$C([0,1],\mathbb{R})$ is complete, and on it the uniform metric and the supremum metric induce the same topology

Let $I:=[0,1]$ carry the metric $d(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ inherited from $\mathbb{R}$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset) and let $\mathbb{R}$ carry the same metric. Write $C(I,\mathbb{R})$ for the continuous real functions on $I$, $\bar{\rho }$ for the uniform metric of For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$ and $d_{\mathrm{\infty }}$ for the supremum metric of The supremum metric $d_{\mathrm{\infty }}(f,g)={\sup }_x\mathrm{\mid }f(x)-g(x)\mathrm{\mid }$ is a metric on the bounded real-valued functions on a nonempty set. Then:

1. every $f\in C(I,\mathbb{R})$ is bounded, so $d_{\mathrm{\infty }}$ is defined on $C(I,\mathbb{R})$;
2. $\bar{\rho }$ and $d_{\mathrm{\infty }}$ are uniformly equivalent on $C(I,\mathbb{R})$, hence induce the same topology there (Topologically, uniformly and Lipschitz equivalent metrics on a set, Lipschitz equivalence implies uniform equivalence implies topological equivalence);
3. $(C(I,\mathbb{R}),\bar{\rho })$ is a complete metric space (Complete metric space: every Cauchy sequence converges in the space).

Claim 2 is this page's guarantee that no second notion of convergence has been created. For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar{\rho }(f,g)={\sup }_x\min \{d(f(x),g(x)),1\}$ is a metric on $Y^X$ mints a metric on $Y^X$ that is not the published supremum metric — it truncates distances at $1$ and needs no boundedness hypothesis — and a reader who has met $d_{\mathrm{\infty }}$ first is entitled to ask whether "uniform convergence" now means two things. On the set where both are defined it does not: the two metrics take different values but have the same open sets, so they have the same convergent sequences, the same continuous functions and the same closed sets.

• The two metrics really are different functions. Take $f$ constant $0$ and $g$ constant $2$: then $d_{\mathrm{\infty }}(f,g)=2$ while $\bar{\rho }(f,g)=1$. What claim 2 says is that this difference is invisible to the topology, not that it does not exist. In particular an assertion about the value of the distance — a diameter, a Lipschitz constant, a radius — must name which metric it means.

• Completeness is inherited from $\mathbb{R}$ and from closedness, in that order. $\mathbb{R}$ is complete, so all the real functions on $I$ are complete in the uniform metric; the continuous ones form a closed subset by the uniform limit theorem (A uniform limit of continuous functions is continuous, so $C(X,Y)$ is closed in $Y^X$ under the uniform metric); and a closed subset of a complete space is complete. Completeness in $d_{\mathrm{\infty }}$ follows as well, since uniformly equivalent metrics have the same Cauchy sequences and the same convergent sequences, both conditions being expressed with $\epsilon$ and $\delta$ alone.

• Compactness of $I$ is used only for claim 1. Boundedness of every continuous function is what makes $d_{\mathrm{\infty }}$ defined at all, and that is the extreme value theorem. On a non-compact domain the supremum metric is unavailable on all of $C(X,\mathbb{R})$, while the uniform metric $\bar{\rho }$ remains defined; that is the whole reason this page mints the truncated metric.