From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: compact metric spaces Date: 5 Mar 1999 22:13:58 GMT Newsgroups: sci.math Keywords: If X is bounded under all compatible metrics then X is compact In article <7bo2oe$3bl$1@nnrp1.dejanews.com>, wrote: >we know that for a compact metric space (X,d) any other compatible (i.e., >giving the same open sets ) metric w.r.t d leaves X bounded,now is the >converse true? i.e., (X,d) is a metric space such that any metric on X >compatible with d leaves X bounded,is X compact? It always struck me as a little odd to change metrics, so let me recast the question first. You know that X compact implies every f: X -> R is bounded. It is natural to ask whether the converse is true. (Note that _this_ question can be asked about compact spaces in general, not just metric spaces.) First, I should observe that for metric spaces, my question is equivalent to yours. For if there is an equivalent but unbounded metric d' on X, then we can define f : X -> R by f(x) = sup(d(x,y), y in X) and get an unbounded function on X. Conversely, if f : X -> R is unbounded, then g : X -> X x R given by g(x) = (x, f(x)) embeds X as an unbounded subset of a metric space, so that we can pull back a product metric on X x R to an unbounded but equivalent metric on X. Now, among metric spaces, compactness can be characterized as complete + totally bounded (A metric space X is "totally bounded" if for every e>0 , X can be covered by a finite set of balls of radius e. Of course X is "complete" if every Cauchy sequence converges.) So what I will do is to assume X is a metric space which is not compact and produce a continuous, unbounded real-valued function on X By the previous paragraph, I need only do this for metric spaces which are not complete, and for metric spaces which are not totally bounded. If X is not complete, let X* be a completion and pick P in X* - X. Define f : X -> R by f(x) = log(d(x,P)). If X is not totally bounded, there is some e>0 such that no finite collection of balls of radius e covers X. So I can pick x1 in X, x2 in X-B(x1,e), x3 in X - B(x1,e) - B(x2,e), and so on. Each d(x_i,x_j) is at least e and so there are no nonconstant Cauchy sequences among the x_i, and so the set of them is closed in X and also discrete. Define f(x_i) = i, and then use the Tietze extension theorem to extend to f : X -> R. There are (non-metric) spaces X which are not compact but for which all f : X -> R are bounded; for example, there are spaces on which there are no non-constant real-valued functions. (Obviously they are far from normal, according to the Tietze theorem.) dave ============================================================================== From: "Brian M. Scott" Subject: Re: compact metric spaces Date: Fri, 05 Mar 1999 19:55:07 -0500 Newsgroups: sci.math arinchaudhuri@my-dejanews.com wrote: > we know that for a compact metric space (X,d) any other compatible (i.e., > giving the same open sets ) metric w.r.t d leaves X bounded,now is the > converse true? i.e., (X,d) is a metric space such that any metric on X > compatible with d leaves X bounded,is X compact? Yes. Suppose that (X,d) is not compact. Then it is not countably compact, and consequently it contains a countably infinite, closed, discrete subset {x(n) : n in N}. Define f(x(n)) = n for each n in N, and use the Tietze Extension Theorem to extend f to a continuous function F : X --> R. Clearly F is unbounded on X. Now define h : X --> X x R by h(x) = (x, F(x)); using the fact that F is continuous, it's not hard to see that h is a homeomorphism of X onto h[X]. But it's also clear that h[X] is unbounded in any of the natural metrics on the product X x R (e.g., the metric e given by e((x,a), (y,b)) = d(x,y) + |a - b|). Brian M. Scott ============================================================================== From: Boudewijn Moonen Subject: Re: compact metric spaces Date: Mon, 08 Mar 1999 09:45:54 +0100 Newsgroups: sci.math arinchaudhuri@my-dejanews.com wrote: > we know that for a compact metric space (X,d) any other compatible (i.e., > giving the same open sets ) metric w.r.t d leaves X bounded,now is the > converse true? i.e., (X,d) is a metric space such that any metric on X > compatible with d leaves X bounded,is X compact? > > -----------== Posted via Deja News, The Discussion Network ==---------- > http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own Here is a Tietzeless variant. oo< ---------------- snip -------------------- %************************* PREAMBLE **************************** % % % \NeedsTeXFormat{LaTeX2e} \documentclass[12pt]{article} \usepackage{a4} \usepackage{amsmath,amssymb,amsthm} % \newtheorem*{thm}{Theorem} % % % %************************** MAIN TEXT **************************** % % % \begin{document} % A topological space is called \emph{metrizable} if it has a metric such that the given topology coincides with the topology defined by the metric. % \begin{thm} For a metrizable topological space $X$, the following properties are equivalent: % \begin{enumerate} % \item[(i)] $X$ is compact % \item[(ii)] Every metric on $X$ inducing the given topology is bounded % \item[(iii)] Every continuous (real valued) function on $X$ is bounded. % \end{enumerate} % \end{thm} % Before embarking on the proof, some terminology and results. As general reference we use the book \textsc{J. L. Kelley}, \emph{General Topology}. Let $X$ be a metric space. A \emph{Cauchy sequence} in $X$ is a sequence $x_1, x_2, \dots$ such that for every given real number $\varepsilon > 0$ the mutual distances $d(x_m,x_n)$ become eventually less than $\varepsilon$, where $d$ is the given metric on $X$. A metric space $X$ is called \emph{precompact} if every sequence in $X$ has a subsequence which is Cauchy. It is called \emph{complete} if every Cauchy sequence converges. The main standard result characterizing compact metric spaces is then that a metric space is compact if and only if it is precompact and complete, i.e. if and only if every sequence in $X$ has a convergent subsequence. In somewhat fancier language, call a metric space \emph{totally bounded} if for every given real number $\varepsilon > 0$ it has a finite covering by $\varepsilon$-balls; then, as is easily seen, it is precompact if and only if it is totally bounded. So a metric space is compact if and only if it is totally bounded and complete. Note that the property of compactness depends only on the underlying topology, while the properties of totally boundedness and completeness do depend on the particular metric. A further standard result is that on a compact space every continuous function is bounded. We now prove the theorem according to the pattern \vspace{2ex} \centerline{(i) $\implies$ (ii) $\implies$ (iii) $\implies$ (i)}. \emph{(i) $\implies$ (ii)}:\,\, If $X$ is compact, so is $X \times X$. Any metric $d$ on $X$ inducing the given topology on $X$ is a continuous function on $X \times X$, whence bounded. \vspace{2ex} \emph{(ii) $\implies$ (iii)}:\,\, Let $f$ be a continuous function on $X$. We then push points of $X$ apart at distances bounded from below by $f$ using the following standard technique. Consider the \emph{graph space} $Z$ of $f$: % \begin{equation*} % Z := \{ (x,f(x)) \,|\, x \in X \} \subseteq X \times \mathbb{R}. % \end{equation*} % The map $i : X \longrightarrow Z$ given by $x \mapsto (x,f(x))$ is then a homeomorphism of $X$ onto $Z$, its inverse being given by the restriction of the first projection $p : X \times \mathbb{R} \longrightarrow X$ to $Z$. The space $X \times \mathbb{R}$ with the product topology is metrizable; e.g. one may take the metric $D$ defined as % \begin{equation*} % D((x,s),(y,t)) := d(x,y) + |t - s| \,. % \end{equation*} % Pulling this metric back to $X$ using the map $i$ therefore equips $X$ with a metric $d'$ inducing the topology given by $d$, which therefore by assumption is bounded, by a constant $B > 0$, say. Now by construction % \begin{equation*} % d'(x,y) = d(x,y) + |f(y) - f(x)| \,, % \end{equation*} % so $d'$ being bounded by $B$ implies % \begin{equation*} % |f(y)| \le |f(x)| + B % \end{equation*} % for all $x, y \in X$. Choosing $x$ to be a fixed arbitrary point of $X$ shows $f$ is bounded. \vspace{2ex} \emph{(iii) $\implies$ (i)}:\,\, We show that on any noncompact metrizable space $X$ there exists a continuous unbounded function. Let $d$ be any metric on $X$ inducing the given topology and let $X'$ be the completion of $X$ with respect to $d$. We distinguish the cases $X'$ being compact and being not so. \emph{Case 1: $X'$ compact.}\,\, Since $X$ is assumed to be noncompact, $X \not= X'$ whence $X' \setminus X$ is not empty. Let $x_{\infty}$ be a point in $X'\setminus X$. Since $X$ is dense in $X'$, the function $f$ defined by $f(x) := 1/d(x,x_{\infty})$ is then a continuous function on $X$ which is not bounded. \emph{Case 2: $X'$ noncompact.}\,\, If $f$ is a continuous unbounded function on $X'$, its restriction to $X$ is a continuous unbounded function on $X$. So we may assume $X$ itself is complete. According to the standard characterization of compactness stated above $X$ cannot be totally bounded since it is assumed to be noncompact. So there is a real number $\varepsilon > 0$ such that $X$ cannot be covered by finitely many closed $\varepsilon$-balls. This allows the following recursive construction of a sequence $x_1, x_2, \dots$ of points in $X$ and real numbers $r_1 \ge r_2 \ge \dots > 0$: Let $x_1$ be any point in $X$ and put $r_1 := \varepsilon$. Then the closed ball $B(x_1;r_1)$ does not cover $X$. So there is $x_2$ in $X \setminus B(x_1;r_1)$. The latter complement being open there is $r_2$ with $r_1 \ge r_2 > 0$ such that $B(x_2;r_2) \subseteq X \setminus B(x_1;r_1)$. The balls $B(x_1;r_1)$ and $B(x_2;r_2)$ together do not cover $X$, so there are $x_3$ and $r_3$ with $B(x_3;r_3) \subseteq X \setminus B(x_1;r_1) \cup B(x_2;r_2)$ and $r_1 \ge r_2 \ge r_3 > 0$. Continuing this way one obtains the said sequences $x_1, x_2, \dots$ and $r_1 \ge r_2 \ge \dots > 0$; they have the property that the balls $B(x_k;r_k)$ are mutually disjoint. Now put % \begin{equation*} % f(x) := \sum_{k=1}^{\infty} k \cdot \frac{d(x,X \setminus B(x_k;r_k))} {d(x,x_k) + d(x,X \setminus B(x_k;r_k))} \,. % \end{equation*} % Since the balls $B(x_k;r_k)$ are mutually disjoint, this gives a well-defined continuous function $f$ on $X$. Since $f(x_k) = k$, $f$ is not bounded. This finishes the proof. % % \end{document} oo< ---------------- snip -------------------- -- Boudewijn Moonen Institut fuer Photogrammetrie der Universitaet Bonn Nussallee 15 D-53115 Bonn GERMANY e-mail: Boudewijn.Moonen@ipb.uni-bonn.de Tel.: GERMANY +49-228-732910 Fax.: GERMANY +49-228-732712