Definitions Before Conclusions
Many errors in metric convergence and continuity arise from applying a familiar-looking statement without checking its quantifiers or hypotheses. A continuity estimate may work at each point without providing one estimate for all points. A sequence may satisfy the Cauchy condition without having a limit in its space. And an expression built from a sequence may converge even when the original sequence does not. Precise definitions help separate these claims.
Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces. A sequence \((x_n)\) in \(X\) converges to \(x\in X\) if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d_X(x_n,x)<\varepsilon\) whenever \(n\ge N\). It is Cauchy if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d_X(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). Convergence compares terms with a point in the space; the Cauchy condition compares terms with one another.
The quantifiers show the difference. Continuity at every point means that \(\delta\) may depend on both \(\varepsilon\) and the point \(a\). Uniform continuity requires a single \(\delta\) for the chosen \(\varepsilon\), working for every pair \(x,y\). Moving a quantifier past another can therefore change the claim.
Point-Dependent Estimates Are Not Uniform Estimates
Worked Example: The Square Function
Consider \(f:\mathbb{R}\to\mathbb{R}\), \(f(x)=x^2\), with the usual metric on \(\mathbb{R}\). At an arbitrary \(a\in\mathbb{R}\), let \(\varepsilon>0\) and choose \(\delta=\min\{1,\varepsilon/(2|a|+1)\}\). If \(|x-a|<\delta\), then \(|x-a|<1\), so \(|x+a|\le |x-a|+2|a|<2|a|+1\). Therefore \(|x^2-a^2|=|x-a||x+a|<\delta(2|a|+1)\le\varepsilon\). This proves continuity at every \(a\).
The displayed choice depends on \(a\), so it does not establish uniform continuity. In fact, \(f\) is not uniformly continuous. Fix \(\varepsilon=1\). For any proposed \(\delta>0\), choose \(n\in\mathbb{N}\) with \(n>1/\delta\), and set \(x=n\), \(y=n+1/n\). Then \(|x-y|=1/n<\delta\), but \[ |f(y)-f(x)|=(n+1/n)^2-n^2=2+1/n^2>1. \] Thus every proposed uniform \(\delta\) fails for this same tolerance \(\varepsilon=1\).
Check: the failure uses two inputs whose distance is strictly less than \(\delta\), while their output distance is strictly greater than \(1\). It disproves uniform continuity without contradicting continuity at any individual point.
A common quantifier error is to find a suitable \(\delta\) for each fixed point and then conclude that one suitable \(\delta\) works across the whole domain. The square function shows why that inference fails. Conversely, when proving continuity at one point, it is enough to provide a \(\delta\) that works at that point; it need not work at all other points.
What Convergence Does and Does Not Guarantee
Proof. Suppose \(x_n\to x\) in \((X,d_X)\). Given \(\varepsilon>0\), convergence supplies \(N\) such that \(d_X(x_n,x)<\varepsilon/2\) for every \(n\ge N\). If \(m,n\ge N\), the triangle inequality gives \(d_X(x_m,x_n)\le d_X(x_m,x)+d_X(x,x_n)<\varepsilon/2+\varepsilon/2=\varepsilon\). Hence \((x_n)\) is Cauchy.
Now suppose \(x_n\to x\) and \(x_n\to y\). If \(x\ne y\), then \(r=d_X(x,y)>0\). Convergence to \(x\) gives an index after which \(d_X(x_n,x)<r/3\); convergence to \(y\) gives an index after which \(d_X(x_n,y)<r/3\). Choose \(n\) at least as large as both indices. The triangle inequality then gives \(r=d_X(x,y)\le d_X(x,x_n)+d_X(x_n,y)<2r/3\), which contradicts \(r>0\). Therefore \(x=y\), proving uniqueness. \(\square\)
The first implication cannot be reversed in every metric space. The distinction is whether the candidate limit belongs to the space. Here is a Cauchy sequence in \(\mathbb{Q}\) that has no limit in \(\mathbb{Q}\).
Worked Example: Cauchy in \(\mathbb{Q}\), Not Convergent There
Use \(\mathbb{Q}\) with the metric \(d(x,y)=|x-y|\). For \(n\in\mathbb{N}\), define \(q_n=\lfloor 10^n\sqrt{2}\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer less than or equal to \(t\). Each \(q_n\) is rational. The defining property of the floor gives \(\lfloor 10^n\sqrt{2}\rfloor\le 10^n\sqrt{2}<\lfloor 10^n\sqrt{2}\rfloor+1\), hence \(0\le\sqrt{2}-q_n<10^{-n}\).
For \(m,n\in\mathbb{N}\), the triangle inequality in \(\mathbb{R}\) now gives \[ |q_m-q_n|\le |q_m-\sqrt{2}|+|q_n-\sqrt{2}|<10^{-m}+10^{-n}. \] Given \(\varepsilon>0\), choose \(N\) so that \(2\cdot 10^{-N}<\varepsilon\). If \(m,n\ge N\), then \(10^{-m}\le10^{-N}\) and \(10^{-n}\le10^{-N}\), so \(|q_m-q_n|<\varepsilon\). Thus \((q_n)\) is Cauchy in \(\mathbb{Q}\).
If it converged to some \(q\in\mathbb{Q}\), it would also converge to \(q\) in \(\mathbb{R}\), since the metric is the same. But the estimate \(|q_n-\sqrt{2}|<10^{-n}\) shows that it converges to \(\sqrt{2}\) in \(\mathbb{R}\): for each \(\varepsilon>0\), choose \(N\) with \(10^{-N}<\varepsilon\), and then \(n\ge N\) implies \(|q_n-\sqrt{2}|<\varepsilon\). Uniqueness of limits in \(\mathbb{R}\) would force \(q=\sqrt{2}\), impossible because \(q\) is rational and \(\sqrt{2}\) is irrational. So the sequence is Cauchy but does not converge in \(\mathbb{Q}\).
Check: the estimate proves Cauchyness using distances between rational terms. The failure of convergence in \(\mathbb{Q}\) is not a failure of that estimate; it is the absence of the required limit point from \(\mathbb{Q}\).
Do Not Infer a Limit Without Checking the Claim
A limit computation needs a stated sequence and a justified argument that it converges to the proposed value. Algebraic transformations can also lose information: a convergent expression derived from a sequence need not imply that the original sequence converges.
Worked Example: Convergent Squares Do Not Force Convergence
In \(\mathbb{R}\) with the usual metric, let \(x_n=(-1)^n\). Then \(x_{2k}=1\) and \(x_{2k+1}=-1\) for every \(k\in\mathbb{N}\). The sequence does not converge. Indeed, if it converged to \(L\), its even-indexed terms would converge to \(L\), but every even-indexed term equals \(1\), so \(L=1\). Its odd-indexed terms would also converge to \(L\), but every odd-indexed term equals \(-1\), so \(L=-1\), a contradiction.
However, \(x_n^2=1\) for every \(n\), so \((x_n^2)\) converges to \(1\): for every \(\varepsilon>0\), \(|x_n^2-1|=0<\varepsilon\) for every \(n\). The invalid inference “the squares converge, therefore the original terms converge” is false. The squaring operation identifies \(1\) and \(-1\), concealing the alternation.
Correction: before applying a limit rule, verify that its hypotheses hold and that its conclusion is the one needed. A computation establishing convergence of \(x_n^2\) establishes a claim about the squared sequence, not automatically about \((x_n)\).
Division gives another frequent trap. A quotient limit rule requires the denominator sequence to converge to a nonzero limit. For example, in \(\mathbb{R}\), set \(a_n=1/n\) and \(b_n=1/n^2\). Both sequences converge to \(0\), but \(b_n\ne0\) for every \(n\), and \[ \frac{a_n}{b_n}=\frac{1/n}{1/n^2}=n. \] This quotient does not converge in \(\mathbb{R}\): for any proposed limit \(L\), choose \(N>|L|+1\); then for every \(n\ge N\), \(|n-L|\ge n-|L|>1\), which prevents convergence to \(L\) with tolerance \(1\). The denominator limit is zero, so the nonzero-limit hypothesis for the quotient rule is absent.
A Practical Error Check
When reviewing a proof about a metric limit or continuity, ask the following questions before accepting its conclusion:
- Are the quantifiers in the right order? For uniform continuity, the chosen \(\delta\) must work for all input pairs.
- Does the proposed \(\delta\) depend on a point? If so, it may establish pointwise continuity, but not uniform continuity.
- Is the sequence shown to converge, or only shown to be Cauchy? In an incomplete space, that distinction matters.
- Does a limit rule have the necessary hypotheses, such as a nonzero limit for a denominator?
- Does the calculation prove convergence of the original sequence, or only of a transformed sequence?
These checks do not replace a proof. They identify exactly what the proof must establish and where a counterexample may expose a missing hypothesis.
Check Your Understanding
For each question, state the relevant quantifiers or hypothesis before giving your argument.
- In the proof that \(f(x)=x^2\) is continuous at \(a\), which part of the chosen \(\delta\) may depend on \(a\)? Why does that choice alone not prove uniform continuity?
- Prove directly from the definition that every convergent sequence in a metric space is Cauchy.
- Can a Cauchy sequence in a metric space have two different limits? Identify the metric property that rules this out.
- Why does the sequence \((1/n)\) in \(\mathbb{Q}\) not provide a counterexample to “Cauchy implies convergent” in a complete metric space?
- Give a sequence \((x_n)\) for which \((x_n^2)\) converges but \((x_n)\) does not, and verify both claims.
- What additional condition is needed before using a quotient limit rule, and what goes wrong for \(a_n=1/n\), \(b_n=1/n^2\)?