Scalar-valued continuous functions are stable under the usual algebraic operations. The focus keyword continuous scalar functions refers to complex- or real-valued mappings whose continuity is preserved by sums, scalar multiples, products, moduli, and reciprocals away from zero. This algebra is used constantly when constructing examples and operators in metric spaces. :::definition[Algebraic Operations on Functions] Let $X$ be nonempty, let $f,g:X\to\mathbb C$, and let $\alpha\in\mathbb C$. Define $(f+g)(x)=f(x)+g(x)$, $(\alpha f)(x)=\alpha f(x)$, $(fg)(x)=f(x)g(x)$, and $|f|(x)=|f(x)|$. If $f(x)\neq0$ for every $x\in X$, define $(1/f)(x)=1/f(x)$. ::: :::theorem[Algebra of Continuous Scalar-Valued Functions] Let $(X,d)$ be a metric space, and let $f,g:X\to\mathbb C$ be continuous. Then $f+g$, $\alpha f$, $fg$, and $|f|$ are continuous. If $f(x)\neq0$ for every $x\in X$, then $1/f$ is continuous. ::: :::proof Given that $f$ and $g$ are continuous. To prove the algebraic permanence properties. Let $x_n\to x$ in $X$. Then $f(x_n)\to f(x)$ and $g(x_n)\to g(x)$. The algebra of limits in $\mathbb C$ gives convergence of sums, scalar multiples, and products. The modulus map is continuous, so $|f(x_n)|\to |f(x)|$. If $f$ has no zeros, then reciprocals converge as $1/f(x_n)\to1/f(x)$. Hence, by the sequential criterion, all stated functions are continuous. $\square$ ::: :::example[Complex Polynomial-Type Function] The function $f:\mathbb R\to\mathbb C$ defined by $f(x)=x+ix^2$ is continuous because it is formed from continuous scalar functions by addition and multiplication. ::: :::example[Dirichlet-Type Function] The function $f(x)=1$ for $x\in\mathbb Q$ and $f(x)=0$ for $x\notin\mathbb Q$ is continuous nowhere on $\mathbb R$. ::: :::proof Given any $a\in\mathbb R$. To prove discontinuity at $a$. Choose rational $r_n\to a$ and irrational $s_n\to a$. Then $f(r_n)=1$ and $f(s_n)=0$ for all $n$. The two image sequences cannot both converge to $f(a)$. Hence $f$ is not continuous at $a$. Since $a$ was arbitrary, $f$ is continuous nowhere. $\square$ ::: :::example[Point Evaluation Functional] On $C[0,1]$ with the uniform metric, the functional $F(x)=x(0)$ is continuous because $|x_n(0)-x(0)|\leq\sup_{0\leq t\leq1}|x_n(t)-x(t)|$. ::: :::example[Integral Functional] On $C[0,1]$ with the uniform metric, the functional $F(x)=\int_0^1x(t)\,dt$ is continuous because $|F(x_n)-F(x)|\leq\sup_{0\leq t\leq1}|x_n(t)-x(t)|$. ::: :::example[Distance from a Fixed Point] For a fixed $x_0\in X$, the function $x\mapsto d(x,x_0)$ is continuous since $|d(x,x_0)-d(y,x_0)|\leq d(x,y)$. ::: :::exercise 1. Prove $f^2$ is continuous if $f$ is continuous. 2. Explain why $1/f$ requires $f$ to avoid zero. 3. Show $x\mapsto|x^2-1|$ is continuous. ::: :::answer 1. Use the product theorem with $g=f$. 2. Division by zero is undefined. 3. Polynomials and modulus are continuous. :::