A homeomorphism is a reversible continuous mapping whose inverse is also continuous. The focus keyword homeomorphism describes the correct notion of topological sameness for metric spaces. Homeomorphisms preserve open-set structure, continuity, and convergence, but not necessarily distances, boundedness, or completeness. :::definition[Homeomorphism] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. A function $f:X\to Y$ is called a homeomorphism if $f$ is one-to-one, onto, continuous, and $f^{-1}:Y\to X$ is continuous. ::: :::definition[Homeomorphic Metric Spaces] Two metric spaces are called homeomorphic if there exists a homeomorphism from one space onto the other. ::: :::example[Homeomorphic Closed Intervals] The spaces $[0,1]$ and $[0,2]$ are homeomorphic under $f(x)=2x$. The inverse is $f^{-1}(y)=y/2$, and both mappings are continuous. ::: :::example[Logarithmic Homeomorphism] The function $f:(0,\infty)\to\mathbb R$ defined by $f(x)=\ln x$ is a homeomorphism with inverse $f^{-1}(y)=e^y$. ::: :::example[Homeomorphism Not Preserving Completeness and Boundedness] Let $X=\mathbb N$ and $Y=\{1/n:n\in\mathbb N\}$ with the usual metric. The map $f(n)=1/n$ is a homeomorphism, but $X$ is complete and unbounded while $Y$ is bounded and incomplete. ::: :::solved-problem Show that $f:\mathbb R\to(-1,1)$ defined by $f(x)=x/(1+|x|)$ is a homeomorphism and is uniformly continuous. ::: :::solution For $y\in(-1,1)$, define $g(y)=y/(1-y)$ when $y\geq0$ and $g(y)=y/(1+y)$ when $y<0$. Direct substitution gives $g(f(x))=x$ and $f(g(y))=y$. Thus $g=f^{-1}$. Both maps are continuous, so $f$ is a homeomorphism. Also $|f(x)-f(y)|\leq|x-y|$, so $f$ is uniformly continuous. ::: :::solved-problem Show that the image of a complete metric space under a homeomorphism need not be complete. ::: :::solution The space $\mathbb R$ is complete, and $x\mapsto x/(1+|x|)$ is a homeomorphism from $\mathbb R$ onto $(-1,1)$. But $(-1,1)$ is not complete, since $x_n=1-1/n$ is Cauchy in $(-1,1)$ and converges to $1\notin(-1,1)$. ::: :::definition[Open and Closed Mappings] A mapping $f:X_1\to X_2$ is called open if $f(G)$ is open whenever $G$ is open. It is called closed if $f(F)$ is closed whenever $F$ is closed. ::: :::theorem[Homeomorphism by Open or Closed Mapping] Let $f:X_1\to X_2$ be one-to-one and onto. Then $f$ is a homeomorphism if and only if $f$ is continuous and either open or closed. ::: :::proof Given that $f$ is bijective. To prove the characterization. If $f$ is a homeomorphism, then $f^{-1}$ is continuous, so images of open and closed sets under $f$ are inverse images under $f^{-1}$. Hence $f$ is open and closed. Conversely, if $f$ is continuous and open, then for every open $G\subseteq X_1$, $(f^{-1})^{-1}(G)=f(G)$ is open in $X_2$, so $f^{-1}$ is continuous. The closed case is analogous. $\square$ ::: :::exercise 1. Why must a homeomorphism be bijective? 2. Give a homeomorphism from $\mathbb R$ to $(0,\infty)$. 3. Give a property not preserved by homeomorphism. ::: :::answer 1. Its inverse must exist on the whole codomain. 2. $x\mapsto e^x$. 3. Boundedness and completeness need not be preserved. :::