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Questions tagged [second-countable]

For questions about second-countable topological spaces, i.e., space with countable base.

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The countability of the Cantor-Bendixson rank in Polish spaces

I've been studying the Cantor-Bendixson theorem and have some questions about the proof of the countability of the Cantor-Bendixson rank in Polish spaces. I would greatly appreciate your insights on ...
Chau Long's user avatar
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1 answer
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Find an example of connected topology for finite space $\{1,...,n\}$ that conforms with the intuition of a commoner.

The most common topology for finite space is the discrete topology, which is clearly not connected. There are some other discussion on this site, which give many example topology of finite space and $\...
dodo's user avatar
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Countable many copies of $[0,1]$

Is $[0,1]^{\mathbb{N}}$ is separable? I’m trying to construct a countable basis from a countable basis of $[0,1]$ to show is separable but i’m not sure if i understand correctly the space. My teacher ...
benjv's user avatar
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Can any non-empty open subset of $\mathbb R^n$ be written as a countable union of open balls whose centres belong to the open set itself?

Consider an arbitrary non-empty open set $\Omega \subset \mathbb R^n$. Is it true that one can find a countable collection of open balls $(B(x_l,r_l))_{l \in \mathbb N}$, where $x_l \in \Omega$ and $...
xyz's user avatar
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Proof about a property of $\mathbb{R}$

The result that I want to prove is the following: Let's associate to each $x\in \mathbb{R}$ a number $r_{x}>0$. Then there is $M\in \mathbb{N}$ and a sequence $ \left( x_{n} \right)_{n\in \mathbb{N}...
HeyHéctor's user avatar
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Is a $\sigma$-locally finite collection of open sets locally countable?

Problem I encountered this statement on nLab, which says that weakly Lindelöf spaces with a $\sigma$-locally finite basis are second-countable. The original proof given below the statement is Proof. ...
Ricky's user avatar
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If a topological space X is not first countable, is it second countable?

I know that topological space is first countable if each point has a countable neighborhood basis. A neighborhood basis at a point. Consider the following topological space, $X=R$ with $\mathcal T=\{A\...
Anonymous's user avatar
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Support of function

As reported in https://en.wikipedia.org/wiki/Support_(mathematics), for a function $f:X \rightarrow \mathbb{R}$ we can define some notion of $supp(f)$, in particular: If $X$ is only a set, we define ...
Manuel Bonanno's user avatar
4 votes
2 answers
156 views

Does there exist a second-countable locally connected space with no countable basis of connected sets?

Space $X$ is called locally connected if it has a basis consisting of connected sets. It's called second-countable if it has a countable basis. If $X$ is both locally connected and second-countable, ...
Jakobian's user avatar
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Is every second countable locally compact Hausdorff space normal?

I know that there are locally compact Hausdorff spaces that aren't normal. Are there examples which are second countable?
Nik Weaver's user avatar
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Why is a Countable Basis Needed in This Proof?

My question is regarding the Theorem in Munkres that states: Every Regular Space with a Countable Basis is Normal. Before reading the proof in Munkres, I tried to prove it myself and came up with a &...
Kenneth Winters's user avatar
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For non second-countable TVS, is the sum of measurable functions again measurable? [duplicate]

Let $(\Omega, \mathcal A)$ be a measurable space and $E$ a topological vector space. Let $f,g:\Omega \to E$ be measurable. I already proved that Theorem $E$ is second-countable, then $f+g$ is ...
Akira's user avatar
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Prove that the following properties are all finitely productive

The question goes as follows: Prove that the following properties are all finitely productive (1) $T_0$ and $T_1$ (2) Separable (3) First Countable (4) Second Countable (5) Finite (i.e., the ...
Ryukendo Dey's user avatar
4 votes
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Locally compact 2nd countable Hausdorff space and complete metrizability

I was recently trying to verify certain things in the setting of Locally compact 2nd countable Hausdorff spaces. I thought that this is a natural collection of spaces more general than metric spaces, ...
Keen-ameteur's user avatar
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Quotient topology, product topology and subspace topology [closed]

Let $S$ be the set $\mathbb R^2\setminus\{(0, y) \mid y \neq 0\}$. Let $\tau_1$ denote the subspace topology on $S$ induced from the usual topology of $\mathbb{R}^2$. Now, consider the surjective map $...
Mamun's user avatar
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Arbitrary union of measure $0$ open sets has measure $0$ in a separable metric space

Let $X$ be a separable metric space, with $(X, \tau)$ the corresponding topological space, and let $\mu$ be a (positive) measure on $(X, \mathcal{B}(X))$, where $\mathcal{B}(X)$ is the Borel $\sigma-$...
gandalfbalrogslayer's user avatar
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Prove that $S^2$ has a countable basis $\{U_n\}$ implies that $P^2$ has a countable basis $\{p(U_n)\}$.

The Problem: Let $p: S^2\to P^2$ be the quotient map from the $2$-sphere $S^2$ to its projective plane $P^2$. Then $S^2$ has a countable basis $\{U_n\}$ implies that $P^2$ has a countable basis $\{p(...
Dick Grayson's user avatar
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How many second-countable $T_1$ spaces are there? [duplicate]

How many second-countable $T_1$ spaces, up to homeomorphism, are there? Let $X$ be a second-countable $T_1$ space. Since $X$ is second-countable, there are at most $\beth_1$ open sets in $X$. And ...
Dannyu NDos's user avatar
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Are first and second countability preserved under intersection of topologies?

For a given set $X$ endowed with two topologies $\mathcal{T}$ and $\mathcal{T}'$, i.e. such that $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ are two topological spaces defined on the same $X$, it is easy ...
K. Makabre's user avatar
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Proof verifying that separable metric space is second countable

Show that a separable metric space $X$ is second countable. I’m trying to figure out whether I got this proof correct. Since $X$ is separable there exists a countable subset $D$ such that the ...
Yoshi's user avatar
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Proof verification for separable metric space implies second countable.

Let $(X,d)$ be a separable metric space and let $Y \subset X$ be its countable dense subset. Take the basis to be $$\mathcal{B}=\{B_{\frac{1}{n}}(x): x \in Y, n \in \Bbb{Z}^+\}$$ I claim for an ...
homosapien's user avatar
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Let $X$ be a topological space that satisfies the second axiom of countability. Show that if ...

Let $X$ be a topological space that satisfies the second axiom of countability. Show that if $\cal{B}$ is a basis for $X$ then there is $\cal{B'} \subseteq \cal{B}$ countable such that $\cal{B'} $ is ...
Gab's user avatar
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Compact Hausdorff Space - X Second Countable iff C(X) separable

I recently stumbled across a property of compact Hausdorff spaces which is supposedly well-known, namely: If $X$ is a compact Hausdorff space, then $X$ is second countable if and only if $C(X)$ is ...
LSK21's user avatar
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The Borel $\sigma$-algebra generated by the product topology coincides with the product of Borel $\sigma$-algebras: where did I get wrong?

Let $(\Omega_n, \tau_n)_n$ be a sequence of metrizable topological spaces. Let $\sigma (\tau_n)$ be the Borel $\sigma$-algebra on $\Omega_n$. Let $\Omega :=\prod_{n =1}^\infty \Omega_n$ and $\pi_n: \...
Analyst's user avatar
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1 vote
1 answer
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Confusion about a proof of Lusin's theorem

I'm reading about Lusin's theorem in textbook Optimal Transport for Applied Mathematicians Let us be more precise: take a topological space $X$ endowed with a finite regular measure $\mu$ (i.e. any ...
Akira's user avatar
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Is the Alexandroff extension of a locally compact, second-countable space second-countable?

If $X$ is a locally compact, second-countable topological space, then is its Alexandroff extension $X^*$ also second-countable? Our definition of locally compact is that for every $x$ in $X$, we have ...
shoteyes's user avatar
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$(\Bbb R, \tau)$ with $\tau$ defined as: $U\in \tau \iff U=\emptyset \vee 0 \in U$ is not second-countable.

Let $(\Bbb R,\tau)$ be a topological space with $tau$ is defined as: $$U\in \tau \iff U=\emptyset \vee 0 \in U.$$ Show that $\tau$ is not second-countable. Attempt: The goal is to show that for any ...
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Borel $\sigma$-algebra with $\sigma$-finite measure: does a function vanish almost everywhere outside its essential support?

Let $(X, \mathcal B(X), \mu)$ be a $\sigma$-finite measure space with $\mathcal B(X)$ the Borel $\sigma$-algebra of a topological space $X$. The essential support of $f:X \to \mathbb R$ is defined as $...
Akira's user avatar
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4 votes
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Why does satisfying the countability conditions make topological spaces so nice?

Motivation for the question: While I can appreciate the significance of Theorem 30.1 below, and also the Urysohn metrization theorem, I can't seem to understand why either of them is rooted in these ...
Cathartic Encephalopathy's user avatar
3 votes
1 answer
276 views

Concerning topological manifolds: Are paracompact and connected locally euclidian Hausdorff spaces always second-countable?

There are different definitions for topological manifolds, sometimes second-countability or paracompactness are added to being locally euclidian Hausdorff. (Sometimes Hausdorff is also left out, but I ...
Samuel Adrian Antz's user avatar
1 vote
2 answers
304 views

Exercise 4, Section 30 of Munkres’ Topology

Show that every compact metrizable space $X$ has a countable basis. [Hint: Let $\mathscr{A}_n$ be a finite covering of $X$ by $1/n$-balls.] My attempt: Approach(1): $B_n =\{ B(x, \frac{1}{n})| x\in X\...
user264745's user avatar
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1 vote
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Exercise 5, Section 30 of Munkres’ Topology

(a) Show that every metrizable space with a countable dense subset has a countable basis. (b) Show that every metrizable Lindelof space has a countable basis. My attempt: (a) Since $X$ is separable, $...
user264745's user avatar
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1 vote
1 answer
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Definition of Countability in Munkres’ Topology

$X$ have a countable basis at $x$, if $\exists \{U_n \in \mathcal{N}_x |n\in \Bbb{N}\}$ with the following property: $\forall U\in \mathcal{N}_x, \exists m\in \Bbb{N}$ such that $U_m\subseteq U$. If $...
user264745's user avatar
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Theorem 30.3 of Munkres’ Topology

Suppose that $X$ has a countable basis. Then: (a) Every open covering of $X$ contains a countable subcollection covering $X$. (b) There exists a countable subset of $X$ that is dense in $X$. ...
user264745's user avatar
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1 answer
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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
249 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
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1 answer
80 views

On second-countable space and its uncountable subset

This is Munkres Chapter 4, Section 30, Problem Number 3. Let $X$ have a countable basis; let $A$ be an uncountable subset of $X$. Show that uncountably many points of $A$ are limit points of $A$. My ...
SU Lee's user avatar
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1 answer
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Topology counterexamples without ordinals

I am looking for three counterexamples in general topology, namely: A set which is sequentially closed, but not closed; A set which is sequentially compact, but not compact; A set which is compact ...
Khalid Wenchao Yjibo's user avatar
2 votes
3 answers
2k views

Proof that Metric Spaces are Second Countable?

Metric spaces can be equipped with the topology given by the open sets (which in turn are defined with the help of open balls) such that metric spaces are topological spaces. Now, the definition of ...
Hermi's user avatar
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1 vote
2 answers
248 views

Let $(E,d)$ be a metric space. Then $X$ is second-countable if and only if $X$ is Lindelöf if and only if $X$ is separable

In proving every subspace of a separable metric space is separable, I need below result. Could you check if my proof is fine? Theorem: Let $(E,d)$ be a metric space. Then $X$ is second-countable if ...
Akira's user avatar
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1 vote
2 answers
202 views

$f:I\rightarrow X$ where $X$ is hausdorff show that $X$ is metrizable.

This question comes from section 44 problem 4 of Munkres. Let $X$ be a Hausdorff space. Let $I=[0,1]$. Show that if there is a continuous surjective map $f : I \rightarrow X$, then $X$ is compact, ...
Killaspe's user avatar
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2 votes
2 answers
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Example of a topological space that has countably (but not finitely) many elements, is regular, but not 2nd countable

I'm trying to find an example of a topological space that has countably (but not finitely) many elements, is regular, but not 2nd countable. It seems to me that if a space has countably many elements, ...
A4F's user avatar
  • 61
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0 answers
47 views

Chain of compact sets in locally compact space

When a topological Hausdorff space X is locally compact and second-countable (has countable weight), can we find a chain of compact sets $\{K_i: n \in \mathbb{N}\}$, where $K_0 = \emptyset$ and $K_n \...
Tereza Tizkova's user avatar
2 votes
3 answers
1k views

How to tell if a space is second-countable

A topological space is called second-countable iff it has a countable basis. How to prove or at least make an assumption about whether a space does, or does NOT have countable basis? Which properties ...
Tereza Tizkova's user avatar
4 votes
2 answers
248 views

Does every Lie group have at most countably many connectected components?

Some proofs in a lecture I took were motivated by this statement that "some people don't assume second countability when they define a topological manifold, but for Lie groups we get this ...
Nolan King's user avatar
3 votes
2 answers
198 views

Example of a Hausdorff topology that is homeomorphic to $\mathbb R^n$ but isn't second countable?

The definition of a topological manifold $M$ I have is: $M$ is Hausdorff. Each point of $M$ has a neighborhood that is homeomorphic to an open subset of $\mathbb{R}^n$. $M$ is second countable. ...
Makogan's user avatar
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0 votes
1 answer
57 views

Why the the new basis is countable?

I was thinking about the following question: If $p: X \rightarrow Y $ a continuous, closed, and surjective map with the property that for each $y \in Y$ we have $p^{-1}(\{y\})$ is compact. Prove that ...
Janbazif's user avatar
  • 433
0 votes
2 answers
167 views

Does second countablity implies closed (or open) set in $X$ is a countable intersection of open sets in $X$?

Is $X$ is second countable implies or equivalent to Every closed(or open) set in $X$ is a countable intersection of open sets in $X$? I know, metric space implies this by Closed set as a countable ...
phy_math's user avatar
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1 vote
1 answer
221 views

On showing that every separable metric space has a countable base

Theorem: Every separable metric space $(M, d)$ has a countable base I've glanced over a couple of proofs for this theorem (for example see the one provided here), but I'm still stuck at the following:...
Epsilon Away's user avatar
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0 votes
1 answer
130 views

Second Countability Implies Separability

Munkres states explicitly that: Theorem 30.3: Suppose that $X$ has a countable basis. Then: (b) There exists a countable subset of $X$ that is dense in $X$. The following proof he gives is ...
SunRoad2's user avatar
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