Comparing Terms Instead of Choosing a Limit
Convergence in a norm measures each term's distance from a proposed limit. A Cauchy sequence is defined differently: its terms become close to one another, without first specifying any limit in the space. This distinction matters in function spaces, where it may be straightforward to estimate the distance between two approximations even when no candidate limit has yet been identified.
Let \(V\) be a real vector space with norm \(N\), and let \((x_n)\) be a sequence in \(V\). The Cauchy condition concerns every pair of sufficiently late terms. Both indices must be large; it is not enough to compare the terms only with one fixed term.
The index \(n_0\) may depend on \(\varepsilon\), but it must work for every pair \(m,n\geq n_0\). There is no proposed \(x\) in this definition, and the sequence need not have a limit in \(V\). As with convergence, changing finitely many initial terms does not affect whether the sequence is Cauchy: one can choose a later index that excludes all those terms from the tail.
Convergent Sequences Are Cauchy
Norm convergence gives a direct way to prove the Cauchy condition. If two terms are both close to the same limit, the triangle inequality makes them close to each other. The limit supplies the comparison point, but it does not appear in the definition of Cauchy behavior.
Proof. Suppose \(x_n\to x\) in norm, and let \(\varepsilon>0\). By convergence, there is an index \(n_0\) such that $$ k\geq n_0\quad\Longrightarrow\quad N(x_k-x)<\frac{\varepsilon}{2}. $$ For any \(m,n\geq n_0\), the triangle inequality gives $$ N(x_n-x_m) =N\bigl((x_n-x)+(x-x_m)\bigr) \leq N(x_n-x)+N(x_m-x) <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ Thus every pair of terms beyond \(n_0\) is less than \(\varepsilon\) apart, which is precisely the Cauchy condition. \(\square\)
The converse is a different question. To infer a limit from the Cauchy condition, one needs to know that the space contains the limit that the sequence is approaching. The next tutorial studies the spaces in which this is guaranteed. For now, the theorem gives a necessary condition for norm convergence, not a general guarantee of convergence.
Worked Examples in Normed Spaces
Worked Example: A Cauchy Sequence in the Plane
In \(\mathbb{R}^2\) with the Euclidean norm, define $$ x_n=\left(2+\frac{(-1)^n}{n},\ 1-\frac{1}{n}\right). $$ For any positive integers \(m,n\), subtracting coordinates gives $$ x_n-x_m =\left(\frac{(-1)^n}{n}-\frac{(-1)^m}{m},\ \frac{1}{m}-\frac{1}{n}\right). $$ The absolute value of the first coordinate is at most \(1/n+1/m\), because \(|(-1)^k|=1\) for every integer \(k\). The absolute value of the second coordinate is also at most \(1/n+1/m\). Since the Euclidean norm of a vector \((a,b)\) is at most \(|a|+|b|\), we obtain $$ \|x_n-x_m\|_2 \leq 2\left(\frac{1}{n}+\frac{1}{m}\right). $$ Given \(\varepsilon>0\), choose an integer \(n_0>4/\varepsilon\). If \(m,n\geq n_0\), then $$ \|x_n-x_m\|_2 \leq 2\left(\frac{1}{n}+\frac{1}{m}\right) \leq\frac{4}{n_0} <\varepsilon. $$ Therefore \((x_n)\) is Cauchy. This verifies the pairwise condition directly, without needing to begin by naming a limit.
Worked Example: Partial Sums in the Supremum Norm
On \(C[0,1]\), use the supremum norm and define $$ f_n(t)=\sum_{k=0}^{n}\left(\frac{t}{3}\right)^k,\qquad 0\leq t\leq1. $$ For \(m>n\), the difference of the two polynomials is $$ f_m(t)-f_n(t)=\sum_{k=n+1}^{m}\left(\frac{t}{3}\right)^k. $$ Every term in this sum is nonnegative, and \(0\leq t/3\leq1/3\). Hence, for every \(t\in[0,1]\), $$ |f_m(t)-f_n(t)| \leq\sum_{k=n+1}^{m}\left(\frac{1}{3}\right)^k \leq\sum_{k=n+1}^{\infty}\left(\frac{1}{3}\right)^k =\frac{(1/3)^{n+1}}{1-1/3} =\frac{1}{2\cdot3^n}. $$ Taking the supremum over \(t\) preserves this bound. If \(n>m\), interchange the indices to get the same bound with the smaller index in the exponent. Thus, for all \(m,n\geq n_0\), $$ \|f_m-f_n\|_\infty\leq\frac{1}{2\cdot3^{n_0}}. $$ Given \(\varepsilon>0\), choose \(n_0\) large enough that \(1/(2\cdot3^{n_0})<\varepsilon\). The sequence \((f_n)\) is Cauchy in the supremum norm. The key estimate controls the entire interval at once, as required by this norm.
Worked Example: A Bounded Sequence That Is Not Cauchy
In \(C[0,1]\) with the supremum norm, let \(g_n(t)=(-1)^n\) for every \(t\in[0,1]\). Each \(g_n\) has norm \(1\), so the sequence is bounded in norm. But whenever \(m\) and \(n\) have opposite parity, \(g_n-g_m\) is the constant function \(2\) or \(-2\). In either case, $$ \|g_n-g_m\|_\infty=2. $$ For every proposed index \(n_0\), there are indices \(m,n\geq n_0\) with opposite parity. Taking, for example, \(\varepsilon=1\), the Cauchy condition fails for that pair. Thus boundedness alone does not imply that a sequence is Cauchy.
Cauchy Sequences Are Bounded
Although boundedness does not imply the Cauchy condition, every Cauchy sequence in a normed space is bounded. The proof uses the definition with one fixed positive tolerance to control the tail, then accounts for the finitely many earlier terms separately.
Proof. Let \((x_n)\) be Cauchy. Apply the definition with \(\varepsilon=1\). There is an index \(n_0\) such that $$ m,n\geq n_0\quad\Longrightarrow\quad N(x_n-x_m)<1. $$ Fix \(m=n_0\). For every \(n\geq n_0\), the triangle inequality yields $$ N(x_n)\leq N(x_n-x_{n_0})+N(x_{n_0})<1+N(x_{n_0}). $$ The terms before \(n_0\) form a finite set, so their norms have a finite maximum (if there are no such terms, no additional bound is needed). Choose \(M\) at least as large as that maximum and \(1+N(x_{n_0})\). Then \(N(x_n)\leq M\) for every \(n\), as required. \(\square\)
The estimate uses the tail condition only after fixing one term in the tail as a reference. It does not claim that the distances between all terms are small; early terms may be far from one another. Their norms are nevertheless finite in number, so they can be included in a single bound.
Operations and Norms of Cauchy Sequences
The same estimates that are useful for norm convergence also apply when comparing pairs of terms. In particular, addition and fixed scalar multiplication preserve the Cauchy property.
Proof. Let \(\varepsilon>0\). Since \((x_n)\) and \((y_n)\) are Cauchy, there are indices \(N_x\) and \(N_y\) such that $$ m,n\geq N_x\Longrightarrow N(x_n-x_m)<\frac{\varepsilon}{2}, \qquad m,n\geq N_y\Longrightarrow N(y_n-y_m)<\frac{\varepsilon}{2}. $$ For \(m,n\geq\max(N_x,N_y)\), the triangle inequality gives $$ N\bigl((x_n+y_n)-(x_m+y_m)\bigr) \leq N(x_n-x_m)+N(y_n-y_m) <\varepsilon. $$ Thus the sum sequence is Cauchy. For scalar multiplication, if \(c=0\), the sequence \((cx_n)\) is constantly zero. If \(c\neq0\), choose \(N_x\) so that \(N(x_n-x_m)<\varepsilon/|c|\) whenever \(m,n\geq N_x\). Homogeneity then gives $$ N(cx_n-cx_m)=|c|N(x_n-x_m)<\varepsilon. $$ Therefore \((cx_n)\) is Cauchy as well. \(\square\)
A related consequence connects a vector-valued Cauchy sequence to an ordinary real sequence. The Reverse Triangle Inequality for Norms, established in Norm Axioms, states that \(|N(u)-N(v)|\leq N(u-v)\). Applying it to two terms gives $$ |N(x_n)-N(x_m)|\leq N(x_n-x_m). $$ Thus, if \((x_n)\) is Cauchy in \(V\), then \((N(x_n))\) is a Cauchy sequence of real numbers: for any \(\varepsilon>0\), an index that makes the right-hand side less than \(\varepsilon\) also makes the left-hand side less than \(\varepsilon\). This conclusion concerns the scalar sizes of the terms; it does not establish that the terms themselves have a limit in \(V\).
What the Cauchy Condition Does—and Does Not—Say
The Cauchy condition is an internal test: it asks whether terms approach one another, not whether a particular candidate is their limit. In a normed function space, it often lets us estimate differences between approximations without yet knowing what function, if any, they approach. The partial-sum example used exactly this strategy by bounding the remaining terms uniformly over the whole domain.
Two common errors are to check only a few pairs of terms, or to check distances from a fixed term rather than controlling all pairs in a sufficiently late tail. The quantifiers require a single index \(n_0\) for each \(\varepsilon\), after which the estimate must hold for every \(m,n\geq n_0\). A second error is to infer convergence just from being Cauchy. The definition gives no element of \(V\) as a limit, and whether such a limit exists depends on the space. Completeness, the subject of the next tutorial, supplies precisely the additional property needed to guarantee convergence for every Cauchy sequence.
Fix an arbitrary \(\varepsilon>0\), as required by the definition.
Estimate \(N(x_n-x_m)\) using both indices, often by a triangle inequality or a tail bound.
Ensure the estimate is less than \(\varepsilon\) for every pair \(m,n\) beyond that index.
Check Your Understanding
Use the definition and results in this tutorial to answer the following questions.
- What must one show, for each \(\varepsilon>0\), to prove that \((x_n)\) is Cauchy?
- In the proof that norm convergence implies the Cauchy property, why are both distances to the limit bounded by \(\varepsilon/2\)?
- Why does the proof that a Cauchy sequence is bounded treat the early terms separately from the tail?
- Can a bounded sequence fail to be Cauchy? Give the example from this tutorial and identify the distance that prevents the Cauchy condition.
- What does the Reverse Triangle Inequality imply about the real sequence \((N(x_n))\) when \((x_n)\) is Cauchy?
- Does the Cauchy definition itself guarantee a limit in \(V\)? What additional topic addresses that question?