Plan Around the Gap Between Assumptions and Conclusion
Choosing the Right Theorem emphasized matching a theorem’s hypotheses and conclusion to the job a proof must do. Proof Strategy Mastery takes the next step: once a useful tool has been identified, how should the argument itself be organized? A reliable proof has a clear route from the assumptions to the target, and each step closes a specific gap along that route.
Begin by writing down the exact conclusion to be proved. If it says “for every,” choose an arbitrary object and keep it arbitrary. If it says “there exists,” identify what object will serve as a witness and verify its required properties. If it asks for an estimate, record the estimate in the form needed and identify which available bounds could produce it. These habits prevent a common failure: proving a nearby statement that is easier, but not the statement that was asked.
Next, decide whether the implication is best handled directly, by contraposition, by contradiction, or by cases. Earlier tutorials developed these individual strategies. The aim here is to combine them deliberately. For instance, a direct proof may become straightforward after establishing an auxiliary estimate; a case split may make an absolute value manageable; and a contradiction may be useful when the negation of the desired conclusion supplies a concrete obstruction.
State exactly what must be shown, including its quantifiers and any required uniformity.
For a universal claim, take an arbitrary object. For a case argument, verify that the cases cover every possibility relevant to the assumptions.
Identify an estimate, intermediate claim, or theorem that connects what is known to the target.
Check that the proof establishes the original statement, not a weaker local, pointwise, or restricted version.
Case Analysis Is a Proof, Not a Guess
A case split is valid only when its cases exhaust the possibilities. It does not suffice to handle several convenient examples or to prove the target under an extra assumption that may fail. The following principle makes the logical requirement explicit.
Proof. Take any \(x\in X\) satisfying \(P(x)\). Since \(E_1\cup\cdots\cup E_m=X\), there is at least one index \(i\) for which \(x\in E_i\). The hypothesis for that case says that an element of \(E_i\) satisfying \(P\) also satisfies \(Q\). Therefore \(Q(x)\) holds. Since \(x\) was arbitrary among the elements satisfying \(P\), the implication is proved for every \(x\in X\). \(\square\)
The cases in this principle need not be disjoint: overlap causes no logical problem. They must, however, cover all of \(X\). In a proof about a real variable, the cases \(x\geq0\) and \(x<0\) cover \(\mathbb R\), while the cases \(x>0\) and \(x<0\) omit \(x=0\). That omitted boundary value must either be handled separately or included in one of the cases.
Worked Example: Proving a Set Identity by Following an Arbitrary Element
Let \(A,B,C\) be subsets of a set \(X\). We prove
For the inclusion from left to right, take \(x\in A\setminus(B\cap C)\). Then \(x\in A\), and \(x\notin B\cap C\). By the definition of intersection, \(x\notin B\cap C\) means that \(x\notin B\) or \(x\notin C\). If \(x\notin B\), then \(x\in A\setminus B\), so \(x\in(A\setminus B)\cup(A\setminus C)\). If \(x\notin C\), then \(x\in A\setminus C\), so again \(x\in(A\setminus B)\cup(A\setminus C)\). Thus the first inclusion holds.
For the reverse inclusion, take \(x\in(A\setminus B)\cup(A\setminus C)\). By the definition of union, either \(x\in A\setminus B\) or \(x\in A\setminus C\). In the first case, \(x\in A\) and \(x\notin B\); hence \(x\notin B\cap C\), so \(x\in A\setminus(B\cap C)\). In the second case, \(x\in A\) and \(x\notin C\); this also implies \(x\notin B\cap C\), so \(x\in A\setminus(B\cap C)\). Both inclusions are proved, and therefore the sets are equal.
The proof follows the form of the target. Equality of sets requires two inclusions, and each inclusion begins with an arbitrary element of the set on its left. The “or” in the membership condition then calls for a case split. This is a small example of a general planning rule: let the definitions of the target determine the proof’s structure.
Use Cases to Control Absolute Values
Expressions involving absolute values often invite a case split because their formulas depend on sign. A useful example is the function \(f:\mathbb R\to\mathbb R\) defined by \(f(x)=x/(1+|x|)\). The denominator is positive for every real \(x\), so the function is defined everywhere. We will prove a global estimate by separating the cases where two inputs have the same sign and where they have opposite signs.
Worked Example: A Case Split Gives a Global Lipschitz Bound
Theorem: For all \(x,y\in\mathbb R\),
Proof. First suppose \(x\) and \(y\) have the same sign, allowing either input to be zero. Put \(a=|x|\) and \(b=|y|\). If both inputs are nonnegative, then \(x=a\), \(y=b\), and
The denominator is at least \(1\), giving the inequality. If both inputs are nonpositive, then \(x=-a\), \(y=-b\). In this case,
Now suppose the inputs have opposite signs. If \(x\geq0\) and \(y\leq0\), then
Here each denominator is at least \(1\), and \(x-y=x+|y|\). If \(x\leq0\) and \(y\geq0\), then
These cases cover all pairs \((x,y)\): either the inputs have the same sign, or they have opposite signs. In both situations the required inequality follows. Thus \(f\) is globally Lipschitz with constant \(1\). \(\square\)
Notice how the proof avoids trying to manipulate one complicated absolute-value expression all at once. It first chooses cases that make the signs explicit, then uses denominators at least \(1\). The final coverage check matters: without it, an argument might prove the estimate only for nonnegative inputs and leave other pairs untreated.
Intermediate Claims Should Close a Specific Gap
A proof often becomes clearer when divided into stages. An auxiliary claim is useful when it supplies a fact needed later, but it should be neither unrelated nor stronger than necessary. For example, to prove a product is bounded, it may be enough to first establish separate bounds on its factors and then combine them. To prove a set equality, the two inclusions are the natural intermediate goals. To prove a uniform estimate, first identify what will give one constant that works for every permitted input.
One practical test for an intermediate claim is to ask: “If this claim were proved, which exact line of the main argument could use it?” If there is no answer, the claim may be a distraction. If the main argument uses a fact that has not yet been justified, that fact should be stated as a claim and proved before it is used. This makes dependencies visible and prevents an apparently smooth proof from hiding a gap.
Worked Example: A Direct Estimate with a Deliberate Intermediate Step
For every \(x\in\mathbb R\), we claim
The proof has two targets, so it is useful to handle them separately. First, \(x^2\geq0\), and \(1+x^2>0\). Dividing a nonnegative number by a positive number gives
For the upper bound, the intermediate claim is \(x^2<1+x^2\). This follows because \(1>0\), so adding \(x^2\) to both sides gives the strict inequality. Since the denominator \(1+x^2\) is positive, division yields
Both parts hold for every real \(x\), including \(x=0\), where the expression equals \(0\). The proof is short, but its structure is worth noticing: positivity of the denominator justifies the division, and the strict upper bound follows from a separate strict inequality. Naming those steps makes the argument easier to check.
Check the Hypotheses Before You Commit to a Strategy
A proof strategy cannot compensate for a false statement. Before trying several methods, test whether the conclusion is plausible and whether the hypotheses rule out simple counterexamples. This is not a substitute for proof; it is a way to detect when the claim needs an additional assumption or a narrower conclusion.
Worked Example: A Counterexample Detects a Missing Hypothesis
Consider the claim: “Every continuous real-valued function is bounded.” The statement does not specify a compact domain, so test it on \(f:\mathbb R\to\mathbb R\), \(f(x)=x\). This function is continuous, but it is not bounded above. Indeed, for any proposed bound \(M\in\mathbb R\), choose \(x=M+1\). Then \(x>M\), so \(f(x)>M\). Thus continuity alone does not imply boundedness on an arbitrary domain.
The point is not that a contradiction proof or a covering argument failed. Rather, the claim itself is false as stated. Earlier in the course, compactness supplied a route from suitable local bounds to a global bound. The counterexample shows why a domain hypothesis can be essential. When a proposed theorem has a missing condition, the right repair is to add a justified hypothesis or weaken the conclusion—not to force a proof strategy onto a false claim.
A final proof audit can be brief but systematic. Check that every arbitrary object was genuinely arbitrary; every selected witness satisfies all required conditions; every case is covered; every division or inequality uses the needed sign information; and the final statement has exactly the quantifiers and strength of the original target. The Subsequence Strategy, Compactness Strategy, and other earlier methods remain available, but the proof must still explain why their hypotheses apply. In the next tutorial, this same discipline will be used to organize a basic epsilon-\(N\) limit proof.
Check Your Understanding
For each question, identify the proof structure that would make the argument complete.
- In the Finite Case-Splitting Principle, what condition on the sets \(E_1,\ldots,E_m\) ensures that no possible input is omitted?
- Why does proving only one inclusion not suffice to establish equality of two sets?
- In the Lipschitz estimate for \(f(x)=x/(1+|x|)\), which sign cases are used, and why must they cover all pairs of real inputs?
- What intermediate inequality gives the strict upper bound in the estimate for \(x^2/(1+x^2)\), and why is the denominator’s sign relevant?
- How does the function \(f(x)=x\) show that continuity alone is insufficient for boundedness on an arbitrary domain?