Definitions Tell You What a Proof Must Do
The previous tutorial focused on recognizing a proof strategy from the logical shape of a claim. The next step is to turn that recognition into an argument: write the relevant definitions in quantified form, then discharge their obligations in order. A definition is not merely terminology. It specifies exactly what must be shown, including which objects are arbitrary and which may be chosen.
For instance, convergence of a real sequence to \(L\) means \(\forall\varepsilon>0\ \exists N\in\mathbb{N}\ \forall n\geq N,\ |a_n-L|<\varepsilon\). This order matters. We begin with an arbitrary positive \(\varepsilon\), choose one threshold \(N\) that may depend on \(\varepsilon\), and then prove the estimate for every \(n\geq N\). A proof that finds a different threshold for each \(n\), or that chooses \(N\) before knowing \(\varepsilon\), has not met the definition.
This is a planning method, not a replacement for proof. Writing “by the definition” is not enough if the required witness or estimate is missing. The useful work is to make the choices explicit and show why they satisfy the conditions.
Read the Quantifiers as a Proof Plan
Different definitions create different proof tasks. The following table records several common patterns. In each case, the wording of the definition tells you what the proof must produce or check.
| Definition pattern | What to do in a proof | What to verify |
|---|---|---|
| \(\forall x\in A,\ P(x)\) | Fix an arbitrary \(x\in A\). | Derive \(P(x)\) without adding assumptions about \(x\). |
| \(\exists y,\ Q(y)\) | Choose or construct a candidate \(y\). | Show that this candidate satisfies \(Q(y)\). |
| \(\forall\varepsilon>0\ \exists N\ \forall n\geq N,\ P(n,\varepsilon)\) | Fix \(\varepsilon>0\), then choose \(N\). | Prove \(P(n,\varepsilon)\) for every \(n\geq N\). |
| \(\forall x\in U\ \exists\delta>0\ \forall y,\ R(x,y,\delta)\) | Fix \(x\in U\), then choose \(\delta\). | Check the required property for every eligible \(y\). |
| \(\exists M>0\ \forall n,\ |a_n|\leq M\) | Give one bound \(M\). | Show the same \(M\) works for every index. |
A frequent source of errors is choosing objects in the wrong order. In a limit proof, \(N\) can depend on \(\varepsilon\), but it cannot depend on the later index \(n\). In an openness proof, \(\delta\) can depend on the fixed point \(x\), but it must work for every \(y\) sufficiently close to \(x\). In a boundedness proof, the bound must work simultaneously for all terms.
Worked Examples: Carrying Out the Definition
Worked Example: Proving a Sequence Converges
Let \(a_n=(5n-2)/(n+3)\) for \(n\geq1\). We prove from the definition that \(a_n\to5\). Let \(\varepsilon>0\) be arbitrary. First simplify the error:
Choose an integer \(N\in\mathbb{N}\) such that \(N>17/\varepsilon\), which is possible by the Archimedean property of the real numbers. For every \(n\geq N\), we have \(n+3\geq N+3>N>17/\varepsilon\). Since these quantities are positive, taking reciprocals gives \(17/(n+3)<\varepsilon\). Therefore \(|a_n-5|<\varepsilon\) for every \(n\geq N\). We have chosen \(N\) after fixing \(\varepsilon\) and verified the estimate for all later indices, exactly as the definition requires.
Worked Example: Proving an Interval Is Open
A subset \(U\) of \(\mathbb{R}\) is open if for every \(x\in U\) there is a \(\delta>0\) such that \((x-\delta,x+\delta)\subseteq U\). We show directly that \(U=(1,4)\) is open. Fix an arbitrary \(x\in(1,4)\), and choose \(\delta=\min(x-1,4-x)/2\). Both \(x-1\) and \(4-x\) are positive, so \(\delta>0\).
Suppose \(|y-x|<\delta\). Then \(y>x-\delta\) and \(y<x+\delta\). Since \(\delta\leq(x-1)/2<x-1\), we have \(x-\delta>1\), so \(y>1\). Since \(\delta\leq(4-x)/2<4-x\), we have \(x+\delta<4\), so \(y<4\). Thus \(y\in(1,4)\). The point \(x\) was arbitrary, and for each such point we produced a positive \(\delta\) with the required property. Hence \(U\) is open.
Worked Example: Proving a Sequence Is Bounded
Consider \(b_n=(-1)^n/(n+1)\) for \(n\geq1\). A sequence is bounded if there is a real number \(M>0\) such that \(|b_n|\leq M\) for every \(n\). We propose \(M=1\). For each \(n\geq1\), \(|(-1)^n|=1\) and \(n+1\geq2\), so
The same positive bound \(M\) works for every index, so the sequence is bounded. Notice that the proof does not need to find the smallest possible bound. The definition asks for one bound, not the optimal bound.
Definitions Can Prove Properties of Limits
The definition-expansion method is especially effective when a theorem compares objects that satisfy the same definition. For example, if a sequence is said to converge to two numbers, expand both convergence statements. Each gives an estimate for every sufficiently late term. Choose a term late enough for both estimates to hold, and compare the proposed limits through that term.
Proof. Suppose \((a_n)\) converges to both \(L\) and \(M\). We show that \(L=M\). If \(L\neq M\), then \(d=|L-M|>0\). Apply convergence to \(L\) with tolerance \(d/3\): there is an integer \(N_1\) such that \(n\geq N_1\) implies \(|a_n-L|<d/3\). Apply convergence to \(M\) with the same tolerance: there is an integer \(N_2\) such that \(n\geq N_2\) implies \(|a_n-M|<d/3\).
Choose \(n\geq\max(N_1,N_2)\). Both estimates hold for this same term. By the triangle inequality,
This is impossible because \(d>0\), so \(2d/3<d\). The assumption \(L\neq M\) must be false. Therefore \(L=M\), and the limit is unique. \(\square\)
The proof’s key move follows directly from the definitions: choose one index beyond both thresholds. The two estimates then apply to the same \(a_n\), allowing the triangle inequality to connect \(L\) and \(M\). If the proof used two unrelated indices, this comparison would not follow.
Proof. We first establish the inequality needed for the definition-based estimate. For real numbers \(u,v\), the triangle inequality gives \(|u|\leq|u-v|+|v|\), hence \(|u|-|v|\leq|u-v|\). Interchanging \(u\) and \(v\) gives \(|v|-|u|\leq|u-v|\). Together these inequalities imply \(\big||u|-|v|\big|\leq|u-v|\).
Now let \(\varepsilon>0\). Since \(a_n\to L\), there is an \(N\) such that \(n\geq N\) implies \(|a_n-L|<\varepsilon\). For every such \(n\), apply the inequality just proved with \(u=a_n\) and \(v=L\):
This is precisely the definition of \(|a_n|\to|L|\). \(\square\)
Worked Example: Applying the Absolute-Value Limit Theorem
The sequence \(c_n=2-1/n\) converges to \(2\). Indeed, for \(n\geq1\), \(|c_n-2|=1/n\); given \(\varepsilon>0\), choose an integer \(N>1/\varepsilon\), and then \(n\geq N\) gives \(1/n\leq1/N<\varepsilon\). The Absolute-Value Limit Theorem therefore gives \(|c_n|\to|2|=2\). Here \(c_n>0\) for every \(n\geq1\), so \(|c_n|=c_n\) as well. The example shows how expanding a definition can establish a basic limit first, after which a proved theorem transfers it to a related sequence.
Common Pitfalls When Expanding Definitions
A definition can be correctly recalled and still be used incorrectly. The most common problems are failures to respect the order of choices, omissions of universal cases, and estimates that assume more than the hypotheses provide.
- Choosing a witness that depends on a later arbitrary variable. In a convergence proof, \(N\) may depend on \(\varepsilon\), but the proof must then work for every \(n\geq N\).
- Giving a candidate without checking it. If the definition says “there exists,” naming a candidate is only the start; verify every required property.
- Checking one case instead of all cases. A universal statement requires an argument for an arbitrary eligible object. A calculation for one example cannot establish it.
- Using an estimate outside its justified range. If a factor is bounded only near a fixed point, first impose a neighborhood restriction that ensures the bound.
- Proving a stronger-looking but irrelevant statement. Keep the final line tied to the exact definition and conclusion requested.
A useful final audit is to read the proof while pointing to each quantifier in the definition: Where was the arbitrary object fixed? Where was the witness chosen? What shows it is admissible? Where was the property verified for every remaining case? If each question has a specific answer in the argument, the definition has been converted into a proof rather than merely repeated.
Check Your Understanding
For each question, identify the proof obligation created by the relevant definition.
- In the definition of sequence convergence, which quantity is chosen after \(\varepsilon\), and which quantity must remain arbitrary afterward?
- Why must the \(\delta\) in the proof that \((1,4)\) is open work for every \(y\) satisfying \(|y-x|<\delta\)?
- In the uniqueness-of-limits proof, why is it important to choose an index at least as large as both thresholds?
- What two inequalities combine to give \(\big||u|-|v|\big|\leq|u-v|\)?
- For the sequence \((-1)^n/(n+1)\), why does proving \(|b_n|\leq1\) for every \(n\) establish boundedness?