Completeness of Bounded Continuous Function Spaces

The Sup Metric on Bounded Continuous Functions

Completeness asks whether every Cauchy sequence in a metric space converges to a point of that same space. For spaces of functions, the metric matters: the sup metric measures the largest difference between two functions anywhere in their domain. A Cauchy sequence for this metric is uniformly Cauchy, which gives enough control to construct a limit that is both bounded and continuous.

Let \((X,\mathcal{T})\) be a topological space. A function \(f:X\to\mathbb{R}\) is bounded if there is a real number \(M\ge 0\) such that \(|f(x)|\le M\) for every \(x\in X\). It is continuous if the preimage of every open subset of \(\mathbb{R}\) is open in \(X\). Write \(C_b(X)\) for the set of all bounded continuous functions \(f:X\to\mathbb{R}\).

The metric is zero exactly when the functions agree at every point. Symmetry follows from \(|f(x)-g(x)|=|g(x)-f(x)|\). For any bounded \(f,g,h\) and every \(x\in X\), the real triangle inequality gives \(|f(x)-h(x)|\le |f(x)-g(x)|+|g(x)-h(x)|\le d_\infty(f,g)+d_\infty(g,h)\). Taking the supremum over \(x\) proves the triangle inequality for \(d_\infty\).

If \(X=\varnothing\), there is exactly one function from \(X\) to \(\mathbb{R}\), and it is bounded and continuous. In that case \(C_b(X)\) is a one-point metric space, with distance zero, so it is complete. In the rest of the tutorial, assume \(X\ne\varnothing\).

Uniform Limits Preserve Continuity

A sequence of functions \((f_n)\) converges uniformly to \(f\) if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in X\) whenever \(n\ge N\). The same \(N\) must work at every point \(x\). In the sup metric, this is exactly convergence \(d_\infty(f_n,f)\to 0\).

Proof. Fix \(x_0\in X\) and \(\varepsilon>0\). By uniform convergence, choose \(n\in\mathbb{N}\) such that \(|f_n(x)-f(x)|<\varepsilon/3\) for every \(x\in X\). Since \(f_n\) is continuous at \(x_0\), there is an open neighborhood \(U\) of \(x_0\) such that \(|f_n(x)-f_n(x_0)|<\varepsilon/3\) for every \(x\in U\). For such \(x\), the triangle inequality gives \(|f(x)-f(x_0)|\le |f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|<\varepsilon\). Thus the inverse image under \(f\) of each open interval about \(f(x_0)\) contains an open neighborhood of \(x_0\). Since this holds at every \(x_0\), \(f\) is continuous. \(\square\)

This argument uses only the definition of continuity and the uniform bound on the differences. It does not require \(X\) to be a metric space, or to satisfy a separation condition.

Completeness of \(C_b(X)\)

Proof. The case \(X=\varnothing\) was established above, so assume \(X\ne\varnothing\). Let \((f_n)\) be a Cauchy sequence in \(C_b(X)\). By the definition of the sup metric, for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(\sup_{x\in X}|f_n(x)-f_m(x)|<\varepsilon\) whenever \(n,m\ge N\). In particular, for each fixed \(x\in X\), the real sequence \((f_n(x))\) is Cauchy. The real numbers are complete, so this sequence has a limit. Define \(f:X\to\mathbb{R}\) by \(f(x)=\lim_{n\to\infty}f_n(x)\).

We first show that convergence to \(f\) is uniform. Fix \(\varepsilon>0\), and use the Cauchy condition to choose \(N\) so that \(d_\infty(f_n,f_m)<\varepsilon/2\) for all \(n,m\ge N\). Fix \(n\ge N\) and \(x\in X\). For every \(m\ge N\), \(|f_n(x)-f_m(x)|<\varepsilon/2\). Letting \(m\to\infty\) and using \(f_m(x)\to f(x)\) gives \(|f_n(x)-f(x)|\le\varepsilon/2\). This bound holds for every \(x\in X\), so \(\|f_n-f\|_\infty\le\varepsilon/2<\varepsilon\). Thus \(f_n\to f\) uniformly.

Next, \(f\) is bounded. Apply the uniform-convergence conclusion with tolerance \(1\), and choose \(N\) such that \(|f(x)-f_N(x)|\le1/2\) for every \(x\in X\). The function \(f_N\) is bounded, so there is \(M\ge0\) such that \(|f_N(x)|\le M\) for all \(x\). Therefore, for every \(x\in X\), \(|f(x)|\le |f(x)-f_N(x)|+|f_N(x)|\le 1/2+M\). This is a bound independent of \(x\), proving that \(f\) is bounded.

Every \(f_n\) is continuous, and the convergence \(f_n\to f\) is uniform. The proposition therefore implies that \(f\) is continuous. Hence \(f\in C_b(X)\), and the uniform convergence already proved is convergence in the sup metric. Every Cauchy sequence in \(C_b(X)\) converges to an element of \(C_b(X)\), so this metric space is complete. \(\square\)

The proof has two separate closure checks. Uniform convergence keeps the limit continuous, while comparison with one fixed bounded function keeps the limit bounded. The Cauchy condition in the sup metric supplies the uniform convergence needed for both arguments.

A Common Pitfall: Pointwise Convergence Is Not Enough

Continuity need not be preserved by pointwise convergence. Nor does pointwise convergence alone make a sequence Cauchy in the sup metric. The uniform control in the theorem is essential.

For a concrete example, use \(I=[0,1]\) with its usual subspace topology and define \(g_n:I\to\mathbb{R}\) by \(g_n(x)=\min(nx,1)\). Each \(g_n\) is bounded and continuous: it equals \(nx\) on \([0,1/n]\), equals \(1\) on \([1/n,1]\), and the two formulas agree at \(x=1/n\). For \(x=0\), \(g_n(x)=0\) for every \(n\); for any fixed \(x>0\), \(nx\ge1\) once \(n\ge1/x\), so \(g_n(x)=1\) from then on. The pointwise limit is therefore \(g(0)=0\) and \(g(x)=1\) for \(x>0\), which is not continuous at \(0\).

The sequence is not Cauchy in the sup metric. For each \(n\in\mathbb{N}\), take \(x=1/(2n)\), which lies in \(I\). Then \(g_n(x)=\min(1/2,1)=1/2\), while \(g_{2n}(x)=\min(1,1)=1\). Hence \(d_\infty(g_n,g_{2n})\ge1/2\) for every \(n\). Since this distance does not tend to zero, the sequence cannot be Cauchy in that metric.

A tempting but incorrect argument is: “Each function is continuous, so its pointwise limit is continuous.” The calculation above shows why this fails. The valid continuity result uses uniform convergence, and the completeness theorem first establishes that uniform convergence from the Cauchy condition.