[proofplan] We exhibit a concrete homeomorphism between $\mathbb{R}$ (which is complete) and $(0,1)$ (which is not complete), thereby showing that completeness is not preserved by topological equivalence. The logistic sigmoid $f(x) = 1/(1 + e^{-x})$ serves as the homeomorphism, with its inverse given by $f^{-1}(y) = \ln(y/(1-y))$. [/proofplan] [step:Construct a homeomorphism between $\mathbb{R}$ and $(0,1)$] Define $f: \mathbb{R} \to (0,1)$ by \begin{align*} f(x) = \frac{1}{1 + e^{-x}}. \end{align*} The [function](/page/Function) $f$ is differentiable with $f'(x) = e^{-x}/(1 + e^{-x})^2 > 0$ for all $x \in \mathbb{R}$, so $f$ is strictly increasing and hence injective. As $x \to -\infty$, $e^{-x} \to \infty$ gives $f(x) \to 0^+$; as $x \to +\infty$, $e^{-x} \to 0$ gives $f(x) \to 1^-$. By the [intermediate value theorem](/theorems/629), $f$ maps $\mathbb{R}$ onto $(0,1)$. The inverse is $f^{-1}: (0,1) \to \mathbb{R}$ given by $f^{-1}(y) = \ln(y/(1-y))$, which is continuous on $(0,1)$ as a composition of the continuous maps $y \mapsto y/(1-y)$ and $\ln$. Since both $f$ and $f^{-1}$ are continuous bijections, $f$ is a homeomorphism. [guided] We need a bijection between $\mathbb{R}$ and $(0,1)$ that is continuous in both directions. The logistic sigmoid is a standard choice, but any strictly monotone continuous function from $\mathbb{R}$ onto $(0,1)$ would work (e.g., $(2/\pi)\arctan(x) + 1/2$). Define $f(x) = 1/(1 + e^{-x})$. Why is $f$ a homeomorphism? **Injectivity**: $f'(x) = e^{-x}/(1 + e^{-x})^2$. Since $e^{-x} > 0$ and $(1 + e^{-x})^2 > 0$, the [derivative](/page/Derivative) is strictly positive everywhere. A function with strictly positive derivative is strictly increasing, hence injective. **Surjectivity onto $(0,1)$**: as $x \to -\infty$, $e^{-x} \to +\infty$, so $f(x) = 1/(1 + e^{-x}) \to 0^+$. As $x \to +\infty$, $e^{-x} \to 0$, so $f(x) \to 1^-$. Since $f$ is continuous and strictly increasing, the intermediate value theorem guarantees that $f$ takes every value in $(0,1)$. **Continuous inverse**: solving $y = 1/(1 + e^{-x})$ for $x$ gives $1 + e^{-x} = 1/y$, so $e^{-x} = (1-y)/y$, hence $x = -\ln((1-y)/y) = \ln(y/(1-y))$. The function $y \mapsto \ln(y/(1-y))$ is a composition of continuous functions on $(0,1)$ (the quotient $y/(1-y)$ is continuous and positive on $(0,1)$, and $\ln$ is continuous on $(0,\infty)$), so $f^{-1}$ is continuous. [/guided] [/step] [step:Verify that $\mathbb{R}$ is complete and $(0,1)$ is not] The space $(\mathbb{R}, |\cdot|)$ is complete: every [Cauchy sequence](/page/Cauchy%20Sequence) in $\mathbb{R}$ converges (this is the completeness axiom of $\mathbb{R}$). The space $((0,1), |\cdot|)$ is not complete. The sequence $x_n = 1/n$ for $n \geq 2$ lies in $(0,1)$. It is Cauchy: for $m, n \geq N$, $|x_m - x_n| \leq 1/N \to 0$. However, $x_n \to 0 \notin (0,1)$, so the sequence does not converge in $(0,1)$. [/step] [step:Conclude that completeness is not a topological invariant] The spaces $\mathbb{R}$ and $(0,1)$ are homeomorphic (via the map $f$ above) but differ in completeness. Since homeomorphisms preserve all topological properties, completeness is not a topological property — it depends on the specific metric, not just the [topology](/page/Topology) it induces. [/step]