The notion of $2^S$ in topological space.
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Definition: A topological space is a set $S$ together with a collection $u$ of subsets of $S$ (that is, $u$ is a subset of $2^S$ ) satisfying the following conditions: ...
This is a definition of topological space in "Lecture Notes on Elementary Topology and Geometry" on page 6. I'm quite confused about the words in the bracket. Since here, $2^S$ denotes the collection of all the subsets of $S$, how to understand the subset of the collection of all subset of $S$? Does it mean that $u in 2^S$ rather than $u subset 2^S$?
general-topology definition
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
$endgroup$
add a comment |
$begingroup$
Definition: A topological space is a set $S$ together with a collection $u$ of subsets of $S$ (that is, $u$ is a subset of $2^S$ ) satisfying the following conditions: ...
This is a definition of topological space in "Lecture Notes on Elementary Topology and Geometry" on page 6. I'm quite confused about the words in the bracket. Since here, $2^S$ denotes the collection of all the subsets of $S$, how to understand the subset of the collection of all subset of $S$? Does it mean that $u in 2^S$ rather than $u subset 2^S$?
general-topology definition
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
$endgroup$
1
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
1
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59
add a comment |
$begingroup$
Definition: A topological space is a set $S$ together with a collection $u$ of subsets of $S$ (that is, $u$ is a subset of $2^S$ ) satisfying the following conditions: ...
This is a definition of topological space in "Lecture Notes on Elementary Topology and Geometry" on page 6. I'm quite confused about the words in the bracket. Since here, $2^S$ denotes the collection of all the subsets of $S$, how to understand the subset of the collection of all subset of $S$? Does it mean that $u in 2^S$ rather than $u subset 2^S$?
general-topology definition
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
$endgroup$
Definition: A topological space is a set $S$ together with a collection $u$ of subsets of $S$ (that is, $u$ is a subset of $2^S$ ) satisfying the following conditions: ...
This is a definition of topological space in "Lecture Notes on Elementary Topology and Geometry" on page 6. I'm quite confused about the words in the bracket. Since here, $2^S$ denotes the collection of all the subsets of $S$, how to understand the subset of the collection of all subset of $S$? Does it mean that $u in 2^S$ rather than $u subset 2^S$?
general-topology definition
general-topology definition
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
edited Dec 27 '18 at 16:58
J.Guo
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
asked Dec 27 '18 at 16:41
J.GuoJ.Guo
4249
4249
1
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
1
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59
add a comment |
1
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
1
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59
1
1
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
1
1
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59
add a comment |
1 Answer
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$begingroup$
No. It means that $usubset2^S$. In other words, $u$ consists of subsets of $S$. Asserting that $uin2^S$ would mean that $u$ is a subset of $S$ instead.
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
add a comment |
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1 Answer
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$begingroup$
No. It means that $usubset2^S$. In other words, $u$ consists of subsets of $S$. Asserting that $uin2^S$ would mean that $u$ is a subset of $S$ instead.
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
add a comment |
$begingroup$
No. It means that $usubset2^S$. In other words, $u$ consists of subsets of $S$. Asserting that $uin2^S$ would mean that $u$ is a subset of $S$ instead.
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
add a comment |
$begingroup$
No. It means that $usubset2^S$. In other words, $u$ consists of subsets of $S$. Asserting that $uin2^S$ would mean that $u$ is a subset of $S$ instead.
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
$endgroup$
No. It means that $usubset2^S$. In other words, $u$ consists of subsets of $S$. Asserting that $uin2^S$ would mean that $u$ is a subset of $S$ instead.
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
answered Dec 27 '18 at 16:43
José Carlos SantosJosé Carlos Santos
168k23132236
168k23132236
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
add a comment |
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
So , an element in $u$ is a 'set' of $2^S$ rather than an element of $S$ , is it right ?
$endgroup$
– J.Guo
Dec 27 '18 at 16:48
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
$begingroup$
I'd rather say that an element of $u$ is a subset of $S$, which is the same thing as asserting that it is an element of $2^S$.
$endgroup$
– José Carlos Santos
Dec 27 '18 at 16:53
add a comment |
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1
$begingroup$
It means $usubseteq 2^S$, as it states. You could write that as $uin 2^{2^S}$ if your prefer.
$endgroup$
– Lord Shark the Unknown
Dec 27 '18 at 16:42
$begingroup$
Is it right that if $a$ is an element of $u$ , then $a$ is an element of $2^S$ rather than an element of $S$
$endgroup$
– J.Guo
Dec 27 '18 at 16:45
1
$begingroup$
I'm voting to close this question as off-topic because it is not a question about the definition of topological spaces, as stated in the title of the question. Rather, it is a question about the use of notation 2^S.
$endgroup$
– user126154
Dec 27 '18 at 16:54
$begingroup$
Thank you , I've just edited it.
$endgroup$
– J.Guo
Dec 27 '18 at 16:59