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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 $X\times X$...

10

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. ...

10

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 $$\... 8 No. In the positive direction, Bourbaki proves in Topologie Générale, Chapitre IX, §9, Proposition 4 the following: Let G be a metrizable topological group. If H is a closed normal subgroup then G/H is metrizable. If G is complete (in one of the one-sided uniformities) then so is G/H. This entails a positive ... 8 This answer is preliminary. Now I am looking in the papers and soon I shall extend the answer. I am related with the subjects, but I am not a specialist. I looked into papers and can say you the following. If you are so interested and want to have a perfect answer, you may ask Hans-Peter Künzi (use this e-mail: Hans-peter.Kunzi (at) uct.ac.za). So, IMHO, ... 7 As you note, the topology of pointwise convergence on Y^X is simply the product topology on the product of |X| copies of Y indexed by X. For x\in X let e_x:Y^X\to Y:f\mapsto f(x) be the evaluation map at x, and let \tau be the coarsest topology on Y^X making each e_x continuous. For each finite F\subseteq X and open V\subseteq Y let$$...

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

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 ...

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 ...

6

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 ...

5

Have a look at this: http://www.math.wm.edu/~vinroot/PadicGroups/519probset1.pdf" Problem 2 is what you want.

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 $a,b\... 5 Yes to all questions (as implicit in the comments, so I put cw). Indeed the solenoid defined as the inverse limit of the sequence of surjective endomorphisms$\mathbf{R}/\mathbf{Z}$given by multiplication by 2 is a compact, metrizable, connected and not path-connected group. 5 First, in order to make your theorem well-defined, the following theorem is needed (II.27 theorem 1 in Bourbaki): Therem 1: Let$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$. In fact, your theorem then follows easily, but the proof ... 4 Let$\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}$is locally finite. (... 4 The notation in the Springer definition is pretty appalling: The product of uniform spaces$(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 $$\Big\{\big(\{x_t\},\{t_t\}\big):(x_{t_i},y_{t_i})\in V_{t_i},i=1,... 4 The product uniformity is by definition the smallest \mathcal D uniformity on \prod_i X_i, which makes all projections \pi_i\colon \prod_i X_i \to X_i uniformly continuous. So for each entourage E_i \in \mathcal D_i, we want to have (\pi_i\times\pi_i)^{-1}[E_i] \in \mathcal D. Taking finite intersections of these sets gives us a basis for \... 4 HINT: Yes, such examples exist. I’m assuming that these are diagonal uniformities. Let X=[0,1], and let \Delta be the diagonal in X\times X. Let a,b\in(0,1) with a\ne b, and let$$\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 X:\Delta\... 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 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 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 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 Willard’... 4 Given a subset of$A\subseteq X\times X$, let$A_x$denote the slice $$A_x = \{ y\in Y \,|\, (x,y)\in A\}.$$ We have an embedding$i_x\colon X\to X\times X$given by$y\mapsto(x,y)$, which allows us to express$A_x$as$i_x^{-1}(A)$. For $$\mathbf S = \left\{\, A\subseteq X\times X \,\middle|\, \text{A is open, containing \Delta}\,\right\}$$ I claim ... 4 The maps in question appear to be uniformly open surjections. I believe that uniformly open maps were introduced in E. Michael, ‘Topologies on spaces of subsets’, Trans. Amer. Math. Soc.$\mathbf{71}$($1951$),$152$-$182$. If$\langle X,\mathscr{U}\rangle$and$\langle Y,\mathscr{V}\rangle$are uniform spaces, a map$f:X\to Y$is uniformly open if for each$...

4

These uniform covers $\mathfrak{T}_U$ only form a base for the covering uniformity. Quote from Willard, General Topology (p. 245) Although, as we have said, coverings and surroundings should be used in the same way as one uses open and closed sets in a topological space, i.e. interchangeably, we should comment that the passage back and forth is not ...

3

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 $i \in I$. ...

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.

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