The Supremum Metric on Continuous Functions
Completeness is a property of a metric space: every Cauchy sequence must converge to a point of that same space. For a space of functions, the choice of metric determines what “Cauchy” means. The supremum metric controls the difference between two functions at every point of the interval at once, so its completeness requires the limit to remain continuous.
Let \(I=[0,1]\) with its usual topology, and let \(C([0,1])\) be the set of continuous functions \(f:I\to\mathbb{R}\), where \(\mathbb{R}\) has its standard topology. Since \(I\) is compact, the extreme value theorem implies that every such \(f\) is bounded. Thus the following supremum is finite.
The supremum norm measures the largest absolute difference over the whole interval, not the difference at one selected input. The metric axioms follow from the corresponding properties of absolute value: \(d_\infty(f,g)\geq0\), symmetry holds, and \(d_\infty(f,g)=0\) means \(|f(x)-g(x)|=0\) for every \(x\), hence \(f=g\). The triangle inequality follows because \(|f(x)-h(x)|\leq |f(x)-g(x)|+|g(x)-h(x)|\) at every \(x\); taking suprema gives \(d_\infty(f,h)\leq d_\infty(f,g)+d_\infty(g,h)\).
Worked Example: Distance Between Two Polynomials
Let \(f(x)=x\) and \(g(x)=x^2\) on \(I\). Their distance is the supremum of \(|x-x^2|\). Since \(0\leq x\leq1\), we have \(x-x^2=x(1-x)\geq0\). Completing the square gives
The square is nonnegative, so \(x(1-x)\leq 1/4\), with equality at \(x=1/2\), which belongs to \(I\). Therefore \(d_\infty(f,g)=1/4\). Substitution checks the value: \(|f(1/2)-g(1/2)|=|1/2-1/4|=1/4\), and the completed-square inequality shows no point gives a larger value.
Completeness: Cauchy Sequences Have Continuous Limits
Proof. Let \((f_n)\) be a Cauchy sequence in \(C([0,1])\). By the definition of a Cauchy sequence, for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(m,n\geq N\) implies \(\|f_n-f_m\|_\infty<\varepsilon\). In particular, for every \(x\in I\), \(|f_n(x)-f_m(x)|<\varepsilon\). Thus \((f_n(x))\) is a Cauchy sequence of real numbers for each fixed \(x\). Completeness of \(\mathbb{R}\) gives a limit; define \(f(x)=\lim_{n\to\infty}f_n(x)\).
We next prove convergence uniformly on \(I\). Given \(\varepsilon>0\), choose \(N\) so that \(m,n\geq N\) implies \(\|f_n-f_m\|_\infty<\varepsilon/2\). Fix \(n\geq N\) and \(x\in I\). For every \(m\geq N\), \(|f_n(x)-f_m(x)|<\varepsilon/2\). Taking \(m\to\infty\) in this real inequality yields \(|f_n(x)-f(x)|\leq\varepsilon/2<\varepsilon\). This holds for every \(x\), so taking the supremum gives \(\|f_n-f\|_\infty\leq\varepsilon/2<\varepsilon\). In particular, \(f\) is bounded: for one such \(n\), \(|f(x)|\leq |f_n(x)|+\varepsilon\) for all \(x\), and \(f_n\) is bounded.
It remains to show that \(f\) is continuous. Fix \(x_0\in I\) and \(\varepsilon>0\). The uniform convergence just proved gives an index \(N\) with \(\|f-f_N\|_\infty<\varepsilon/3\). Continuity of \(f_N\) at \(x_0\), using the subspace topology on \(I\), gives \(\delta>0\) such that \(x\in I\) and \(|x-x_0|<\delta\) imply \(|f_N(x)-f_N(x_0)|<\varepsilon/3\). For these \(x\), the triangle inequality yields \[ |f(x)-f(x_0)|\leq |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\varepsilon. \] Therefore \(f\) is continuous at \(x_0\). Since \(x_0\) was arbitrary, \(f\in C([0,1])\), and the uniform convergence already shown is convergence in \(d_\infty\). Hence the metric space is complete. \(\square\)
The proof uses two distinct facts about the limit. Completeness of \(\mathbb{R}\) supplies a pointwise limit, while the supremum Cauchy condition gives uniform convergence and ensures that the limit is continuous. Pointwise convergence alone does not provide that conclusion.
Worked Example: A Cauchy Sequence of Partial Sums
For \(n\geq0\), define \(s_n(x)=\sum_{k=0}^{n}(x/3)^k\) on \(I\). Each \(s_n\) is a polynomial, hence belongs to \(C([0,1])\). If \(m>n\), then \[ s_m(x)-s_n(x)=\sum_{k=n+1}^{m}(x/3)^k. \] Every term is nonnegative and increases as \(x\) increases on \(I\), so the largest difference occurs at \(x=1\). The finite geometric sum gives \[ \|s_m-s_n\|_\infty=\sum_{k=n+1}^{m}3^{-k} =\frac{1}{2\cdot 3^n}\left(1-\frac{1}{3^{m-n}}\right) \leq\frac{1}{2\cdot 3^n}. \] For any \(\varepsilon>0\), choose \(n\) sufficiently large that \(1/(2\cdot3^n)<\varepsilon\). Then every \(m>n\) has \(\|s_m-s_n\|_\infty<\varepsilon\); reversing the indices gives the same bound when \(n>m\). Thus \((s_n)\) is Cauchy.
Its limit is the geometric sum \(s(x)=1/(1-x/3)\). Indeed, for each \(x\in I\), the finite-sum formula gives \[ s(x)-s_n(x)=\frac{(x/3)^{n+1}}{1-x/3}. \] Since \(1-x/3\geq2/3\) and \(x\leq1\), this difference is nonnegative and at most \((1/3)^{n+1}/(2/3)=1/(2\cdot3^n)\). Equality holds at \(x=1\), so \(\|s-s_n\|_\infty=1/(2\cdot3^n)\to0\). Also \(s\) is continuous on \(I\), since its denominator is at least \(2/3\) and never vanishes. This verifies the limit and the supremum-distance calculation.
A Baire-Category Consequence
A subset is dense if every nonempty open set intersects it. A set is nowhere dense if its closure has empty interior; a set is meager if it is a countable union of nowhere-dense sets. A set containing a countable intersection of open dense sets is called residual. The Baire theorem connects completeness to these notions.
We apply it to a specific family in \(C([0,1])\). Let \((q_n)\) enumerate the rational numbers in \(I\), and define \(U_n=\{f\in C([0,1]):f(q_n)\ne0\}\). We verify that each \(U_n\) is open and dense before using the theorem.
Worked Example: Functions Nonzero at Every Rational Input
Fix a rational \(q\in I\), and write \(U_q=\{f:f(q)\ne0\}\). If \(f\in U_q\) and \(g\in C([0,1])\) satisfies \(d_\infty(f,g)<|f(q)|/2\), then \(|g(q)-f(q)|<|f(q)|/2\). The reverse triangle inequality gives \(|g(q)|\geq |f(q)|-|g(q)-f(q)|>|f(q)|/2>0\). Hence the open ball of radius \(|f(q)|/2\) around \(f\) lies in \(U_q\), proving that \(U_q\) is open.
To prove density, take any \(f\in C([0,1])\) and any radius \(r>0\). Choose \(t\) from the two numbers \(r/2\) and \(-r/2\) so that \(t\ne-f(q)\); at most one of these two distinct numbers can equal \(-f(q)\). Let \(g(x)=f(x)+t\). Then \(g\) is continuous, \(d_\infty(f,g)=|t|=r/2<r\), and \(g(q)=f(q)+t\ne0\). Thus every open ball around every \(f\) meets \(U_q\), so \(U_q\) is dense.
Apply the complete-metric Baire theorem to the countable family \((U_n)\). The intersection \(\bigcap_{n=1}^{\infty}U_n\) is dense in \(C([0,1])\). Its members are exactly the continuous functions that are nonzero at every rational point of \(I\). Equivalently, the complement is the union of the closed sets \(C_n=\{f:f(q_n)=0\}\). Each \(C_n\) has empty interior because \(U_n\) is dense, so each is nowhere dense; their union is meager. This gives an explicit residual set and shows what completeness contributes: it turns a countable list of dense open conditions into a single dense intersection.
Common pitfall. Pointwise convergence is not convergence in the supremum metric. Consider \(f_n(x)=x^n\). For each \(x\in[0,1)\), \(f_n(x)\to0\), while \(f_n(1)=1\) for every \(n\); thus the pointwise limit is \(0\) on \([0,1)\) and \(1\) at \(1\). A tempting but incorrect calculation is to say \(\sup_{x\in I}x^n=0\), by looking only at \(x<1\). The endpoint is part of \(I\): substituting \(x=1\) gives \(x^n=1\), and \(x^n\leq1\) throughout \(I\), so \(\|f_n\|_\infty=1\), not \(0\). In fact the pointwise limit is discontinuous at \(1\), and the completeness proof explains why a supremum-metric limit of continuous functions cannot be this discontinuous function.
Check Your Understanding
Use the supremum metric and the definitions of open, dense, and Cauchy in your reasoning.
- Compute the supremum distance between the constant functions \(f(x)=3\) and \(g(x)=-2\).
- For \(f_n(x)=x/n\), determine \(\|f_n\|_\infty\) and the limit in the supremum metric.
- Explain why the pointwise limit of \(x^n\) on \(I\) is not a limit in the supremum metric.
- For a fixed \(q\in I\), prove directly that \(\{f:f(q)=0\}\) is closed in \(C([0,1])\).
- State precisely which completeness and density hypotheses allow the Baire theorem to imply that a countable intersection is dense.