Questions tagged [separable-spaces]

For questions about separable spaces, i.e., topological spaces containing a countable dense set.

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Seperability for the collections of all non-empty compact subsets of $\mathbb{R}^2$ with Hasudorff metric

Let $X$ be the collections of all non-empty compact subsets of $\mathbb{R}^2$, which has the Euclidean metric. Let $(X,d)$ be a metric space, where $d$ is the Hasudorff metric. Is $X$ separable? ...
Paul H.Y. Cheung's user avatar
3 votes
1 answer
66 views

Proving a Proposition about separable Hausdorff spaces that are locally euclidean

Let X be a separable Hausdorff space such that for every $x \in X$ there exists an open neighborhood $U$ of $x$ such that $U$ is homeomorphic to an open subset of $\mathbb{R}^n$. Show that: (i) $X$ is ...
Philip's user avatar
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2 answers
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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
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33 views

Show that Banach space valued function is measurable.

I am given a seperable Banach space and an interval $J$. Pettis' theorem tells that for such a space $Y$, we have: $f : J \rightarrow Y$ is measurable iff for any bounded linear functional $g \in Y^*$,...
julian2000P's user avatar
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Strong Seperation between hypreplane and an affine subspace

Let $h \in \mathbb{R}^n\backslash \{0\}$ and $r \in \mathbb{R}$. A sure that $M$ is an affine subspace of $\mathbb{R}^n$ with $H(h, r) \cap M=\phi$. I tried to use Banach separation theorem to show ...
fezaninsonu's user avatar
1 vote
1 answer
60 views

There exists the converse of this corollary from Brezis?

In Brezis's Functional Analysis, there is a corollary Corollary 3.30. Let $E$ be a separable Banach space and let $\left(f_{n}\right)$ be a bounded sequence in $E^*$. Then there exists a subsequence $...
Francesca's user avatar
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A separable normed space that is continuously embedded in a non-separable normed space implies that this embedding isn't dense.

As a preliminary I introduce the definition of denseness I am using: Definition (dense subsets of metric spaces). Suppose $(M,d)$ is a metric space. A subset $S \subset M$ is called dense in $M$ if ...
xyz's user avatar
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1 vote
1 answer
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Separability of codomains of Borel functions taking values in completely regular spaces

I am looking for a reference (or a counterexample) to the following statement. Let $X$ be a separable metric space. Suppose that $Y$ is a completely regular topological space and $f\colon X\to Y$ is a ...
Tomasz Kania's user avatar
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3 votes
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Adapting a proof of the non-separability of Morrey Spaces for a different definition.

In the article "Morrey spaces, their duals and preduals", by Marcel Rosenthal and Hans Triebel, for every $1 \leqslant p < \infty$ and $-\frac{n}{p} < r < 0$ the Morrey Spaces are ...
xyz's user avatar
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Are dual spaces to separable normed spaces first-countable?

In [1] I found the following theorem (roughly translated by me) Satz 13.10 If $X$ is a separable normed $k$-vector space ($k = \mathbb R$ or $\mathbb C$) with continuous dual space $X'$, then the ...
red_trumpet's user avatar
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Alternative Idea to show $l^{\infty}$ is not seperable. [duplicate]

We want to show ${l}^{\infty}$ not separable. (whereas usual ${l}^{\infty}$ denotes the set of all bounded sequences $f:\mathbb{N}\rightarrow \mathbb{R}$ with $\sup\limits_{n\in \mathbb{N}}|f(n)|<\...
Wjvr46's user avatar
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1 vote
0 answers
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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
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Does a separable, $US$, sequential space have cardinality at most the continuum?

Let $X$ be a separable sequential space with unique sequential limits ($US$). Can we prove that $X$ has cardinality at most $\mathfrak c=2^{\aleph_0}$? Context. If $X$ were Fréchet-Urysohn instead of ...
M W's user avatar
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3 votes
2 answers
257 views

Characterization of distance between points in separable space

I want to prove that if $X$ is a separable metric space, then for all $x,y \in X$ and any dense set $\{x_i\}$, we can write: $$d(x,y) = \sup_{i \in \mathbb{N}} |d(x,x_i) - d(y,x_i) |.$$ Some attempts. ...
Porcupine's user avatar
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A separable Hilbert space has a countable orthonormal basis. Is this true also for Banach spaces?

In this conversation, it was proved that an infinite-dimensional Hilbert space H has a countable orthonormal basis iff H is separable (has a countable dense subset). Is there any analogous statement ...
Michael_1812's user avatar
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How to prove that an absolutely continuous function space on a finite closed interval is a separable space in this norm sense?

$\quad$ Let a finite closed interval be $[a,b]$. Consider the absolutely continuous function space $\mathrm{AC}([a,b])$ on it, with a norm as follows: $$\|f\|_{\mathrm{AC}}=\sup_{x\in[a,b]}|f(x)|+\|f'\...
daidaitx's user avatar
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Let $X$ be a normed space. Show that if $X^*$ is separable, then $X$ is separable and show that the reverse does not hold.

Show that if $X^*$, the dual space of a normed space $X$, is separable, then $X$ is also separable. My Solution: Since $X^*$ is separable, there exists a countable dense subset $D \subseteq X^*$. To ...
CanDoMajoringMath's user avatar
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$C(X, \mathbb{C})$ separable [duplicate]

Let $(X, d)$ be a compact metric space. Is $C(X, \mathbb{C})$ separable? I've already checked an answer, but I think that my problem is that my continuous functions take value in $\mathbb{C}$. Does ...
Nick's user avatar
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1 answer
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Difficulty interpreting and solving Baby Rudin Problem 2.25, and how to optimally progress through Baby Rudin?

Context I am trying to solve Problem 2.25 in Baby Rudin. Here is the problem statement: (Rudin Problem 2.25)- Prove that every compact metric space has a countable base and that $K$ is therefore ...
Brendan Chamberlain's user avatar
1 vote
0 answers
133 views

How to prove that $l^p$ is separable? [duplicate]

I want to show that $l^p$ is separable. I got the hint to use the set $E:=\{e_n: n \in \mathbb{N}\}$, where $e_n$ denotes the sequence with a 1 in the n-th position and else zeros. There is a ...
Walli's user avatar
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6 votes
3 answers
159 views

Must a closed set with empty interior in a separable compact space of cardinality at most $2^{\aleph_{0}}$ be separable?

Consider a separable compact Hausdorff space $X$ with $|X| \leq 2^{\aleph_{0}}$. If $A\subseteq X$ is a closed set with empty interior, must $A$ be separable? Each example I've been able to come up ...
xote's user avatar
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A particular covering made up by disjointed families of Closed Balls

Proposition. Let $(X,d)$ a metric space and let $A\subseteq X$ a subset of $X$ such that $A$ is separable (then $A$ has a dense -with respect the subspace topology of $A$ - and countable subset $D$). ...
Grace53's user avatar
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0 answers
31 views

Whether the function $(x,x')\mapsto\rho\big(f(x),f(x')\big)$ is Borel for a Borel map $f\colon (X,d)\to (Y,\rho)$ with $(Y,\rho)$ not being separable?

Let $(X,d)$ be a compact metric space, let $(Y,\rho)$ be an arbitrary metric space. Let $\mathcal{B}(\mathbb{R}),\mathcal{B}(X),\mathcal{B}(Y),\mathcal{B}(X\times X), \mathcal{B}(Y\times Y)$ denote ...
Rafael's user avatar
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Space of smooth differential forms with compact support on $U \subset \mathbb{R}^n$ is separable

I am interested in how to prove that the space of smooth differential forms with compact support on some open $U \subset \mathbb{R}^n$, denoted $\mathcal{D}^n(U)$, is separable, i.e. it has a ...
Sean Kim's user avatar
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A question of countable product and uncountable product

It is intuitive and easy to show that a countable product of separable spaces is separable. It is tempting to think that an uncountable product of separable spaces will not be separable (a) Show that ...
Yanbo PAN's user avatar
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0 answers
40 views

An inner product space that every closed subspace is complemented, is it must be separable?

In a comment of this answer someone said that if an inner product space satisfy that every closed subspace is complemented then it has an (countable) orthonormal basis and this answer claim that H has ...
anyon's user avatar
  • 369
2 votes
1 answer
37 views

Infinite intersection of closed Hilbert subspaces whose finite intersection is infinite-dimensional

I'm working in the setting of complex, separable, infinite-dimensional Hilbert spaces and I came out with the following statement which, intuitively, seems to be true to me, though I haven't found a ...
Manuel Norman's user avatar
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1 answer
114 views

Continuous tensor products [closed]

Is there a notion of infinite tensor product Hilbert spaces? Let $I$ be an infinite set, for example taking $I= \mathbb{R}$, then is the space $\mathcal{H} = \bigotimes_{i \in I} \mathcal{H}_{I}$ well-...
QuantumFieldMedalist's user avatar
4 votes
0 answers
65 views

$H=\bigoplus H_n$ is separable?

Let $(H_n)$ a collection enumerable of Hilbert spaces. Consider $$ H=\bigoplus_{n=1}^\infty H_n=\left\{(x_n):\sum_{n=1}^\infty\|x_n\|^2<\infty\right\}. $$ If each $H_n$ is separable, then $H$ also ...
Kempa's user avatar
  • 648
2 votes
1 answer
119 views

A metric space is separable if and only if it is homeomorphic to a totally bounded metric space

In the wikipedia page: https://en.wikipedia.org/wiki/Totally_bounded_space I found the following theorem: A metric space is separable if and only if it is homeomorphic to a totally bounded metric ...
User271828's user avatar
0 votes
0 answers
58 views

Separable space, the countable dense sequence, and it's index family

The question is not directly related to seperabilty, but is more about the "limits" of countability. There is a known proof by Cantor which establishes that the cardinality of the real ...
user12456's user avatar
4 votes
2 answers
130 views

Is $C^\infty([0,1]^n)$ a separable Frechet space?

Gaussian Processes (GP) are widely viewed as practical ways to implement Gaussian Measures (GM) numerically. In fact, in many contexts it seems that to each GP corresponds a GM and vice versa, see, e....
Sobolev's user avatar
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0 answers
74 views

Prove that, an atomic/purely atomic measure on a separable metric space is supported by a countable set

Let $X$ be a separable metric space and $\mathcal{F}$ be the borel $\sigma$-algebra of $X$. Let $\mu$ be a $\sigma$-finite atomic or purely atomic measure on $\mathcal{F}$ i.e. every measurable set of ...
MathBS's user avatar
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1 vote
0 answers
38 views

Why the metric space $(\mathbb{D}(\mathbb{R}^d),\delta_{lu})$ is not separable? [duplicate]

Let $\mathbb{D}(\mathbb{R}^d)$ be the space of all cadlag functions $\mathbb{R}_{+}\rightarrow\mathbb{R}^d$. Consider the local uniform topology $$ \delta_{lu}(\alpha,\beta)=\sum_{n=1}^{\infty}2^{-n}...
AlmostSureUser's user avatar
2 votes
1 answer
136 views

If $X$ is inseparable, must $Y$ be inseparable?

Let $X$ and $Y$ be normed vector spaces. Suppose that there is linear isometry $T: X \rightarrow Y$, namely that $\Vert Tx\Vert =\Vert x\Vert$ for all $x\in X$. If $X$ is inseparable, must $Y$ be ...
Tony Weng's user avatar
1 vote
1 answer
25 views

$\forall r>0,j=1,2,\ldots:\mu(\partial B(x_j,r))=0$ for a Borel probability measure $\mu$ over a metric space $X$ where $x_j$ are countably dense

Let $(X, d)$ be a separable metric space and $S:=\{x_1,x_2,\dots,\}$ be a countably dense subset of $X$. Let $\mu$ be a Borel probability measure of $X$. A proof I am reading states the following ...
Cartesian Bear's user avatar
2 votes
2 answers
201 views

Alternate proof of "$\ell^\infty$ is not separable".

I am self-studying Functional Analysis from Kreyszig's Introductory Functional Analysis with Applications. In section-1.3, he proves that the sequence space $\ell^\infty$ with the metric $d_\infty(x,y)...
math-physicist's user avatar
0 votes
1 answer
105 views

Subspace of Euclidean space

Consider the euclidean space $\mathbb{R}^n$. Consider a closed compact subset $A\subset\mathbb{R}^n$. For example, take $A=[a,b]^n$, with $0<a<b<\infty$. It is well know that $\mathbb{R}^n$ ...
chaki chaki's user avatar
8 votes
2 answers
88 views

Prove that $X := \{ f: [0,1] \to [0,1] : f \text{ is continuous and } f(1) = 0 \}$ with the given distance is neither connected nor separable

Prove that $X := \{ f: [0,1] \to [0,1] : f \text{ is continuous and } f(1) = 0 \}$ with $d(f,g) := \inf \{r \geq 0 : f(t) = g(t) \forall r ≤ t ≤ 1 \}$ is neither connected nor separable. Here is my ...
Squirrel-Power's user avatar
1 vote
1 answer
101 views

Topology question on first countable.

This question was asked in the GATE MA 2023 paper: Q.44. Let $(\mathbb{R},\tau)$ be a topological space, where the topology $\tau$ is defined as $$\tau = \{U \subset \mathbb{R}: U = \emptyset \ or \ 1 ...
Gajendra Basavaraju's user avatar
2 votes
1 answer
58 views

Density of functions in "increasing pointwise" sense

Suppose that $(E,d)$ is a locally compact metric space with the metric topology. I am given a particular countable dense subset $S$ of $(C_c(E), \|\cdot\|_{\infty})$ i.e. in the supremum norm. By ...
Sarvesh Ravichandran Iyer's user avatar
4 votes
2 answers
157 views

Is it possible to define a topology on the real line such that 0 and non-zero integers are dense but no finite subset of non-zero integers is dense?

Is it possible to define a topology $\mathcal{T}$ on $\mathbb{R}$ such that $\{0\}$ and $\mathbb{Z}^*=\mathbb{Z}\setminus\{0\}$ are dense but no finite $F\subseteq \mathbb{Z}^*$ is dense? This is what ...
K. Makabre's user avatar
  • 1,760
0 votes
1 answer
208 views

Proof that the dual space $(C[0,1])^*$ is not separable.

Let $X=C[0,1]$ equipped with the maximum norm and define a bounded linear functional $\phi_x(f)=f(x)$ for $x\in [0,1]$. I want to show that $X^*$ is not separable. There was a hint in this exercise, ...
user1294729's user avatar
  • 2,008
0 votes
0 answers
68 views

Follow up to an answer: Why is $l^{\infty}$ not separable?

In this question thread, user940 answered and said that, "Assume that $A⊂l^∞$ is countable..." I want to understand: why can we assume something like this? can assuming something like this ...
ali's user avatar
  • 184
1 vote
1 answer
98 views

Separability vs. metrizability in compact Hausdorff

Exercise #4 in Section 34 of Munkres' Topology (2nd ed) states the following: Let $X$ be a compact Hausdorff space. Show that $X$ is metrizable if and only if it has a countable basis. However, even ...
Nick F's user avatar
  • 1,219
0 votes
1 answer
98 views

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
  • 11
1 vote
1 answer
78 views

Show that $L(X)$ is not separable if $X= c_0(\mathbb{N})$ and $l^p(\mathbb{N})$

The following problem was left as an exercise in my lecture of Functional Analysis course and I am not able to solve 2 parts of them. Show that the spaces $c_0(\mathbb{N}) $ is separable and $l^p(\...
user avatar
1 vote
0 answers
120 views

Countable union of separable subsets of a metric space is separable

This is a question from Irving Kaplansky's Set theory and metric spaces. Let $A_1, A_2, A_3, \cdots$ be subsets of a metric space $M$. Suppose each $A_i$ is separable. Prove that $\cup_i A_i$ is ...
Balkys's user avatar
  • 749
2 votes
0 answers
174 views

Let $(X, d)$ be a separable metric space. Then the space of all real-valued bounded uniformly continuous functions on $X$ is separable

It's well-known that Theorem 1 A metric space is compact IFF its space of bounded, continuous, real-valued functions is separable in the uniform topology. I would like to prove its complementary ...
Analyst's user avatar
  • 5,657
3 votes
3 answers
134 views

Continuous operators on $C[0,1] $ are not separable

Taking class in functional analysis, I was told that continuous operators $$A:C[0,1]\rightarrow C[0,1]$$ are not separable. I'm trying to prove it by finding uncountable family of operators $A_\alpha$ ...
Big Coconut's user avatar

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