Why Limit Laws Need Proof
Convergence is defined by asking that terms eventually lie within every prescribed positive distance of a point. That definition does not immediately say that a sequence cannot approach two different points, or that combining two convergent sequences preserves convergence. The metric axioms supply the estimates that answer both questions. These results let us calculate with limits while keeping track of the hypotheses that justify each calculation.
Let \((X,d)\) be a metric space. A sequence \((x_n)\) in \(X\) converges to \(x\in X\), written \(x_n\to x\), if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). A metric has \(d(u,v)\ge0\), \(d(u,v)=0\) exactly when \(u=v\), symmetry, and the triangle inequality \(d(u,w)\le d(u,v)+d(v,w)\).
Proof. Suppose \(x_n\to x\) and \(x_n\to y\). We show that \(d(x,y)=0\). Let \(\varepsilon>0\). Convergence to \(x\) gives an index \(N_1\) such that \(d(x_n,x)<\varepsilon/2\) for \(n\ge N_1\). Convergence to \(y\) gives an index \(N_2\) such that \(d(x_n,y)<\varepsilon/2\) for \(n\ge N_2\). Choose \(n\ge\max(N_1,N_2)\). The triangle inequality and symmetry give \(d(x,y)\le d(x,x_n)+d(x_n,y)<\varepsilon/2+\varepsilon/2=\varepsilon\). This holds for every \(\varepsilon>0\). A nonnegative real number smaller than every positive real number must be zero, so \(d(x,y)=0\). The identity axiom for a metric now gives \(x=y\). \(\square\)
The choice of \(\varepsilon/2\) matters: the two errors, one from each limit, must add to less than the desired bound. Uniqueness is a statement about limits in metric spaces; it does not say that a sequence must converge in the first place.
Limit Laws in Normed Vector Spaces
To speak of sums and scalar multiples, the space must also have vector operations. Let \(V\) be a vector space over \(\mathbb{R}\) with a norm \(\|\cdot\|\). A norm satisfies \(\|v\|\ge0\), \(\|v\|=0\) exactly when \(v=0\), \(\|cv\|=|c|\|v\|\), and \(\|u+v\|\le\|u\|+\|v\|\). It defines the metric \(d(u,v)=\|u-v\|\). The following limit laws use convergence for this metric; they do not apply to an arbitrary metric space with no vector operations.
Proof. For the sum, let \(\varepsilon>0\). Convergence gives \(N_1,N_2\in\mathbb{N}\) such that \(n\ge N_1\) implies \(\|x_n-x\|<\varepsilon/2\), and \(n\ge N_2\) implies \(\|y_n-y\|<\varepsilon/2\). For \(n\ge\max(N_1,N_2)\), the norm triangle inequality gives \(\|(x_n+y_n)-(x+y)\|=\|(x_n-x)+(y_n-y)\|\le\|x_n-x\|+\|y_n-y\|<\varepsilon\). This proves the sum converges to \(x+y\).
For scalar multiplication, if \(c=0\), then \(cx_n=0=cx\) for every \(n\), so the sequence is constant and converges to \(cx\). If \(c\ne0\), let \(\varepsilon>0\). Choose \(N\) so that \(n\ge N\) implies \(\|x_n-x\|<\varepsilon/|c|\), which is a positive tolerance. Homogeneity of the norm then gives \(\|cx_n-cx\|=|c|\|x_n-x\|<\varepsilon\). Thus \(cx_n\to cx\) in both cases. \(\square\)
The real line with its usual absolute-value norm and Euclidean space \(\mathbb{R}^m\) with its Euclidean norm are examples of normed vector spaces. Subtraction also follows from these laws because \(x_n-y_n=x_n+(-1)y_n\), so \(x_n-y_n\to x-y\).
Worked Example: Sum and Scalar Multiple in \(\mathbb{R}\)
Use the usual metric \(d(s,t)=|s-t|\) on \(\mathbb{R}\). Let \(a_n=(3n-1)/(n+2)\) and \(b_n=(2n+5)/(n+1)\), for \(n\in\mathbb{N}\). Find the limits of \(a_n+b_n\) and \(-2a_n\).
First compare each term with its proposed limit: \(a_n-3=(3n-1-3n-6)/(n+2)=-7/(n+2)\), so \(|a_n-3|=7/(n+2)<7/n\). Also \(b_n-2=(2n+5-2n-2)/(n+1)=3/(n+1)\), so \(|b_n-2|=3/(n+1)<3/n\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>10/\varepsilon\). For \(n\ge N\), both errors are positive and \(|(a_n+b_n)-5|\le |a_n-3|+|b_n-2|<7/n+3/n=10/n\le10/N<\varepsilon\). Thus \(a_n+b_n\to5\).
For the scalar multiple, \(-2a_n=-2(3-7/(n+2))=-6+14/(n+2)\), and hence \(|(-2a_n)-(-6)|=14/(n+2)<14/n\). Given \(\varepsilon>0\), any \(N>14/\varepsilon\) makes this error less than \(\varepsilon\) for \(n\ge N\). Therefore \(-2a_n\to-6\).
Check: Directly combining the fractions gives \(a_n+b_n=(5n^2+11n+9)/(n^2+3n+2)\). Subtracting \(5\) gives \((-4n-1)/(n^2+3n+2)\), which tends to zero; the numerator and denominator here are the result of the displayed fraction calculation. Also, substituting the formula for \(a_n\) into \(-2a_n\) gives \((-6n+2)/(n+2)\); subtracting \(-6\) gives \(14/(n+2)\), as above. Both direct checks agree with the limit laws.
Worked Example: Vector Limits in \(\mathbb{R}^2\)
Use the Euclidean norm \(\|(s,t)\|=\sqrt{s^2+t^2}\) on \(\mathbb{R}^2\). Let \(u_n=(1+2/n,-3+1/n)\) and \(v_n=(4-1/n,2+3/n)\). Compute the limits of their sum and of \(2u_n\).
The distance from \(u_n\) to \(u=(1,-3)\) is \(\|u_n-u\|=\|(2/n,1/n)\|=\sqrt{(2/n)^2+(1/n)^2}=\sqrt{5}/n\). For every \(\varepsilon>0\), choosing \(N>\sqrt{5}/\varepsilon\) makes this distance less than \(\varepsilon\) whenever \(n\ge N\). Thus \(u_n\to(1,-3)\). For \(v=(4,2)\), \(\|v_n-v\|=\|(-1/n,3/n)\|=\sqrt{(-1/n)^2+(3/n)^2}=\sqrt{10}/n\). Choosing \(N>\sqrt{10}/\varepsilon\) proves \(v_n\to(4,2)\).
The sum is exactly \(u_n+v_n=(5+1/n,-1+4/n)\). Its distance to \((5,-1)\) is \(\|(1/n,4/n)\|=\sqrt{1/n^2+16/n^2}=\sqrt{17}/n\), which tends to zero. Hence \(u_n+v_n\to(5,-1)\). Also \(2u_n=(2+4/n,-6+2/n)\), and its distance to \((2,-6)\) is \(\|(4/n,2/n)\|=\sqrt{16/n^2+4/n^2}=\sqrt{20}/n\). Therefore \(2u_n\to(2,-6)\).
Check: At \(n=2\), \(u_2=(2,-5/2)\) and \(v_2=(7/2,7/2)\), so \(u_2+v_2=(11/2,1)\). The difference from \((5,-1)\) is \((1/2,2)\), whose norm is \(\sqrt{1/4+4}=\sqrt{17}/2\), matching \(\sqrt{17}/n\) at \(n=2\). Also \(2u_2=(4,-5)\); its difference from \((2,-6)\) is \((2,1)\), of norm \(\sqrt{5}\), matching \(\sqrt{20}/2=\sqrt{5}\).
Limits of Distances
Distances can be compared even when the metric space has no addition or scalar multiplication. The key estimate follows by applying the triangle inequality in both directions.
Proof. Applying the triangle inequality along the points \(x_n,x,y,y_n\) gives \(d(x_n,y_n)\le d(x_n,x)+d(x,y)+d(y,y_n)\), so \(d(x_n,y_n)-d(x,y)\le d(x_n,x)+d(y_n,y)\), using symmetry for the last term. Interchanging the roles of the pairs \((x_n,y_n)\) and \((x,y)\) gives \(d(x,y)\le d(x,x_n)+d(x_n,y_n)+d(y_n,y)\), and therefore \(d(x,y)-d(x_n,y_n)\le d(x_n,x)+d(y_n,y)\). Together these inequalities yield \(|d(x_n,y_n)-d(x,y)|\le d(x_n,x)+d(y_n,y)\).
Now let \(\varepsilon>0\). By the two convergence assumptions, there are \(N_1,N_2\) such that for \(n\ge N_1\), \(d(x_n,x)<\varepsilon/2\), and for \(n\ge N_2\), \(d(y_n,y)<\varepsilon/2\). For \(n\ge\max(N_1,N_2)\), the displayed estimate gives \(|d(x_n,y_n)-d(x,y)|<\varepsilon\). This is precisely convergence in \(\mathbb{R}\) with its usual metric. \(\square\)
Worked Example: Distances Between Two Real Sequences
In \(\mathbb{R}\) with \(d(s,t)=|s-t|\), let \(x_n=1+1/n\) and \(y_n=4+2/n\). Their limits are \(1\) and \(4\), since \(|x_n-1|=1/n\to0\) and \(|y_n-4|=2/n\to0\). For every \(n\in\mathbb{N}\), \(y_n-x_n=4+2/n-(1+1/n)=3+1/n>0\). Thus \(d(x_n,y_n)=|y_n-x_n|=3+1/n\), which converges to \(3=|4-1|=d(4,1)\).
Check: At \(n=2\), \(x_2=3/2\), \(y_2=5\), and \(d(x_2,y_2)=|5-3/2|=7/2\). The formula \(3+1/n\) gives \(3+1/2=7/2\). The distance between the limits is \(|4-1|=3\), and the difference is \(7/2-3=1/2\), equal to \(1/n\) at \(n=2\).
A Common Pitfall and Why the Estimates Matter
Common pitfall. The distance limit law does not say that \(d(x_n,y_n)\) equals the sum of the distances from the terms to their limits. For the last example, at \(n=2\), the distance between the terms is \(7/2\), whereas the sum of their errors is \(|x_2-1|+|y_2-4|=1/2+1=3/2\). So the incorrect calculation \(d(x_2,y_2)=|x_2-1|+|y_2-4|\) would give \(3/2\), not \(7/2\). The correction is the inequality \(|d(x_n,y_n)-d(x,y)|\le d(x_n,x)+d(y_n,y)\): the errors bound the change in distance; they are not the distance itself.
These laws are useful because they turn convergence statements into controlled calculations. The triangle inequality explains why sums preserve limits, norm homogeneity explains scalar multiplication, and the two-sided distance estimate works in every metric space. In each case the proof identifies exactly what the assumptions provide, rather than relying on informal arithmetic with limits.
Check Your Understanding
Use the definitions and estimates in this tutorial to justify your answers.
- In a metric space, why is choosing \(\varepsilon/2\) for each of two convergent sequences enough to prove the sum law?
- What separate argument proves the scalar-multiple law when the scalar is zero?
- In \(\mathbb{R}\), let \(r_n=2-5/n\) and \(s_n=-1+1/n\). Find the limits of \(r_n+s_n\) and \(-3s_n\).
- For two convergent sequences in a metric space, state an estimate that bounds \(|d(x_n,y_n)-d(x,y)|\).
- Explain why a general metric space need not have a meaningful sum \(x_n+y_n\), even though the distance limit law still applies.