Which metrizable topological spaces $(X,\tau)$ posses the following property:  Every compatible metric (i.e one which induces the same topology $\tau$) is complete.  Compact metrizable spaces satisfy this.  Are there any non-compact examples?   Note:  It turns out that if you impose additional structure the answer is yes. In particular, if $M$ is a smooth manifold, and every Riemannian metric on $M$ is complete, then $M$ must be compact.  Reference: Nomizu, Katsumi, and Hideki Ozeki. "The existence of complete Riemannian metrics." Proceedings of the American Mathematical Society 12.6 (1961): 889-891.
123902
I have seen that completeness is not a topological property like compactness or connectedness. I have seen some examples also showing that there are two equivalent metrics one of which is complete and the other one is incomplete. I want to know some general result. Consider any metric space $(X,d)$. Let $d$ be complete. Does there exist an equivalent metric $d'$ in $X$ which is incomplete? When $d$ is compact, such $d'$ does not exist (Since compactness implies completeness). So answer me when $d$ is noncompact.
99860
Show that a totally bounded complete metric space $X$ is compact.   I can use the fact that sequentially compact $\Leftrightarrow$ compact.   Attempt: Complete $\implies$ every Cauchy sequence converges. Totally bounded $\implies$ $\forall\epsilon&gt;0$, $X$ can be covered by a finite number of balls of radius $\epsilon$. I'm trying to show that all sequences in $X$ have a subsequence that converges to an element in $X$. I don't see how to go from convergent Cauchy sequences and totally bounded to subsequence convergent $in$ $X$.
74996
I wish to show that any metrizable space $(X,\mathcal{T})$ is Hausdorff  Proof attempt:  Let $d$ be the metric that generates the topology on $X$. Pick two points $x,y \in X$, we wish to produce two disjoint open sets that separates $x,y$.  Let $B_\epsilon(x)$ and $B_\delta(y)$ be two metric balls containing $x,y$ respectively. Suppose that  $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$       Stuck Here: Hmm...How should I adjust $\epsilon, \delta$ so that these   balls are separated?   Idea: Reduce $\epsilon$ by half. If they are still intersecting...reduce $\delta$ by half. Continue ad infinitum         Is there a more satisfying solution i.e. closed form expression for   reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!
54095
Question: Let $(X, d)$ be a metric space. Let $A$ and $B$ be disjoint subsets of $X$ that are closed in the topology induced by $d$. Prove that there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B\subset V$.   Is this proof correct? : By assumption, $A\cap B=\emptyset$, so there is no pair like $(x_A,x_B)$ (i.e., $x_A\in A$ and $x_B\in B$), such that $x_A=x_B$. Because every metric space is Hausdorff, so for every pair $(x_A,x_B)$ there are two disjoint open sets $U_A\ni x_A$ and $V_B\ni x_B$; and thus union of all $U_A$s (which is an open set containing A) are disjoint from union of all $V_B$s (which is an open set containing A).   Thanks a lot for any help.    PS In special case of standard metric topology on $\mathbb{R}$, if $[a,b]$ and $[c,d]$ are disjoint (say $c&gt;b$), intuitively it is apparent that there is a 'space' between $c$ and $b$ so we can have two disjoint open sets, each of them containing $[a,b]$ or $[c,d]$.
9008
Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete.  My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not figure out the exit way. Maybe I am on the wrong track. Could you please help me?  edit: More from my attempt:  Suppose $A$ is a closed set and let $(x_n)$ be a sequence of points $A$ such that $\lim_n x_n\to b$.  Suppose now that $A$ has the property that $b\in A$, whenever $x_n$ converges to $b$. We know that every element of $x_n$ which is convergent in $X$ also converges to a point in $A$. Since $x_n$ is a Cauchy sequence in $A$, it must converge to a point $y\in A$. But the limit of a convergent sequence is unique.   Take $x\in A$ and select an appropriate $n$, which enables $x_n$ to converge to a point $x$ in $X$. Since the limit is unique, it must follow that $x=y$. Thus $x\in A$ and $A$ is closed.   If $A$ is a closed subset of $X$, then any Cauchy sequence of a point in $A$ is convergent in $X$ and hence converges to a point in $A$. Thus $A$ is complete.
38249
Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? The closeed-ness comes from Rudin's proof, and the bounded-ness comes from the fact that for a compact subset $E$, we can take the finite open cover and than this open cover is what bounds $E$? Discussions are welcomed! Thank you!!!
70989
The question is (from Topology without tears) that: Let $(X,d)$ be a metric space and $\tau$ the corresponding topology on $X$. Fix $a \in X$. Prove that the map $f:(X,\tau) \rightarrow \mathbb{R}$ defined by $f(x) = d(a,x)$ is continuous.  My first attempt was that let an open ball, $B_{\epsilon}(a)\subset \mathbb{R}$ and show that $f^{-1}(B_{\epsilon}(a)) \subset \tau$. Since $f(x) = d(a,x)$, $f^{-1}(B_{\epsilon}(a))=B_{\epsilon}(a) \in \tau$.   I am pretty sure the last sentence is wrong. I am not sure if every open ball is an open set in topology induced by metric space.  Please give me some hint and direction. Thanks
120483
Let $X$ is locally compact Hausdorff space .If $C_{c}(X)$ is complet,then $X$ is compact (this is to be proved). I know that $C_{c}(X)$ is dense in $C_{0}(X)$. As $C_{c}(X)$ is complete implies that $C_{c}(X)=C_{0}(X)$. and I know only urysohn's lemma and Tiez's extension theorem on locally compact Hausdorff spaces and other basic things.
2115
Let $X$ be a metric space ; then which of the following is possible ?  1) $X$ has exactly $3$ dense subsets   2)  $X$ has exactly $4$ dense subsets   3)  $X$ has exactly $5$ dense subsets   4)  $X$ has exactly $6$ dense subsets   I know that if $X$ has a proper dense subset then for some $a \in X$ , we should have $X \setminus \{a\}$ is dense in $X$ and then $\{a\}$ is not open in $X$ ; but I can't relate this to no. of dense subsets except that if $X$ has only finitely many dense subsets then the topology of $X$ cannot be discrete . Please help . Thanks in advance
29405
I've got a question: I'm trying to prove that every metric space is compact if and only if the space is sequentially compact. In all the proofs I have found, they used the Bolzano-Weierstrass theorem. Is there a way to prove this fact without using Bolzano-Weierstrass?  Thanks so much!!
80772
Suppose I am asked to show that some topology is not metrizable. What I have to prove exactly for that ?
91073
Please ,how to prove that the space $\mathbb{R}$ endowed with the metric $d(x,y)=|e^x-e^y|$ is  not a complete space?  I don't find a Cauchy sequence   but not convergent   Please  Thank you.
6675
I happen to found  about finding two metric spaces $X$ and $Y$ such that there exists bijective continuous functions $f:X\to Y$ and $g:Y\to X$, and $X$ and $Y$ are not homeomorphic. I could not produce my own example, and I think the reason is that I am thinking of manifolds only.  Suppose we have two second-countable Hausdorff metrisable manifolds $X$ and $Y$ such that there exists bijective continuous functions $f:X\to Y$ and $g:Y\to X$. Can we show that the two manifolds are homeomorphic? Or is there still a counterexample?  Please note that I have taken only one course on metric space and topology, and I don't have much experience with manifolds. So please explain answers in details or post references to the details.  Edit: In case of any ambiguity, I would like to state that I do not restrict attention to manifolds without boundary, i.e. the manifolds may have boundary, if such a counterexample exists.
26068
Let $M$ be a smooth, finite-dimensional manifold. Suppose $M$ is also a metric space, with a given distance function $d: M \times M \rightarrow \mathbb{R}_{+}$, which is compatible with the original (manifold) topology on $M$.  Question: is there a Riemannian metric $g$ on $M$ such that the distance $$d_g(p, q) = \inf_{\gamma \in \Omega(p, q)} L_g(\gamma)$$ coincides with $d$?  I believe that this setting is very standard, but for the sake of completeness: $L_g$ denotes the Riemannian length of the curve $\gamma$, and $\Omega(p, q)$ the set of all piecewise smooth curves $\gamma : [a, b] \rightarrow M$ s.t. $\gamma(a) = p$ and $\gamma(b) = q$. Thanks in advance.
104356
Can someone please verify my proof?     Is the collection $\tau_\infty = \{U:X-U$ is infinite or empty or all of $X\}$ a topology on $X$?   No. Let $X = \mathbb{R}$.  Clearly, $\{x\} \in \tau_\infty$ for all $x \in X$.  Set $\displaystyle{B = \cup_{x \neq 0} \{x\}}$. Clearly, $X - B = \{0\}$, but $\{0\} \notin \tau_\infty$. Since $\tau_\infty$ is not closed under arbitrary unions, it does not define a topology on $X$.
81589
Is $(\mathbb{R},\tau_{co})$ compact where $\tau_{co}$ is the cocountable topology on $\mathbb{R}$?  I have the answer of my teacher but I'd like to see another one so I can understand better how people find the answer intuitively. He says:     $\mathbb{N}-\{n\}$ is infinite countable for all $n\in\mathbb{N}$.   Then $V_n = \mathbb{R}-(\mathbb{N}-\{n\})=(\mathbb{R}-\mathbb{N})\cup\{n\}\in\tau_{co}$ for all $n\in\mathbb{N}$.      Then $U=\{ V_n : n\in\mathbb{N}\}$ is a open covering of $(\mathbb{R},\tau_{co})$.      $U$ does not contain a countable subcover since $V_n\cap\mathbb{N}=\{n\}$ for each $n\in\mathbb{N}$ and $\mathbb{N}$ is infinite.   How would you solve that problem?
3999
Let $(M,d)$ be a metric space, and $A\subset M$. By definition, $A$ is said to be compact if every open cover of $A$ contains a finite subcover.  What is wrong with saying that, in $\mathbb{R}$, if $I=(0,1)$, we can choose $G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}$, which satisfies $I \subset \bigcup_{U\in G} U$, but we can't extract a finite subcover, so $I$ is not compact. Is $G$ a finite subcover of $G$, so it is not a valid cover for proving this? I would take $\cup_{n\in\mathbb{N}} (\frac{1}{n},1)$ in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets $\neq \emptyset$?