Skip to main content

Questions tagged [metrizability]

For questions pertaining to the metrizability of topological spaces and / or metrization theorems.

Filter by
Sorted by
Tagged with
3 votes
1 answer
45 views

Product of uniformizable spaces

I say that a space $X$ is uniformizable if there exists a familily of pseudometrics $ \{d_{\alpha}\}_{\alpha \in A} $ such that induces its topology. I want to prove the following thm: If $\{X_{\mu}\...
Pablo Garcia Pastor's user avatar
6 votes
1 answer
121 views

A metrizable space is realcompact iff it has non-measurable cardinality?

A space is realcompact if its a closed subspace of an arbitrary product of real lines, with product topology. A cardinal $\kappa$ is called measurable if there exists a (countably additive) $\{0, 1\}$-...
Jakobian's user avatar
  • 10.3k
2 votes
2 answers
87 views

If $X$ is a compact separable topological space with a countable family of complex valued continuous functions, then $X$ is metrizable

I am studying measure theory, topology, and functional analysis in mathematics. Let $X$ be a compact topological space. We assume that there exists a countable family $\{f_n\: X \to \mathbb C: n\in\...
love and light's user avatar
4 votes
1 answer
107 views

$L^p_{loc}(\Omega)$ is completely metrizable

Let $\Omega \subset \mathbb{R}^n$ be a (not necessarily bounded) domain and $1 \leq p \leq \infty$. Then define $L^p_{loc}(\Omega)$ to be the set of functions $f: \Omega \rightarrow \mathbb{R}$ such ...
CBBAM's user avatar
  • 6,009
3 votes
0 answers
119 views

Topological space not being metrizable.

Question: Consider $\mathbb{R}^{2}$ with the topology $\mathcal{T} = \{U \subseteq \mathbb{R}^{2} : \mathbb{R}^{2} \setminus U \text{ is countable }\} \cup \{\emptyset\}$. Suppose that there is a ...
ByteBlitzer's user avatar
4 votes
2 answers
102 views

Special Type of Locally Metrizable Space

We say a topological space is locally metrizable if for every $x\in X,$ there is an open set $U$ containing $x$ which is metrizable. That is, the subspace topology on $U$ is the topology induced by ...
Miles Gould's user avatar
0 votes
1 answer
110 views

If $E$ is the strict inductive limit of Frechet spaces $\lbrace E_k\rbrace$, why can't the countable basis of $E_1$ be appropriated for $E$?

Not wishing to dispute that strict LF-spaces are non-metrisable, I am trying to see the flaw in the intuitive sense I have that we could appropriate, for a strict LF space, a countable basis from the ...
demim00nde's user avatar
1 vote
0 answers
47 views

Locally compact Polish space admits a proper metric [duplicate]

If $X$ is locally compact Hausdorff, then the following are all equivalent: $X$ is second countable, $X$ is metrizable and $\sigma$-compact, $X$ is metrizable and separable, $X$ is Polish. I want to ...
subrosar's user avatar
  • 4,682
0 votes
1 answer
76 views

Metrizable topology given by convergence of sequences

Let $\alpha$ be a (Hausdorff) topology on a set $X$. Let $T_{\alpha} = \{ \beta \text{ topology on } X \mid \forall (x_n)_{n \in \mathbb{N}} \text{ in } X, x \in X: x_n \overset{\alpha} {\...
Smiley1000's user avatar
  • 1,293
2 votes
1 answer
84 views

Why is ${\mathbb R}^\infty$ with the colimit topology not metrizable?

To be clear, I mean $$ {\mathbb R}^\infty = \bigcup_{n} {\mathbb R}^n $$ with the colimit topology, where ${\mathbb R}^n\to {\mathbb R}^{n+1}$ is the inclusion with the last coordinate 0. It seems ...
Three aggies's user avatar
  • 5,203
1 vote
1 answer
46 views

Show that the following metric is complete on $L_1(X,Y)$

I am working through the exercises of Kechris' Classical Descriptive Set Theory for personal edification. I've solved Exercise 3.5, but the proof was two pages. Did I miss a more straightforward ...
Eli Johnson's user avatar
3 votes
1 answer
263 views

Is the difference between metric spaces and ultrametric spaces topological?

My question is exactly what is written in the title. A metric space is a set $X$ equipped with a function $d: X \times X \to \mathbb{R}_{\geq 0}$ satisfying $d(x, y) = 0 \iff x = y$, $d(x, y) = d(y, x)...
Joe Stephen's user avatar
2 votes
0 answers
45 views

Applications of Alexandroff-Urysohn metrization theorem

A star is $\text{St}(A, \mathcal{U}) = \bigcup\{U\in \mathcal{U} : U\cap A\neq\emptyset\}$. A development is a sequence of open covers, $(\mathcal{U}_n)_{n\in\mathbb{N}}$ such that $\text{St}(x, \...
Jakobian's user avatar
  • 10.3k
0 votes
0 answers
33 views

Containment of kernels of continuous seminorms

Let $X$ be a locally convex space, whose topology is defined by a family of seminorms $\mathcal{P}$. Fact. For every continuous seminorm $q: X\to \mathbb{F}$ and positive number $\epsilon>0$, the ...
user760's user avatar
  • 1,422
1 vote
1 answer
108 views

Is there a non-metrizable topological space $X$ for which every closed subset of $X$ is the zero set of a continuous real function on $X$?

Let us say a given topological space $X$ has property $P$ if for every closed subset $A \subset X$ there is a continuous function $f : X\rightarrow \Bbb R$ such that $A = f^{-1}(\{0\})$. As I ...
I.A.S. Tambe's user avatar
  • 2,441
0 votes
0 answers
68 views

Examples of topological spaces in which the converse of the sequence lemma holds, but is not metrizable

From Munkres, Let X be a topological space; let $ A \subset X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \overline{A}$; the converse holds if $X$ is metrizable. Of ...
Anthony Lee's user avatar
1 vote
1 answer
44 views

How can the implication metrizable -> first countable be generalized?

I'm making a contribution to the pi-Base to automatically deduce certain spaces are not metrizable based on the lack of first-countability. How can this theorem be generalized to deduce more [non-]...
Steven Clontz's user avatar
1 vote
0 answers
72 views

Does every metric space that topologically embeds into $\mathbb R^n$ isometrically embeds into some metrizable space homeomorphic to $\mathbb R^n$?

Follow up to this question. Motivated by wanting to extend a metric space into one that intuitively "extends indefinitely and has no gaps". And also motivated by the famous result that every ...
Carla_'s user avatar
  • 329
4 votes
1 answer
125 views

Can a topological embedding of a metric space into a metrizable space be extended to an isometric embedding for some metric on the codomain?

Motivation: Consider for example a metric space that is a disjoint union of a point with $\mathbb R$ (with the usual metric on $\mathbb R$). Intuitively, it feels like there is some space "...
Carla_'s user avatar
  • 329
0 votes
0 answers
38 views

Separation axioms from a product to a particular space

I developed a proof of the following statement but I'm not sure if it is correct: Proposition: Let $\lbrace X_\alpha \rbrace_{\alpha \in I}$ a collection of topological spaces and consider $X := \...
Ivan Camilo Ballen Mendez's user avatar
3 votes
0 answers
32 views

Coincidence of Subspace and Metric Topologies

How to show that a subspace of a metrizable space is also metrizable? Let $X$ be a topological space, and let $Y$ be a (non-empty) subset of $X$; let $d$ be a metric on $X$ that induces the topology ...
Saaqib Mahmood's user avatar
4 votes
4 answers
137 views

Construct a metric that induces a given topology

In a topology textbook there was a exercise to determine the topology induced by $$x^2:\mathbb{R}\to\mathbb{R}$$ where the target has the euclidean topology. I am the opinion that $x^2$ induced a kind ...
HyperPro's user avatar
  • 869
1 vote
1 answer
81 views

Denumerable dimensional tvs

I am trying to prove the above. However, I even cannot understand why it is true. For example, $\ell^2$ is a complete normed space which has countable orthonormal basis and so countable dimensional. ...
handa's user avatar
  • 149
1 vote
2 answers
201 views

Why is Niemytzki plane locally metrizable?

The answer https://math.stackexchange.com/a/4411293/32337 includes the assertion that the Niemytzki plane $\Gamma$ (also known as the Moore plane, and as the tangent disk space) is locally metrizable. ...
murray's user avatar
  • 778
1 vote
1 answer
35 views

Apparent contradiction between weak metrizability results

I will fix $E \doteq \ell^2$, an infinite dimensional space, with separable dual. In here, there was a discussion that the weak topology of $E$ is not metrizable. By a result given in here, the disk $...
Gustavo Labegalini's user avatar
0 votes
1 answer
177 views

Prob. 5 (b), Sec. 30, in Munkres' TOPOLOGY, 2nd ed: Every metrizable Lindelof space is second-countable

Here is Prob. 5 (b), Sec. 30, in the book Topology by James R. Munkres, 2nd edition: Show that every metrizable Lindelof space has a countable basis. My Attempt: Let $X$ be a metrizable Lindelof ...
Saaqib Mahmood's user avatar
2 votes
1 answer
242 views

Prob. 5 (a), Sec. 30, in Munkres' TOPOLOGY, 2nd ed: Every metrizabe separable space is second-countable

Here is Prob. 5 (a), Sec. 30, in the book Topology by James R. Munkres, 2nd edition: Show that every metrizable space with a countable dense subset has a countable basis. My Attempt: Let $X$ be a ...
Saaqib Mahmood's user avatar
4 votes
1 answer
156 views

Is the "unit sphere" in $\mathbb{R}^\omega$ metrizable?

Let $\mathbb{R}^\omega$ be the countably infinite product of $\mathbb{R}$ with itself in the product topology. $\mathbb{R}^\omega$ is metrizable but the metric doesn't arise from a norm, so a natural ...
Compact_Ball's user avatar
0 votes
1 answer
87 views

Baire space which doesn't contain a dense completely metrizable subspace

Let us work with separable metrizable spaces. It's known that if $X$ contains a dense completely metrizable subspace then $X$ is Baire. Moreover, if we assume that $X$ embeds into any space as a Borel ...
Jakobian's user avatar
  • 10.3k
4 votes
1 answer
82 views

How to prove that the following metric induces the subspace topology?

I am trying to follow Theorem (3.11) of Kechris's Classical Descriptive Set Theory. In this part of the proof he shows that a $G_{\delta}$-subspace Y of a completely metrizable space $(X,d)$ is ...
a_hayler's user avatar
0 votes
1 answer
80 views

Strongly zero-dimensinal spaces that are homeomorphic to other spaces under certain conditions

According to a theorem from van Engelen: Homogeneous zero-dimensional absolute Borel sets (1986): Theorem: If $X$ is a separable metrizable zero-dimensional absolute $F_{\sigma\delta}$ that is nowhere ...
Tereza Tizkova's user avatar
1 vote
0 answers
257 views

Question 3.8 from Brezis' book of Functional Analysis

QUESTION: Let $E$ be an infinite-dimensional Banach space. Our purpose is to show that $E'$ equipped with the weak $^\star$ topology is not metrizable. Suppose, by contradiction, that there is a ...
Alana Tavares's user avatar
3 votes
1 answer
172 views

If $X$ is locally compact Hausdorff and $C_0(X)$ is separable, then $X$ is metrizable and separable

Let $X$ be a locally compact Hausdorff space and suppose that $C_0(X)$ is separable w.r.t the uniform metric. Then $X$ is metrizable and separable. I'm reading a proof of this statement that goes ...
Gabriel Romon's user avatar
12 votes
2 answers
153 views

Is "ternary metrizability" equivalent to pseudometrizability?

Below, $X$ is always a set with at least three elements to avoid triviality. Say that a ternary metric on a set $X$ is a map $t:X^3\rightarrow\mathbb{R}$ with the following properties: Non-...
Noah Schweber's user avatar
2 votes
1 answer
117 views

Definition of countable dimension of vector space

It's a well known that Hausdorff locally convex space $(X, P)$, where $P$ is a family of seminorms which generate topology on $X$ is metrizable iff $P$ is equivalent to an at most countable subfamily $...
NeoFanatic's user avatar
0 votes
1 answer
111 views

Measure Space without any form of metric

I'm with a difficult question that I was thinking . Any help will be useful We know that the Urysohn's Theorem states that: every Hausdorff second-countable regular space is metrizable. There are ...
Lucas Vasconcellos de Souza's user avatar
2 votes
2 answers
110 views

How did submetrizability come into existence ? What is submetrizability of a topological space used for?

Can anybody tell me how did submetrizability come into existence, and what is its use in topology ? Any examples to make me understand ?
Mir Aaliya's user avatar
1 vote
1 answer
217 views

Showing that three statements are equivalent (Topology)

$(a)$ Assume that $X$ is a compact metrizable space. Prove that the following conditions are equivalent: $(i)$ $T$ is transitive. $(ii)$ The set of transitive points for $T$ is dense in $X$. $(iii)$ ...
JOJO's user avatar
  • 1,080
1 vote
0 answers
90 views

Perfect image of a metrizable space is metrizable.

How can I prove that the perfect image of a metrizable space is metrizable? I know the following three theorems about equivalent conditions of metrizability. A space $X$ is metrizable if and only if ...
Sphere's user avatar
  • 711
-1 votes
1 answer
222 views

Show that $C(\mathbb R, \mathbb R)$ is not connected

$\bar{d}: \mathbb R × \mathbb R → \mathbb R$ denotes the standard bounded metric on $\mathbb R$ defined by $\bar{d}(x, y) := \min\{|y − x|, 1\}$. Given a topological space $X$, let $C(X, \mathbb R)$ ...
JOJO's user avatar
  • 1,080
1 vote
0 answers
151 views

Metrizability of the set of probability measures with the Kantorovich–Rubinstein metric

Let $X$ be a compact metric space and let $\mathcal{M}(X)$ be the set of probability measures. Then the weak topology on $\mathcal{M}(X)$ is metrizable, for example with the Wasserstein-metric and ...
user avatar
0 votes
1 answer
253 views

Prove on metrization of uncountable product [duplicate]

I am given the following problem: Show that an uncountable product of unit intervals is not first countable, and thus not metrizable. My answer would be that a), since the elements of the neighborhood ...
NoneTheBrighter's user avatar
1 vote
1 answer
209 views

Show strong topology on the closed unit ball is metrizable

Let $S$ be the closed unit ball of $B(H)$, the bounded operators on the Hilbert space $H$. I want to show that the relative strong topology on $S$ is metrizable. Attempt: I have already established ...
user avatar
3 votes
1 answer
457 views

Weak topology is not metrizable

I'm now reading some properties of weak topology, I have some problems which may related to the topology property in non-metrizable space ($E $ is a Banach Space): I know that $E^*$ with weak* ...
EggTart's user avatar
  • 385
0 votes
1 answer
305 views

Is it possible to show $\mathbb{R}^\omega$ is not metrizable without using the first countable definition and sequence lemma?

I encountered a question about proving that $\mathbb{R}^\omega$, which is the countably infinite product of $\mathbb{R}$ under the product topology, is not metrizable. I have seen many solutions here ...
Edi's user avatar
  • 349
0 votes
1 answer
226 views

Metrization problem

Can the following problem be solved by Nagata Smirnov metrization theorem, which states that A topological space is metrizable if and only if it is frechet, regular and has a sigma locally finite ...
Charlie's user avatar
  • 25
3 votes
1 answer
124 views

Nagata-Smirnov Metrization Theorem for Pseudometric Spaces

The Nagata-Smirnov Metrization Theorem states that $X$ is metrizable iff it is $T_3$ and has a $\sigma$-locally finite base So, I was wondering if this holds for pseudometric spaces too, if we ...
Ishan Deo's user avatar
  • 3,594
0 votes
1 answer
133 views

If $B_n$ is bounded then $\bigcup_{n=0}^{\infty} \varepsilon_n E_n$ is bounded

Let $E$ be metrizable space, that is, there exists a basis $\mathcal{B}:=\{U_n \subset E \; ; \; n\in \mathbb{N}\}$ of neighborhoods of $0 \in E$. I want to prove that: if $\{B_n \subset E \; ; \; n \...
Guilherme's user avatar
  • 1,591
0 votes
2 answers
374 views

The countable product of Fréchet spaces is a Fréchet space

Let $\{E_n \; ; \; n \in \mathbb{N}\}$ be a family of Fréchet spaces. I want to prove that the product $$E:= \prod_{n=1}^{\infty} E_n$$ is a Fréchet space, that is, $E$ is metrizable (Hausdorff space ...
Guilherme's user avatar
  • 1,591
0 votes
2 answers
39 views

In a metrizable TVS $E$ a point $x$ is an accumulation point of a sequence $S$ if and only if $S$ contains a subsequence which converges to $x$.

Let $E$ be a metrizable space, that is, $E$ is a Hausdorff space and if there is a countable basis of neighborhoods of $0 \in E$ in E. I want to prove: a point $x \in E$ is a an accumulation point of ...
Guilherme's user avatar
  • 1,591