Forcing names and the forcing theorem
Interpret recursively coded names, distinguish truth in a generic extension from a condition’s forcing assertion, and state the theorem as a formula scheme.
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The question for today
What you’ll understand
- Evaluate names recursively and compute canonical names for finite ground sets.
- Distinguish the definability and truth parts of the set-forcing theorem.
- Check a finite atomic example without claiming that every generic filter adds a new set.
A condition describes partial information, but a generic extension contains sets of arbitrary membership depth. Forcing names encode those sets recursively before the generic filter is known. Valuation turns a name into an actual set after a filter has been chosen. The forcing theorem connects this interpretation with a relation definable in the ground model, allowing set-theoretic statements about the extension to be controlled by finite or otherwise ground-model conditions.
Names are syntax with recursively nested members
Continue the external conditional setting of a countable transitive M satisfying ZFC, and let P∈M be a set forcing order with a weakest condition 1. A P-name is a set of pairs whose first coordinates are earlier names and second coordinates are conditions. The well-founded name hierarchy makes valuation possible: retain a pair exactly when its condition belongs to G, evaluate its first coordinate, and collect the resulting values. Different pairs can yield the same value, since the output is a set rather than a multiset.
The canonical name check-x for a ground set x pairs each canonical name check-y for y∈x with 1. Induction on membership gives check-x evaluated by G equal to x. The extension M[G] consists of valuations of all P-names belonging to M. It contains M and G, is transitive, and has the same ordinals as M. Saying that it satisfies all of ZFC requires the forcing theorem and axiom verification; transitivity alone would not establish Power Set or Replacement.
Forcing connects conditions to truth in the extension
For each fixed first-order formula φ, the set-forcing theorem provides a ground-model definable relation p forces φ of the indicated names. Definability is a formula-by-formula scheme; it is not an internal truth predicate for every statement of the whole universe. The truth lemma says that φ holds of the evaluated names in M[G] exactly when some p in G forces φ of the names. A condition forcing a statement therefore ensures it in every relevant generic extension containing that condition, not merely in one arbitrarily chosen branch.
Forcing is persistent under strengthening: once p forces a statement, a stronger q also forces it. A condition need not decide a sentence, but the conditions that decide it form a dense set. Negation is governed by the absence of stronger conditions forcing the opposite statement, not by failure of the original condition to force it. The recursive atomic definitions, quantifier clauses, and their inductive verification are substantial parts of the theorem, cited here rather than replaced by an informal appeal to eventual truth.
An undecided condition can still force a disjunction
Consider a three-condition order with weakest condition 1 and two incompatible atoms p₀ and p₁ below it. The two generic filters are {1,p₀} and {1,p₁}. Each is already an element of M in this finite example, so neither extension adds new sets. This does not contradict the earlier Cohen-real argument: its newness proof used dense disagreement requirements and the possibility of splitting further, while an atom has no two incompatible stronger extensions.
A name can nevertheless take different ground-set values in these two filters. The weakest condition can force a statement true for both values while deciding neither of two incompatible alternatives separately. This is useful for separating the logical meaning of forcing from the vague idea of choosing a single future. The truth lemma uses some condition in the actual filter as a witness, and strengthening reaches a deciding atom here. An undecided top condition does not make the entire forcing relation undefined or undermine definability.
Work it through
Evaluate a two-branch name
In the three-condition atomic order, let τ={(check-0,p₀),(check-1,p₁)}. Compute its values in G₀={1,p₀} and G₁={1,p₁}, and decide whether 1 forces 0∈τ.
Canonical valuation gives check-0 evaluated as 0=∅ and check-1 evaluated as 1={0}. In G₀ only the p₀ pair is retained, so τ evaluated in G₀ is {0}=1.
In G₁ only the p₁ pair is retained, so τ evaluated in G₁ is {1}, a singleton whose member is 1. In particular 0 is not a member of this value.
The atom p₀ forces 0∈τ, while p₁ forces its negation. The weakest condition 1 decides neither, although it forces τ to be nonempty because both valuations are nonempty.
A single name can have different values in different generic filters; failure of 1 to force a statement does not mean that 1 forces its negation.
The living notebook
Explore the model
Interactive lab · probability
Select a branch, then evaluate the name on paper
Pause and Reset, then set p = 0 and n = 1. The exact count law has all mass at K=0; use label zero to select G₀ and predict τᴳ⁰={0}. Reset and set both p = 1 and n = 1; all mass is at K=1, selecting G₁ and value {1}. This degenerate binomial display is only a finite branch selector for the worked valuation. Probabilities are not the definition of genericity, and the instrument does not evaluate arbitrary names.
- Mean · np
- 5
- Variance · np(1−p)
- 2.5
- Total probability
- 1
These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forInterpretation requires a specified branch; a name by itself need not determine one fixed value.
Pause & reason
Check your understanding
Your turn
Try it yourself
Using the worked name τ, let σ={(τ,1),(check-2,p₁)}. Evaluate σ in both generic filters. Why is this finite example insufficient to show that forcing adds a new real?
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Evaluate the first coordinate of each retained pair before placing its value into the output set. Both filters contain 1.
Compare with the solution
In G₀, the pair with condition 1 contributes τᴳ⁰={0}, and the pair with p₁ is discarded. Thus σᴳ⁰={{0}}. In G₁, both pairs are retained: τᴳ¹={1} and check-2 evaluates as 2={0,1}. Therefore σᴳ¹={{1},2}. The two contributed elements are distinct. All these sets, and the two finite filters themselves, already belong to the transitive ground model. The finite atomic order produces different valuations of a name but no new sets, so it cannot replace the atomless Cohen argument for a new real.
Keep the idea
Take it with you
- Names are recursively coded sets whose valuations depend on a chosen generic filter.
- Definability and the truth lemma connect ground-model forcing to extension truth for each formula.
- An undecided condition need not force a negation, and atomic forcing need not add anything new.
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