# Tag Info

The answer is no. That's simply because a uniform limit of homeomorphisms need not be a homeomorphism (for notational convenience, I shall consider $[0,2]$ instead of $[0,1]$): Let $$\varphi_n(x) = \begin{cases} \frac xn & \text{for }x\in \left[0,1\right] \\ 2\left(1-\frac{1}n \right)(x-1)+\frac1 n & \text{for }x\in [1,2] \end{cases}$$ and ...
As t.b. noted in the comments, every metric space is paracompact, so an even stronger result is: Theorem. Every paracompact Hausdorff space is completely uniformizable. Let $\langle X,\tau\rangle$ be a paracompact Hausdorff space. The first step of the proof is to show that the collection $\mathfrak{N}$ of all open nbhds of the diagonal of $X\times ... 8 In a metric space all three properties are equivalent. In a uniform space every Cauchy filter converges iff every Cauchy net converges; the usual equivalence between filters and nets in arbitrary topological spaces preserves the property of being Cauchy in uniform spaces. This is strictly stronger than merely requiring Cauchy sequences to converge. ... 8 No. In the positive direction, Bourbaki proves in Topologie Générale, Chapitre IX, §9, Proposition 4 the following: Let$G$be a metrizable topological group. If$H$is a closed normal subgroup then$G/H$is metrizable. If$G$is complete (in one of the one-sided uniformities) then so is$G/H$. This entails a positive ... 7 Every topological group is automatically a uniform group. In particular$G$becomes a uniform space if we define a subset$V$of$G\times G$to be an entourage if and only if it contains the set$\{ (x, y) : x⋅y^{−1} \in U \}$for some neighborhood$U$of the identity element of$G.$The way to think about it is that uniform spaces have ways to compare ... 6 As you note, the topology of pointwise convergence on$Y^X$is simply the product topology on the product of$|X|$copies of$Y$indexed by$X$. For$x\in X$let$e_x:Y^X\to Y:f\mapsto f(x)$be the evaluation map at$x$, and let$\tau$be the coarsest topology on$Y^X$making each$e_x$continuous. For each finite$F\subseteq X$and open$V\subseteq Y$let ... 6 It seems that the conjecture is well known and can be proved straightforwardly. Let$f:(X,{\cal E})\to (Y,\cal F)$be a surjective uniformly continuous map between uniform spaces and the space$(X,\cal E)$is totally bounded. Let$F\in\cal F$be an arbitrary entourage. Since the map$f$is uniformly continuous, there exists an entourage$E\in\cal E$such ... 5 I had some attempts to obtain elementary proofs of (1) or (2), but I failed. Maybe these proofs should not be easy, for instance, in the case when (1) and (2) elementarily imply that the group$G$is topological. Since I am a specialist in paratopological groups, not semitopological, I propose a sketch of a proof that each locally compact Hausdorff ... 5 For a metric space$X$, the following are equivalent: Every continuous map from$X$to another metric space is uniformly continuous.$X$is complete and almost totally bounded. Definition.$X$is almost totally bounded if for every$r>0$there is a finite set$\{x_1,\dots,x_n\}\in X$and a number$\delta>0$such that for any distinct points ... 5 Here is only a very partial list: Geodesics, betweenness, Lipschitz Functions, Hausdorff dimension, coarse functions, isometries, Menger convexity. Arguably, one can distill those properties of metric spaces that allow one to speak of any given property of metric spaces, and thus define a new abstract class of objects. This is, in some sense, what topology ... 4 HINT: A topological space is uniformizable if and only if it is completely regular, so you need to find a space that is not completely regular. On the other hand, if$X$is a Hausdorff space and$x\in X$, then$\{x\}$is the intersection of the neighborhoods of$x$. (Why?) Thus, it suffices to find a Hausdorff space that is not completely regular, and there ... 4 Let$\mathscr{U}=\{X\setminus\operatorname{cl}F:F\in\mathfrak{F}\}$. Let$\mathscr{V}$be a locally finite open refinement of$\mathscr{U}$. A paracompact Hausdorff space is normal, so$X$has an open cover$\mathscr{W}=\{W_V:V\in\mathscr{V}\}$such that for each$V\in\mathscr{V}$,$\operatorname{cl}W_V\subseteq V$; clearly$\mathscr{W}$is locally finite. ... 4 No. The following are equivalent for a Tychonov space$X$:$X$is locally compact. There is a minimal uniformity on$X$. There is a minimal totally bounded uniformity. The uniformities form a complete lattice. The totally bounded uniformities form a complete lattice. See Shirota On systems of structures of a completely regular space Osaka Math. J. ... 4 Unfortunately not, but (the way I know how) to demonstrate this is not entirely simple Definition: A T$_1$-space$X$is called collectionwise normal if for every discrete family$\{ F_i : i \in I \}$of closed subsets of$X$there is a pairwise disjoint family$\{ W_i : i \in I \}$of open subsets of$X$such that$F_i \subseteq W_i$for all$i \in I$. ... 4 Yes, compact Hausdorff spaces have a unique uniformity compatible with their topology. I assume you already know that the system$\mathcal{D}_0$of all neighborhoods of the diagonal$\Delta = \{(x,x) \in X \times X \mid x \in X\}$of$X \times X$is a uniformity on a compact Hausdorff space$X$. This is not trivial, however, it is not very hard to prove. ... 4 The volume form tells you very little about the metric. Let$V$be an$n$-dimensional vector space, with volume form$v_1\wedge \ldots \wedge v_n$, where$\{v_1,\ldots,v_n\}$are linearly independent elements of$V$(any volume form can be written this way). Now define a metric by the condition that this be an orthonormal set; this metric gives you the ... 4 Every uniformizable space is completely regular, so a Dieudonné complete space is necessarily completely regular, as of course is any closed subspace of a product of metrizable spaces. Moreover, a closed subspace of a product of metrizable spaces is Hausdorff, so the spaces in question must be Tikhonov. The key result is Theorems$39.11$in Stephen ... 4 Have a look at this: http://www.math.wm.edu/~vinroot/PadicGroups/519probset1.pdf" Problem 2 is what you want. 4 Yes to all questions (as implicit in the comments, so I put cw). Indeed the solenoid defined as the inverse limit of the sequence of surjective endomorphisms$\mathbf{R}/\mathbf{Z}$given by multiplication by 2 is a compact, metrizable, connected and not path-connected group. 4 First, in order to make your theorem well-defined, the following theorem is needed (II.27 theorem 1 in Bourbaki): Therem 1: Let$X$be a compact space. Then there exists only one uniform structure on$X$compatible with its topology, namely the neighborhoods of the diagonal$\Delta$in$X \times X$. In fact, your theorem then follows easily, but the proof ... 3 Not in general. The Cantor set minus any one point and the Cantor set itself are a counterexample, as are$[0,1]$and$(0,1)$and many others. 3 The notation in the Springer definition is pretty appalling: The product of uniform spaces$(X_t,\mathfrak{A}_t),\,t\in T$, is the uniform space$(\prod X_t,\prod\mathfrak{A}_t)$, where$\prod\mathfrak{A}_t$is the uniformity on$\prod X_t$with as base for the entourages sets of the form $$\Big\{\big(\{x_t\},\{t_t\}\big):(x_{t_i},y_{t_i})\in ... 3 The product uniformity is by definition the smallest \mathcal D uniformity on \prod_i X_i, which makes all projections \pi_i\colon \prod_i X_i \to X_i uniformly continuous. So for each entourage E_i \in \mathcal D_i, we want to have (\pi_i\times\pi_i)^{-1}[E_i] \in \mathcal D. Taking finite intersections of these sets gives us a basis for ... 3 HINT: Yes, such examples exist. I’m assuming that these are diagonal uniformities. Let X=[0,1], and let \Delta be the diagonal in X\times X. Let a,b\in(0,1) with a\ne b, and let$$\mathscr{D}_1=\big\{D\subseteq X\times X:\Delta\cup\{\langle a,1\rangle,\langle 1,a\rangle\}\subseteq D\big\}$$and$$\mathscr{D}_2=\big\{D\subseteq X\times ... 3 First, notice that$X$is$T_0$: the uniform structure induced on a singleton make it a complete uniform space, so any singleton is closed. Then, Hausdorffness follows from the results: Theorem 1: A topological space is uniformizable iff it is$T_{3 \frac{1}{2}}$. Theorem 2:$T_0+ T_{3 \frac{1}{2}}$is equivalent to$T_2+T_{3 \frac{1}{2}}$. I can ... 3 Recall that a function$f:A \to B$between uniform spaces is uniform precisely when$(f\times f)^{-1}(U)$is an entourage in$A$for all entourages$U$in$B$. So, if you want your$f:X\to Y$to induce a uniform structure on$X$from the one on$Y$, such that it is the smallest uniform structure such that$f$is uniform, then you define the uniform ... 3 Such metric spaces are called UC-spaces (Google Scholar, Google Books) or Atsuji spaces. (Google Scholar, Google Books). Quote from G. Beer: Topologies on Closed and Closed Convex Sets, p.54: 2.3.1 Theorem. Let$\langle X,d \rangle$be a metric space. The following are equivalent: (1) Each continuous function on$\langle X,d\rangle$with values in an ... 3 A topological group is a uniform space in two ways... one by left translation, the other by right translation. Replace$x\cdot y^{-1}$in Owen's answer by$y^{-1}\cdot x\$ to switch. The two are equivalent in abelian groups, compact groups, and in many others. But not all. I have heard it said that the notion of "uniform space" was originally given as a ...