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A topological characterization of the underlying spaces of complete R-trees

Paul Fabel

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Michigan Math. J., Volume 64, Issue 4 (2015), 881-887.

First available in Project Euclid: 18 November 2015

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Primary: 54D05: Connected and locally connected spaces (general aspects) 54F50: Spaces of dimension $\leq 1$; curves, dendrites [See also 26A03]
Secondary: 54E35: Metric spaces, metrizability


Fabel, Paul. A topological characterization of the underlying spaces of complete R-trees. Michigan Math. J. 64 (2015), no. 4, 881--887.

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  • \item[(1)] There exists a compatible metric $d$ such that $(P,d)$ is a complete R-tree.
  • \item[(2)] There exists a compatible metric $d$ such that $(P,d)$ is an R-tree and such that $(P,d)$ is an open subspace of its metric completion $\overline{ (P,d)}$.
  • \item[(3)] $P$ is metrizable, locally path connected, uniquely arcwise connected, and locally interval compact. \endenumerate \endtheorem \sectionPreliminaries, Examples, Remarks, and Lemmas An arc is a single point or a space homeomorphic to $[0,1]$. A $p$ -based topological R-tree $(P,\tau,p,\leq,\hat{\ }\,)$ is a metrizable, uniquely arcwise connected, locally path connected space with $p\in P$ and $ [x,y]\subset P$ denoting the unique arc from $x$ to $y$. The space $P$ enjoys both the associative binary operation $\hat{\ }$ such that $[p,x\,\hat{\ }\, y]=[p,x]\cap[ p,y]$ and the partial order $\leq$ such that $y\leq x$ iff $y\in[ p,x]$. Notationally, we may suppress $\leq$ and $\hat{\ }$ if it is understood that $p$ is the basepoint, and $\tau$ can be replaced by $d$ or $D$ if $P$ is equipped with the particular metric $d$ or $ D$. A metric space $(P,d)$ is complete if each Cauchy sequence has a limit, and we remind the reader that every metric space can be embedded as a dense subspace of a complete metric space [Munkres?], uniquely up to isometry. \beginexample \labelbadLet $P$ denote the planar subspace $([0,1]\times\{0\})\cup (\bigcup_{n=1}^{\infty}\{\frac{1}{n}\}\times[ 0,\frac{1}{n}))$. Note that $P$ is not the underlying space of a complete R-tree since the half open intervals $\{\frac{1}{n}\}\times[ 0,\frac{1}{n})$ would be forced to have infinite geometric length, violating the topological fact that $ x_{n}\rightarrow0$ if $x_{n}\in\{\frac{1}{n}\}\times[ 0,\frac{1}{n} )$. Note that $P$ is a $G_{\delta}$ subspace of the plane (the intersection of countably many open planar sets), and hence $P$ is topologically complete. \endexample The following fact follows easily from the algebraic properties of $(P, \hat{\ },\leq)$. \beginlemma \labelhatSuppose $(P,p,\tau\leq,\hat{\ }\,)$ is a p-based topological R-tree and $[p,z]\cap[ x,y]=\emptyset$. Then $x\,\hat{\ }\,z=y\,\hat{\ }\,z$. \endlemma \beginproof Note that $a\,\hat{\ }\,b\leq b$ since $a\,\hat{\ }\,b\in[ p,b]$ and $ a\leq b\Rightarrow a\,\hat{\ }\,b=a$ since $[p,a]\cap[ p,b]=[p,a]=[p,a\,\hat{\ }\,b]$. Note $\{x\,\hat{\ }\,z,x\,\hat{\ }\,y\}\subset[ p,x]$ and $x\,\hat{\ }\,z<x\,\hat{\ }\,y$ (since otherwise we obtain the contradiction $x\,\hat{\ }\,z\in[ p,z]\cap [ x\,\hat{\ }\,y,x]\subset[ p,z]\cap[ x,y]$). By a symmetric argument we conclude $y\,\hat{\ }\,z<y\,\hat{\ }\,x$. Thus, $ \{x\,\hat{\ }\,z,y\,\hat{\ }\,z\}\subset[ p,x\,\hat{\ }\,y]$. Note that $y\,\hat{\ }\,z\in[ p,x]\cap[ p,z]$ and thus $y\,\hat{\ }\,z\leq x\,\hat{\ }\,z$. By a symmetric argument, $x\,\hat{\ }\,z\leq y\,\hat{\ }\,z$, and thus $x\,\hat{\ }\,z=y\,\hat{\ }\,z$. \endproof The following lemma is also a consequence of the fact that the metric completion of an R-tree is an R-tree [Imrich,Chiswell?]. \beginlemma \labeldivergeSuppose $(P,d,p,\leq,\hat{\ }\,)$ is an incomplete $p$ -based R-tree with metric completion $\overline{(P,d,p,\leq,\hat{\ }\,)} $. Suppose $y\in\partial P=\overline{(P,d)}\setminus P$. There exists an order-preserving isometric embedding $h:[0,d(x,y))\rightarrow(P,d)$ such that $h(0)=p$ and $y=\lim_{t\rightarrow d(x,y)}h(t)$. In particular, since the compactum $h([0,d(x,y)])$ is closed in the metric space $\overline{(P,d)}$, $h([0,d(x,y))=P\cap h([0,d(x,y)])$ is a closed subspace of $P$. \endlemma \beginproof Obtain a sequence $z_{n}\in P$ with $d(z_{n},y)\rightarrow0$. For each $ N\in\{1,2,3,\dots\}$, obtain $M_{N}>N$ such that $[p,z_{N}]\cap[ z_{m},z_{n}]=\emptyset$ if $M_{N}\leq m\leq n$. Define $y_{N}=z_{N}\,\hat{\ }\,z_{M_{N}}$ and note that by Lemma \refhat $y_{N}=z_{N}\,\hat{\ }\,z_{m}\,\hat{\ }\,z_{n}=z_{N}\,\hat{\ }\, z_{m}$ if $M_{N}\leq m\leq n$. Note that $ y_{n}\rightarrow y$ and by construction there exists a subsequence $ y_{k_{1}}<y_{k_{2}}\dots$. Let $h:[0,d(x,y))\rightarrow\bigcup_{k=1}^{\infty }[p,y_{n_{k}}]\subset P$ be the natural isometry mapping $ [d(p,y_{k_{n}}),d(p,y_{k_{n+1}})]$ onto $[y_{k_{n}},y_{k_{n+1}}]\subset P$. By construction, $h$ is continuously extendable at $d(p,y)$. \endproof The following lemma establishes that locally interval compact R-trees are open subspaces of their metric completions. \beginlemma \labelnonclosedSuppose that $(P,d,p)$ is a $p$-based incomplete R-tree and $ \partial P=\overline{(P,d,p)}\setminus P$ is not a closed subspace of the metric completion $\overline{(P,d,p)}$. Then $P$ is not locally interval compact. \endlemma \beginproof Obtain $x\in P\cap\overline{\partial P}$. Suppose $\varepsilon>0$. Obtain $ y\in\partial P$ such that $d(x,y)<\varepsilon$. Obtain by Lemma \refdiverge an isometric embedding $ [ 0,d(p,y) ] \rightarrow\overline {P}$ such that $0\mapsto p$, $d(p,y)\mapsto y$, and $[0,d(x,y))$ is order isometric to a closed subspace $\alpha\subset P$. Let $\delta =\varepsilon -d(x,y)$. Obtain $z\in\alpha$ with $d(z,y)<\delta$. Note that if $z<w$ and $ w\in\alpha$, then $ d(w,x)=d(w,z)+d(z,x)<(\varepsilon -d(x,y))+d(x,y)<\varepsilon$. Thus, $[z,y)$ is a closed subspace of $P$, $ [z,y)$ is homeomorphic to $[0,1)$, $[z,y)\subset\overline {B(x,\varepsilon)}$, and $[z,y)$ is not compact. \endproof \beginremark \labelnowayIf $(P,d,p)$ is a $p$-based R tree and $\alpha\subset P$ is homeomorphic to $[0,1)$, then $(\alpha,d)$ is isometric to a unique finite Euclidean half open interval $[0,R)$ for some $R>0$ or the infinite ray $ [0,\infty)$. If $\alpha$ is closed in $P$ and $(\alpha,d)$ is isometric to the finite interval $[0,R)$, then the preimage of the sequence $R-\frac{1}{n}$ shows that $(P,d,p)$ is incomplete. \endremark The following easy lemma is used in the proof of Lemma \refpromote. \beginlemma \labelLcontSuppose that $(X,D)$ is a metric space and $A\subset X$ and $2^{X}$ denotes the collection of compact subsets of $X$ with the Hausdorff distance. Define $L:2^{X}\rightarrow[ 0,\infty)$ as $ L(C)=\inf_{(c,a)\in C\times A}D(c,a)$. Then $L$ is continuous. If $ (P,d,p,\leq,\hat{\ }\,)$ is an R-tree, then $\lambda$ is continuous if $ \lambda:P\rightarrow2^{P}$ is defined as $\lambda(x)=[p,x]$. \endlemma \beginproof By definition the Hausdorff distance $H(C,B)$ [Munkres?] between compacta $\{B,C\}\subset X$ satisfies $0\leq H(B,C)<\varepsilon$ iff for each $b\in B$, there exists $c\in C$ with $D(b,c)<\varepsilon$ and for each $ c\in C$, there exists $b\in B$ with $D(b,c)<\varepsilon$. If $b\in B$ and $ c\in C$ with $D(b,c)<\varepsilon$, then $\llvert L(C)-L(B)\rrvert <\varepsilon$, and in particular $L$ is continuous. If $\{x,y\}\subset P$ with $d(x,y)<\varepsilon$, then $H([p,x],[p,y])=d(x,y)<\varepsilon$, and in particular $\lambda$ is continuous. \endproof The following lemma and its proof also appear in another preprint of the author [Fabel?]. \beginlemma \labelbijectionSuppose that $(P,p,\tau,\leq,\hat{\ }\,)$ is a p-based topological R-tree. Suppose that the continuous function $l:P\rightarrow[ 0,\infty)$ satisfies $x<y\Rightarrow l(x)<l(y)$. Define $d:P\times P\rightarrow[ 0,\infty)$ as $d(x,y)=l(x)+l(y)-2l(x\,\hat{\ }\,y)$. Then $d$ is a metric on the set $P$, inclusion $\kappa:(P,\tau )\rightarrow (P,d)$ is a continuous bijection, each arc $\kappa[ x,y]\subset(P,d)$ is isometric to the Euclidean segment $[0,d(x,y)]$, and $d(x,x\,\hat{\ }\,x_{m})\rightarrow0\Rightarrow x\,\hat{\ }\,x_{m}\rightarrow x$ in $ (P,\tau)$. \endlemma \beginproof Note that $d(x,x)=0$ since $x\,\hat{\ }\,x=x$ and $y\neq x\Rightarrow x\,\hat{\ }\,y<x$ or $x\,\hat{\ }\,y<y$ and hence $d(x,y)>0$. $d(x,y)=d(y,x)$ since $x\,\hat{\ }\,y=y\,\hat{\ }\,x$. Note that $0\leq2(l(y)-l(x\,\hat{\ }\,y))$ since $x\,\hat{\ }\,y\leq y$. Note that $d(x,z)\leq d(x,y)+d(y,z)$ iff $-2l(x\,\hat{\ }\, z)\leq 2l(y)-2l(x\,\hat{\ }\,y)-2l(x\,\hat{\ }\,z)$ iff $0\leq 2(l(y)-l(x \,\hat{\ }\,y))$. The latter holds since $x\,\hat{\ }\,y\leq y$. Thus, $d$ is a metric on the set $P$. If $x_{m}\rightarrow x$ in $(P,\tau)$, then $ x\,\hat{\ }\,x_{m}\rightarrow x $ in $(P,\tau)$. Thus, since $l$ is continuous at $x$, $ l(x)-l(x_{m})\rightarrow0$ and $l(x)-l(x\,\hat{\ }\,x_{m}) \rightarrow0$. Hence, $(l(x)-l(x\,\hat{\ }\,x_{m}))+(l(x_{m})-l(x))+(l(x)-l(x\,\hat{\ }\,x_{m}))=d(x,x_{m})\rightarrow0$. Note that if $\{w,z\}\subset(P,\tau)$ then $w\leq z$ iff $w=z\,\hat{\ }\,w$, and hence by definition, $d(w,z)=l(z)-l(w)$. Thus, if $\{x,y\}\subset (P,\tau)$, then the natural homeomorphism $h_{x,y}:\kappa[ x\,\hat{\ }\, y,x]\rightarrow[ 0,l(x)-l(x\,\hat{\ }\,y)]$ (defined as $ h_{x,y}(z)=l(z)-l(x\,\hat{\ }\,y))$ is an isometry onto the Euclidean segment since $w<u<z\Rightarrow d(z,w)=l(z)-l(w)=(l(z)-l(u))+(l(u)-l(w))=d(z,u)+d(u,z)$. Pasting at $0$ ($ h_{y,x}^{-1}$ union the reverse of $h_{x,y}^{-1}$) yields the natural isometry $[l(x\,\hat{\ }\,y)-l(x),l(y)-l(x\,\hat{\ }\,y)]\rightarrow \kappa [ x,y]$. Suppose $d(x,x\,\hat{\ }\,x_{m})\rightarrow0$. Then $\{x\,\hat{\ }\,x_{m}\}$ is a sequence in the (metrizable) compact arc $[p,x]\subset(P,\tau)$. Since $\kappa$ is continuous at $y$, if $y\in[ p,x]\subset(P,\tau)$ is a subsequential limit of $\{x\,\hat{\ }\,x_{m}\}$, then $y=\kappa (y)=x$. Hence, $x\,\hat{\ }\,x_{m}\rightarrow x$ in $(P,\tau)$. \endproof The standard fact that a space $U$ is topologically complete if $U$ is an open subspace of some complete metric space $(X,d)$ is often established [Munkres?] via a closed embedding $\phi:U\rightarrow X\times R$ with $ u\mapsto(u,\frac{1}{\partial(u,\partial U)})$. For several reasons, this proof does not work “off the shelf” when trying to obtain a complete R-tree metric for a connected open subspace $P\subset Q$ of a complete R-tree $(Q,D)$. Instead, we build a strictly increasing length function $ l:P\rightarrow[ 0,\infty)$ such that $l(x_{n})\rightarrow\infty$ if $ x_{n}\rightarrow\partial P$, apply Lemma \refbijection, and verify completeness of the metric and continuity of the inverse mapping. \beginlemma \labelpromoteSuppose that $(Q,D)$ is a complete metric space, suppose that the subspace $P\subset Q$ is open, nonempty, and dense, and suppose tthat he metric space $(P,D,p,\leq,\hat{\ }\,)$ is a $p$-based R-tree. There exists a topologically compatible metric $d$ on $P$ such that $(P,d,p)$ is a complete R-tree. \endlemma \beginproof Let $\partial P=Q\setminus P$. Define $L:P\rightarrow[ 0,\infty)$ as $L(x)=\inf\{D(y,z)\mid y\in[ p,x]$ and $z\in\partial P\}$. Note that $L>0$ since $[p,x]$ is compact and $\partial P$ is closed. Note that $y\leq x\Rightarrow L(y)\geq L(x)$ since $[p,y]\subset[ p,x]$. Define $ l:P\rightarrow[ 0,\infty)$ as $l(x)=D(p,x)+\frac{1}{L(x)}$. Note that $l$ is continuous since $D$ is continuous and by Remark \refLcont $L$ is continuous. Observe that $\{x,y\}\subset P$ and $x<y\Rightarrow D(p,x)<D(p,y)$ (since $(P,D)$ is an R-tree) and $\frac{1}{L(x)}\leq\frac{1}{L(y)}$ since $ L(y)\geq L(x)$, and hence $l(x)<l(y)$. Thus, applying Lemma \refbijection, the metric $d(x,y)=l(x)+l(y)-2l(x\,\hat{\ }\,y)$ ensures that the inclusion $\kappa :(P,D)\rightarrow(P,d)$ is a continuous bijection, and $\kappa[ x,y]\subset(P,d)$ is isometric to the Euclidean segment $[0,d(x,y)]$. By definition, $D(x,y)=d(x,y)-l(x)-l(y)\leq d(x,y)$. Hence, $\kappa$ is a homeomorphism. Thus, $(P,d)$ is uniquely arcwise connected, and hence $(P,d)$ is an R-tree. Observe that for real numbers, if $0<t<s$, then $1<\frac{1}{t}-\frac {1}{s}$ iff $ st<s-t$. To obtain a contradiction, suppose that $(P,d)$ is incomplete. Let $\overline{(P,d) }$ denote the metric completion of $(P,d)$. By Lemma \refdiverge obtain $ y\in\overline{(P,d)}\setminus P$, and an isometric embedding $ h:[0,d(p,y)]\rightarrow\overline{(P,d)}$, so that $h(0)=p$, $h(d(p,y))=y$ and $h|[0,d(p,y))$ is an order-preserving embedding into $P$. Let $y_{m}=h( \frac{d(p,y)m}{m+1})$. Note that $\{y_{m}\}$ is Cauchy in $(P,d)$ and hence $ \{y_{m}\}$ is Cauchy in $(P,D)$ since $D\leq d$. Note that for all $m\geq1$ and $k\geq1$, $0<L(y_{m})\leq D(y_{m},y_{m+k})$ since $[p,y_{m}]\subset[ p,y_{m+k}]$. Thus, since $\{y_{m}\}$ is Cauchy in $(P,D)$, the sequence $L(y_{m})\rightarrow0$. Hence (applying the continuity of $\times:R\times R\rightarrow R$ and $-:R\times R\rightarrow R $ (familiar multiplication and substraction of real numbers)), for each $ M\geq1$, we obtain $N_{M}>M$ so that $L(y_{M})\times L(y_{n})<L(y_{M})-L(y_{n})$. Thus, if $n\geq N_{M}>M$, then $y_{M}=y_{M}\,\hat{\ }\,y_{n}$, and hence $ d(y_{n},y_{M})=D(y_{n},y_{M})+(\frac{1}{L(y_{n})}-\frac {1}{L(y_{M})})\geq( \frac{1}{L(y_{n})}-\frac{1}{L(y_{M})})>1$, contradicting the fact that $ \{y_{m}\}$ is Cauchy in $(P,d)$. \endproof \sectionProof of Theorem \protect\refmain For $3\Rightarrow2$, suppose that $(P,\tau)$ is a locally interval complete topological R-tree. Obtain by [Oversteegen?] a topologically compatible metric $d$ such that $(P,d)$ is an R-tree. If $(P,d)=\overline{(P,d)}$, then note that $ \overline{(P,d)}$ is open in $\overline{(P,d)}$. If $(P,d)\neq\overline { (P,d)}$, then Lemma \refnonclosed ensures that $P$ is open in $\overline{(P,d)}$. For $2\Rightarrow1$, suppose that $(P,d)$ is an R-tree, open in its metric completion $\overline{(P,d)}$. Apply Lemma \refpromote. For $1\Rightarrow 3$, suppose that $(P,d)$ is a complete R-tree. 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