How a Cauchy Sequence Detects a Missing Point
A metric space is incomplete when at least one Cauchy sequence in it has no limit in that space. A practical way to find such a sequence is to place the space inside a larger metric space, construct a sequence that approaches a point missing from the smaller space, and check two things separately: its terms are Cauchy, and it cannot converge to any point that the smaller space does contain.
For a metric space \((X,d)\), a sequence \((x_n)\) in \(X\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). It converges in \(X\) to \(x\in X\) if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). The space is complete if every Cauchy sequence in it converges to a point of that space.
Suppose \(A\subseteq Y\), where \((Y,d)\) is a metric space. We give \(A\) the subspace metric \(d_A(a,b)=d(a,b)\) for \(a,b\in A\). Thus all distances between terms of a sequence in \(A\) are exactly the same whether calculated in \(A\) or in \(Y\). The distinction is that a limit in \(Y\) need not belong to \(A\).
Proof. The sequence is Cauchy in \(A\), because \(d_A(x_m,x_n)=d(x_m,x_n)\) for every \(m,n\). Suppose, for a contradiction, that it converges in \(A\) to some \(a\in A\). Then it also satisfies \(d(x_n,a)\to0\), while the assumed convergence in \(Y\) gives \(d(x_n,y)\to0\). The triangle inequality gives \(d(a,y)\le d(a,x_n)+d(x_n,y)\) for every \(n\). Given \(\varepsilon>0\), for all sufficiently large \(n\) both terms on the right are less than \(\varepsilon/2\), so \(d(a,y)<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(d(a,y)=0\), and the metric axiom implies \(a=y\). This contradicts \(a\in A\) and \(y\notin A\). Therefore the Cauchy sequence has no limit in \(A\), so \(A\) is incomplete. \(\square\)
Rational Approximations to an Irrational Number
Worked Example: A Cauchy Sequence in \(\mathbb{Q}\) with No Rational Limit
Give \(\mathbb{Q}\) the metric inherited from \(\mathbb{R}\) with its standard metric \(d(x,y)=|x-y|\). We construct rational decimal truncations of \(\sqrt{2}\). For each \(n\in\mathbb{N}\), let \(k_n\) be the integer satisfying \(k_n\le 10^n\sqrt{2}<k_n+1\), and set \(q_n=k_n/10^n\). Each \(q_n\) is rational. Since \(\sqrt{2}\) is irrational, the first inequality is strict, and hence \(0<\sqrt{2}-q_n<10^{-n}\).
The first terms can be checked directly. We have \(1.4^2=1.96<2<2.25=1.5^2\), so \(q_1=1.4=7/5\). Next, \(1.41^2=1.9881<2<2.0164=1.42^2\), so \(q_2=1.41=141/100\). Finally, \(1.414^2=1.999396<2<2.002225=1.415^2\), so \(q_3=1.414=707/500\).
To prove the sequence is Cauchy, let \(\varepsilon>0\). Choose \(N\in\mathbb{N}\) so that \(2\cdot10^{-N}<\varepsilon\). If \(m,n\ge N\), the triangle inequality and the truncation bounds give \(|q_m-q_n|\le|\sqrt{2}-q_m|+|\sqrt{2}-q_n|<10^{-m}+10^{-n}\le2\cdot10^{-N}<\varepsilon\). This is the Cauchy condition, with both indices beyond the same \(N\).
The same bound shows that \(q_n\to\sqrt{2}\) in \(\mathbb{R}\): for any \(\varepsilon>0\), choose \(N\) with \(10^{-N}<\varepsilon\), and then \(n\ge N\) implies \(|q_n-\sqrt{2}|<10^{-n}\le10^{-N}<\varepsilon\).
It remains to show that no rational number can be the limit in \(\mathbb{Q}\). If \(q_n\) converged in \(\mathbb{Q}\) to \(r\), then \(r\) would also be a real limit. The triangle inequality gives \(|r-\sqrt{2}|\le|r-q_n|+|q_n-\sqrt{2}|\), and both terms on the right tend to zero; therefore \(r=\sqrt{2}\). But \(\sqrt{2}\notin\mathbb{Q}\). Indeed, if \(\sqrt{2}=a/b\) for relatively prime positive integers \(a,b\), then \(a^2=2b^2\), so \(a\) is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is even as well, contradicting that \(a,b\) are relatively prime. Thus \((q_n)\) is Cauchy in \(\mathbb{Q}\) and has no limit there.
Check: the sequence terms are rational, its Cauchy estimate uses only distances between those terms, and its real limit is the irrational number \(\sqrt{2}\). The obstruction is membership of the limit, not failure of the sequence to approach a point.
Sequences Approaching Missing Points
Worked Example: A Missing Endpoint of an Interval
Let \(X=(0,1]\subseteq\mathbb{R}\), with the subspace metric inherited from the standard metric on \(\mathbb{R}\). Take \(x_n=1/n\). Every term belongs to \(X\), since \(n\ge1\) implies \(0<1/n\le1\).
Let \(\varepsilon>0\), and choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). If \(m,n\ge N\), then \(0<1/m\le1/N\) and \(0<1/n\le1/N\). Consequently, \(|x_m-x_n|\le1/N<\varepsilon\). So \((x_n)\) is Cauchy in \(X\). Also, \(x_n\to0\) in \(\mathbb{R}\), because for \(n\ge N\) we have \(|x_n-0|=1/n\le1/N\).
The point \(0\) is not in \(X\), but we must still rule out a different limit in \(X\). If \(x_n\) converged in \(X\) to \(x\in X\), then for every \(n\), \(|x|\le|x-x_n|+|x_n|=|x-x_n|+1/n\). Both terms on the right can be made arbitrarily small, so \(|x|=0\) and \(x=0\), contradicting \(x\in(0,1]\). Therefore \(X\) is incomplete.
Check: \(x_1=1\) is allowed, and all later terms are positive as well. The omitted endpoint is not itself a term of the sequence; it is the real limit that the terms approach.
Worked Example: A Missing Point Inside the Real Line
Let \(X=\mathbb{R}\setminus\{2\}\), again with the subspace metric from the standard metric on \(\mathbb{R}\). Define \(x_n=2+1/n\). Since \(1/n>0\), each \(x_n>2\), so every term belongs to \(X\). For example, \(x_1=3\), \(x_2=5/2\), and \(x_3=7/3\).
For \(m,n\in\mathbb{N}\), \(|x_m-x_n|=|1/m-1/n|\). If \(m,n\ge N\), then both \(1/m\) and \(1/n\) lie in \((0,1/N]\), so their difference in absolute value is at most \(1/N\). Given \(\varepsilon>0\), choose \(N\) with \(1/N<\varepsilon\). Then \(|x_m-x_n|\le1/N<\varepsilon\), proving that \((x_n)\) is Cauchy.
In the ambient real line, \(|x_n-2|=1/n\to0\), so \(x_n\to2\). Suppose instead that \(x_n\) converged in \(X\) to \(x\in X\). The triangle inequality would give \(|x-2|\le|x-x_n|+|x_n-2|=|x-x_n|+1/n\). The right side tends to zero, forcing \(x=2\), which is excluded from \(X\). Thus \(X\) is incomplete. Unlike the preceding example, the missing point is an interior point of the ambient real line, not an endpoint of an interval.
Check: for the first two terms, \(|x_1-x_2|=|3-5/2|=1/2\), and the general difference estimate above proves the required condition for every pair of sufficiently late terms, not just this one pair.
A Common Pitfall: An Ambient Limit Is Not Enough
A common incomplete argument says, “The sequence approaches a point outside the space, so it has no limit in the space.” The conclusion may be right, but the reasoning skips two checks. First, the sequence must be Cauchy in the metric under discussion. Second, one must exclude every possible limit in the space, rather than merely observe that the ambient limit is absent. The ambient-limit test supplies the second step through the triangle inequality, but it applies only after the Cauchy condition has been established.
For example, in \(X=(0,1]\), it would be insufficient to say only that \(1/n\to0\) in \(\mathbb{R}\) and \(0\notin X\). That statement alone does not verify the Cauchy condition. The required calculation is: for \(m,n\ge N\), both \(1/m\) and \(1/n\) lie in \((0,1/N]\), hence \(|1/m-1/n|\le1/N\); choosing \(N\) with \(1/N<\varepsilon\) proves the condition. Then the triangle-inequality argument rules out a different limit in \(X\).
The method is useful because it turns incompleteness into a concrete construction. Look for points the space omits, choose terms in the space whose distances shrink as they approach such a point, and verify the Cauchy estimate using only distances between terms. A sequence’s limit must be a point of the space for completeness; a limit in a larger ambient space does not satisfy that requirement.
Check Your Understanding
For each question, state the metric being used and distinguish the Cauchy calculation from the argument about possible limits.
- In \((0,1]\) with the inherited Euclidean metric, verify directly that \(x_n=1/(n+1)\) is Cauchy and determine whether it has a limit in the space.
- For \(X=\mathbb{R}\setminus\{0\}\) with the inherited Euclidean metric, use \(x_n=(-1)^n/n\) to test whether this sequence is Cauchy and whether it converges in \(X\).
- Explain why the rational truncations of \(\sqrt{2}\) have rational terms even though their real limit is irrational.
- Suppose \(A\subseteq Y\) has the inherited metric and a sequence in \(A\) converges in \(Y\) to a point of \(A\). What additional condition would show that this sequence witnesses incompleteness of \(A\)?
- In the ambient-limit test, where exactly is the triangle inequality used to rule out a second limit?