Mixing
Explain the statemets related to separable spaces.
$begingroup$
- Separable spaces are "simpler" than nonseparable ones-KREYSZIG
- Separable space $X$ is not "too big" in the sense that we can approach each element of space $X$ through a sequence of elements of a countable set-PONNUSAMY
I'm reading the article of Dense subsets and separability from different texts.I got the above statements which I'm not getting.
1.What notions are involved in non-separable spaces so that they are complex than Separable ones?
2.What is the meaning of "too big" in the second statement?
general-topology metric-spaces separable-spaces
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
$endgroup$
add a comment |
$begingroup$
- Separable spaces are "simpler" than nonseparable ones-KREYSZIG
- Separable space $X$ is not "too big" in the sense that we can approach each element of space $X$ through a sequence of elements of a countable set-PONNUSAMY
I'm reading the article of Dense subsets and separability from different texts.I got the above statements which I'm not getting.
1.What notions are involved in non-separable spaces so that they are complex than Separable ones?
2.What is the meaning of "too big" in the second statement?
general-topology metric-spaces separable-spaces
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
$endgroup$
add a comment |
$begingroup$
- Separable spaces are "simpler" than nonseparable ones-KREYSZIG
- Separable space $X$ is not "too big" in the sense that we can approach each element of space $X$ through a sequence of elements of a countable set-PONNUSAMY
I'm reading the article of Dense subsets and separability from different texts.I got the above statements which I'm not getting.
1.What notions are involved in non-separable spaces so that they are complex than Separable ones?
2.What is the meaning of "too big" in the second statement?
general-topology metric-spaces separable-spaces
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
$endgroup$
- Separable spaces are "simpler" than nonseparable ones-KREYSZIG
- Separable space $X$ is not "too big" in the sense that we can approach each element of space $X$ through a sequence of elements of a countable set-PONNUSAMY
I'm reading the article of Dense subsets and separability from different texts.I got the above statements which I'm not getting.
1.What notions are involved in non-separable spaces so that they are complex than Separable ones?
2.What is the meaning of "too big" in the second statement?
general-topology metric-spaces separable-spaces
general-topology metric-spaces separable-spaces
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
asked Jan 22 at 11:26
P.StylesP.Styles
1,467827
1,467827
add a comment |
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2 Answers
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A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.
For example $l_{infty}$ is non-separable because the open balls of radius $frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{infty}$. Intuitively this makes $l_{infty}$ look rather large.
Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $mathbb{R}$ and $l_{infty}$ have the same cardinality so $l_{infty}$ isn't really larger than $mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.
answered Jan 22 at 12:20
Sorin TircSorin Tirc
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Re: 2.
Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.
If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{aleph_0}=|Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $fin F$ let $psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X={psi (f):fin F}$. Therefore $|X|=|{psi(f):fin F}|le |F|le |E|le |Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.
On the other hand if the topology $T$ on $X$ is $T={X,emptyset}$ then any non-empty $Dsubset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.
BTW. A topological definition of "$(x_n)_{nin Bbb N}$ converges to $x$" is: ${x}=cap_{nin Bbb N}overline {S_n},$ where $S_n={x_j:jge n}.$
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
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2 Answers
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2 Answers
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$begingroup$
A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.
For example $l_{infty}$ is non-separable because the open balls of radius $frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{infty}$. Intuitively this makes $l_{infty}$ look rather large.
Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $mathbb{R}$ and $l_{infty}$ have the same cardinality so $l_{infty}$ isn't really larger than $mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.
For example $l_{infty}$ is non-separable because the open balls of radius $frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{infty}$. Intuitively this makes $l_{infty}$ look rather large.
Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $mathbb{R}$ and $l_{infty}$ have the same cardinality so $l_{infty}$ isn't really larger than $mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.
For example $l_{infty}$ is non-separable because the open balls of radius $frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{infty}$. Intuitively this makes $l_{infty}$ look rather large.
Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $mathbb{R}$ and $l_{infty}$ have the same cardinality so $l_{infty}$ isn't really larger than $mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
$endgroup$
A separable space is a space with a countable dense subset. So for a non-separable space, for any countable subset, its closure will not be the entire space.
For example $l_{infty}$ is non-separable because the open balls of radius $frac{1}{2}$ centred at the elements with only 0 and 1 entries form an uncountable collection and any dense set would have to contain an element of each of these balls. This in particular proves that $R$ cannot continuously surject onto $l_{infty}$. Intuitively this makes $l_{infty}$ look rather large.
Having said all these, remember that these are just intuitions in a broad sense and things that are meant to sell the idea of non-separable to you as a new and mostly inexperienced reader. After all, $mathbb{R}$ and $l_{infty}$ have the same cardinality so $l_{infty}$ isn't really larger than $mathbb{R}$ in this sense. An experienced mathematician doesn't really use these intuitions in proofs or reasonings.
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
answered Jan 22 at 12:20
Sorin TircSorin Tirc
1,855213
1,855213
add a comment |
add a comment |
$begingroup$
Re: 2.
Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.
If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{aleph_0}=|Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $fin F$ let $psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X={psi (f):fin F}$. Therefore $|X|=|{psi(f):fin F}|le |F|le |E|le |Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.
On the other hand if the topology $T$ on $X$ is $T={X,emptyset}$ then any non-empty $Dsubset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.
BTW. A topological definition of "$(x_n)_{nin Bbb N}$ converges to $x$" is: ${x}=cap_{nin Bbb N}overline {S_n},$ where $S_n={x_j:jge n}.$
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
Re: 2.
Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.
If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{aleph_0}=|Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $fin F$ let $psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X={psi (f):fin F}$. Therefore $|X|=|{psi(f):fin F}|le |F|le |E|le |Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.
On the other hand if the topology $T$ on $X$ is $T={X,emptyset}$ then any non-empty $Dsubset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.
BTW. A topological definition of "$(x_n)_{nin Bbb N}$ converges to $x$" is: ${x}=cap_{nin Bbb N}overline {S_n},$ where $S_n={x_j:jge n}.$
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
add a comment |
$begingroup$
Re: 2.
Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.
If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{aleph_0}=|Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $fin F$ let $psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X={psi (f):fin F}$. Therefore $|X|=|{psi(f):fin F}|le |F|le |E|le |Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.
On the other hand if the topology $T$ on $X$ is $T={X,emptyset}$ then any non-empty $Dsubset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.
BTW. A topological definition of "$(x_n)_{nin Bbb N}$ converges to $x$" is: ${x}=cap_{nin Bbb N}overline {S_n},$ where $S_n={x_j:jge n}.$
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
$endgroup$
Re: 2.
Not all topologies can be described in terms of sequences. The definition of a topology $T$ on a set $X$ is short and not very restrictive. An important class of topological spaces is the class of sequential spaces, in which the closure of a set can be described in terms of convergent sequences. This class includes all metric spaces.
If $X$ is a sequential space and $D$ is countable dense subset of $X,$ let $E$ be the set of all sequences of members of $D.$ The cardinal $|E|$ of $E$ is at most $2^{aleph_0}=|Bbb R|.$ Let $F$ be the subset of $E$ that contains only the convergent sequences. For any sequence $fin F$ let $psi(f)$ be the limit of the sequence $f.$ Then, since $D$ is dense in $X,$ we have $X={psi (f):fin F}$. Therefore $|X|=|{psi(f):fin F}|le |F|le |E|le |Bbb R|.$ So $X$ is not "too big" in terms of its cardinality.
On the other hand if the topology $T$ on $X$ is $T={X,emptyset}$ then any non-empty $Dsubset X$ is dense, so $X$ is separable. But $X$ can be any set so $|X|$ can be any cardinal. And if $X$ has at least $2$ members then $X$ is not a sequential space.
BTW. A topological definition of "$(x_n)_{nin Bbb N}$ converges to $x$" is: ${x}=cap_{nin Bbb N}overline {S_n},$ where $S_n={x_j:jge n}.$
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
answered Jan 22 at 14:15
DanielWainfleetDanielWainfleet
35.4k31648
35.4k31648
add a comment |
add a comment |
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