive. If the criterion says the stages are one person whenever the tie holds, a missing (a, c) is a contradiction, not a free choice.
Take every pair joined by a chain of ties. This is a criterion of continuity, and it can be transitive because it was built to be.
Check whether one stage can be continuous with two stages that are not continuous with each other. If it can, identity has been asked to follow a fork.
Three stages: the boy, the officer and the general. A pair is a memory of the earlier by the later.
The tie is the weak relation and the closure is the one identity can use. Had the criterion been the tie itself, it would have said the general is not the boy and the officer is both; with the chain it says the three are one.
Given a set of pairs of stages, you should be able to name the pair of pairs whose shortcut is missing, list the pairs the transitive closure adds, and state which stages are continuous without being connected. A claim that a criterion of identity fails, made without the missing pair, has not been checked.
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