Theorem. Let \(x_n \to x\) and \(y_n \to y\) in the space \(\mathbb{R}\). Then
\[ \begin{array}{c} x_n + y_n \to x + y, \\ x_n - y_n \to x - y, \\ x_n y_n \to xy, \end{array} \]
and provided that each \(y_n \neq 0\) and \(y \neq 0\),
\[ x_n / y_n \to x/y. \]
[Hint: Apply Lemma 21.4; recall from the exercises of §19 that if \(x_n \to x\) and \(y_n \to y\), then \(x_n \times y_n \to x \times y\).]
Solution
If \(F\colon X\to Y\) is continuous and \(z_n\to z\) in \(X\), then \(F(z_n)\to F(z)\) in \(Y\): a neighborhood \(V\) of \(F(z)\) has open preimage containing \(z\), which contains \(z_n\) for all large \(n\).
1. \((x_n,y_n)\to(x,y)\) in \(\mathbb R\times\mathbb R\).
A sequence in a product converges if and only if each coordinate sequence converges, by E-AH7RC.
2. The maps \(S(u,v)=u+v\), \(D(u,v)=u-v\), and \(M(u,v)=uv\) from \(\mathbb R\times\mathbb R\) to \(\mathbb R\), and \(Q(u,v)=u/v\) from \(\mathbb R\times(\mathbb R-\{0\})\) to \(\mathbb R\), are continuous.
For \((u,v)\) near \((a,b)\), \[ \abs{(u\pm v)-(a\pm b)}\le\abs{u-a}+\abs{v-b},\qquad \abs{uv-ab}\le\abs u\abs{v-b}+\abs b\abs{u-a}, \] and the right sides tend to \(0\) as \((u,v)\to(a,b)\). For \(b\ne0\) and \(\abs{v-b}<\abs b/2\), \[ \Bigl\lvert\frac uv-\frac ab\Bigr\rvert=\frac{\abs{bu-av}}{\abs{vb}}\le\frac{2\bigl(\abs b\abs{u-a}+\abs a\abs{v-b}\bigr)}{b^2}, \] which also tends to \(0\).
3. Q.E.D.
By steps 1 and 2 and the sequence property of continuous maps, \(S(x_n,y_n)\to S(x,y)\), \(D(x_n,y_n)\to D(x,y)\), and \(M(x_n,y_n)\to M(x,y)\). When every \(y_n\ne0\) and \(y\ne0\), the sequence \((x_n,y_n)\) lies in \(\mathbb R\times(\mathbb R-\{0\})\) and converges there to \((x,y)\), so \(Q(x_n,y_n)\to Q(x,y)\).