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Question about “equivalent” definitions for small inductive dimension of topological spaces
$begingroup$
$DeclareMathOperator{ind}{ind}$I've been reading through this document, trying to get a better handle on the interrelationships between various notions of topological dimension, and I came across something that I suspect is incorrect (or at least making use of an unstated assumption). Here is a (slightly paraphrased) excerpt:
Definition 3.1. The small inductive dimension of $X$ is denoted $ind(X),$ and is defined as follows:
We say that $ind(X)=-1$ iff $X=emptyset$.
$ind(X)le n$ if for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1$ (where $partial V$ is the boundary of $V$).
$ind(X)=n$ if $ind(X)le n$ but $ind(X)notleq n-1$.
$ind(X)=infty$ if for every $n,$ $ind(X)notleq n$.
Remark 3.2. An equivalent condition to condition 2 is:
- The space $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind(partial U)le n-1$.
Now, it is easy to see that condition 2 implies this alternate condition, but it seems to me that, unless we know that the space $X$ is regular--i.e., unless we know that for every point $xin X$ and every open $U$ with $xin U$, there is some open $V$ such that $xin V$ and $overline Vsubseteq U$--we can't conclude that the alternate condition implies condition 2. Am I correct about this, or am I missing something?
general-topology dimension-theory
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{ind}{ind}$I've been reading through this document, trying to get a better handle on the interrelationships between various notions of topological dimension, and I came across something that I suspect is incorrect (or at least making use of an unstated assumption). Here is a (slightly paraphrased) excerpt:
Definition 3.1. The small inductive dimension of $X$ is denoted $ind(X),$ and is defined as follows:
We say that $ind(X)=-1$ iff $X=emptyset$.
$ind(X)le n$ if for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1$ (where $partial V$ is the boundary of $V$).
$ind(X)=n$ if $ind(X)le n$ but $ind(X)notleq n-1$.
$ind(X)=infty$ if for every $n,$ $ind(X)notleq n$.
Remark 3.2. An equivalent condition to condition 2 is:
- The space $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind(partial U)le n-1$.
Now, it is easy to see that condition 2 implies this alternate condition, but it seems to me that, unless we know that the space $X$ is regular--i.e., unless we know that for every point $xin X$ and every open $U$ with $xin U$, there is some open $V$ such that $xin V$ and $overline Vsubseteq U$--we can't conclude that the alternate condition implies condition 2. Am I correct about this, or am I missing something?
general-topology dimension-theory
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
$endgroup$
add a comment |
$begingroup$
$DeclareMathOperator{ind}{ind}$I've been reading through this document, trying to get a better handle on the interrelationships between various notions of topological dimension, and I came across something that I suspect is incorrect (or at least making use of an unstated assumption). Here is a (slightly paraphrased) excerpt:
Definition 3.1. The small inductive dimension of $X$ is denoted $ind(X),$ and is defined as follows:
We say that $ind(X)=-1$ iff $X=emptyset$.
$ind(X)le n$ if for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1$ (where $partial V$ is the boundary of $V$).
$ind(X)=n$ if $ind(X)le n$ but $ind(X)notleq n-1$.
$ind(X)=infty$ if for every $n,$ $ind(X)notleq n$.
Remark 3.2. An equivalent condition to condition 2 is:
- The space $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind(partial U)le n-1$.
Now, it is easy to see that condition 2 implies this alternate condition, but it seems to me that, unless we know that the space $X$ is regular--i.e., unless we know that for every point $xin X$ and every open $U$ with $xin U$, there is some open $V$ such that $xin V$ and $overline Vsubseteq U$--we can't conclude that the alternate condition implies condition 2. Am I correct about this, or am I missing something?
general-topology dimension-theory
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
$endgroup$
$DeclareMathOperator{ind}{ind}$I've been reading through this document, trying to get a better handle on the interrelationships between various notions of topological dimension, and I came across something that I suspect is incorrect (or at least making use of an unstated assumption). Here is a (slightly paraphrased) excerpt:
Definition 3.1. The small inductive dimension of $X$ is denoted $ind(X),$ and is defined as follows:
We say that $ind(X)=-1$ iff $X=emptyset$.
$ind(X)le n$ if for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1$ (where $partial V$ is the boundary of $V$).
$ind(X)=n$ if $ind(X)le n$ but $ind(X)notleq n-1$.
$ind(X)=infty$ if for every $n,$ $ind(X)notleq n$.
Remark 3.2. An equivalent condition to condition 2 is:
- The space $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind(partial U)le n-1$.
Now, it is easy to see that condition 2 implies this alternate condition, but it seems to me that, unless we know that the space $X$ is regular--i.e., unless we know that for every point $xin X$ and every open $U$ with $xin U$, there is some open $V$ such that $xin V$ and $overline Vsubseteq U$--we can't conclude that the alternate condition implies condition 2. Am I correct about this, or am I missing something?
general-topology dimension-theory
general-topology dimension-theory
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
edited Sep 6 '13 at 3:41
Cameron Buie
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
asked Sep 5 '13 at 5:04
Cameron BuieCameron Buie
85.6k772160
85.6k772160
add a comment |
add a comment |
2 Answers
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oldest
votes
$begingroup$
$DeclareMathOperator{ind}{ind}$Of course, if I'd simply thought about it a bit more before posting, I'd not have asked the question in the first place. Ah, well. Hopefully, this will help other users in the future.
To distinguish between the two versions of small inductive dimension, I will let $ind(cdot)$ represent the initially presented version, and $ind'(cdot)$ represent the alternate version.
Proposition: Given an integer $nge-1$, the following are equivalent.
$ind(X)le n$--meaning that for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1.$
- The space $X$ is regular, and $ind'(X)le n$--meaning that $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind'(partial U)le n-1$.
Proof: We proceed inductively on $n$, with the $n=-1$ case immediate. Suppose that $n$ is the least integer greater than or equal to $-1$ for which we have not yet concluded that the proposition holds.
On the one hand, suppose $ind(X)le n,$ and let $$mathcal B:={Vsubseteq X:Vtext{ open and }ind(partial V)le n-1}.$$ By inductive hypothesis, $ind'(partial V)le n-1$ for each $Vinmathcal B,$ and it is readily shown from $ind(X)le n$ that $mathcal B$ is a basis for the topology on $X$. Also, taking any $xin X$ and any open $U$ with $xin U,$ there is an open $V$ with $xin V$ and $overline Vsubseteq U$, and so $X$ is regular. [Poster's note: I can't believe I missed that.]
On the other hand, suppose that $X$ is regular, and $ind'(X)le n.$ Take any $xin X$ and any open $U$ with $xin U$. Since $X$ is regular and $x$ lies in the open set $U,$ then there is some open $W$ such that $xin W$ and $overline Wsubseteq U$. Since $x$ lies in the open set $W$ and $mathcal B$ is a basis for the topology on $X,$ then there is some $Vinmathcal B$ such that $xin Vsubseteq W$. Then $overline Vsubseteqoverline Wsubseteq U$, and since $Vinmathcal B,$ then $ind'(partial V)le n-1.$ By inductive hypothesis, $ind(partial V)le n-1.$ Thus, $ind(X)le n$. $Box$
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
$endgroup$
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
add a comment |
$begingroup$
Yes, you are right. As I know small inductive dimension is usually defined only for regular spaces. Similary large inductive dimension is usually defined only for normal spaces and covering dimension only for completely regular spaces.
Also note that for being zero-dimensional the conditions are always equivalent and it implies regularity.
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
$endgroup$
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$DeclareMathOperator{ind}{ind}$Of course, if I'd simply thought about it a bit more before posting, I'd not have asked the question in the first place. Ah, well. Hopefully, this will help other users in the future.
To distinguish between the two versions of small inductive dimension, I will let $ind(cdot)$ represent the initially presented version, and $ind'(cdot)$ represent the alternate version.
Proposition: Given an integer $nge-1$, the following are equivalent.
$ind(X)le n$--meaning that for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1.$
- The space $X$ is regular, and $ind'(X)le n$--meaning that $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind'(partial U)le n-1$.
Proof: We proceed inductively on $n$, with the $n=-1$ case immediate. Suppose that $n$ is the least integer greater than or equal to $-1$ for which we have not yet concluded that the proposition holds.
On the one hand, suppose $ind(X)le n,$ and let $$mathcal B:={Vsubseteq X:Vtext{ open and }ind(partial V)le n-1}.$$ By inductive hypothesis, $ind'(partial V)le n-1$ for each $Vinmathcal B,$ and it is readily shown from $ind(X)le n$ that $mathcal B$ is a basis for the topology on $X$. Also, taking any $xin X$ and any open $U$ with $xin U,$ there is an open $V$ with $xin V$ and $overline Vsubseteq U$, and so $X$ is regular. [Poster's note: I can't believe I missed that.]
On the other hand, suppose that $X$ is regular, and $ind'(X)le n.$ Take any $xin X$ and any open $U$ with $xin U$. Since $X$ is regular and $x$ lies in the open set $U,$ then there is some open $W$ such that $xin W$ and $overline Wsubseteq U$. Since $x$ lies in the open set $W$ and $mathcal B$ is a basis for the topology on $X,$ then there is some $Vinmathcal B$ such that $xin Vsubseteq W$. Then $overline Vsubseteqoverline Wsubseteq U$, and since $Vinmathcal B,$ then $ind'(partial V)le n-1.$ By inductive hypothesis, $ind(partial V)le n-1.$ Thus, $ind(X)le n$. $Box$
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
$endgroup$
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
add a comment |
$begingroup$
$DeclareMathOperator{ind}{ind}$Of course, if I'd simply thought about it a bit more before posting, I'd not have asked the question in the first place. Ah, well. Hopefully, this will help other users in the future.
To distinguish between the two versions of small inductive dimension, I will let $ind(cdot)$ represent the initially presented version, and $ind'(cdot)$ represent the alternate version.
Proposition: Given an integer $nge-1$, the following are equivalent.
$ind(X)le n$--meaning that for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1.$
- The space $X$ is regular, and $ind'(X)le n$--meaning that $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind'(partial U)le n-1$.
Proof: We proceed inductively on $n$, with the $n=-1$ case immediate. Suppose that $n$ is the least integer greater than or equal to $-1$ for which we have not yet concluded that the proposition holds.
On the one hand, suppose $ind(X)le n,$ and let $$mathcal B:={Vsubseteq X:Vtext{ open and }ind(partial V)le n-1}.$$ By inductive hypothesis, $ind'(partial V)le n-1$ for each $Vinmathcal B,$ and it is readily shown from $ind(X)le n$ that $mathcal B$ is a basis for the topology on $X$. Also, taking any $xin X$ and any open $U$ with $xin U,$ there is an open $V$ with $xin V$ and $overline Vsubseteq U$, and so $X$ is regular. [Poster's note: I can't believe I missed that.]
On the other hand, suppose that $X$ is regular, and $ind'(X)le n.$ Take any $xin X$ and any open $U$ with $xin U$. Since $X$ is regular and $x$ lies in the open set $U,$ then there is some open $W$ such that $xin W$ and $overline Wsubseteq U$. Since $x$ lies in the open set $W$ and $mathcal B$ is a basis for the topology on $X,$ then there is some $Vinmathcal B$ such that $xin Vsubseteq W$. Then $overline Vsubseteqoverline Wsubseteq U$, and since $Vinmathcal B,$ then $ind'(partial V)le n-1.$ By inductive hypothesis, $ind(partial V)le n-1.$ Thus, $ind(X)le n$. $Box$
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
$endgroup$
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
add a comment |
$begingroup$
$DeclareMathOperator{ind}{ind}$Of course, if I'd simply thought about it a bit more before posting, I'd not have asked the question in the first place. Ah, well. Hopefully, this will help other users in the future.
To distinguish between the two versions of small inductive dimension, I will let $ind(cdot)$ represent the initially presented version, and $ind'(cdot)$ represent the alternate version.
Proposition: Given an integer $nge-1$, the following are equivalent.
$ind(X)le n$--meaning that for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1.$
- The space $X$ is regular, and $ind'(X)le n$--meaning that $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind'(partial U)le n-1$.
Proof: We proceed inductively on $n$, with the $n=-1$ case immediate. Suppose that $n$ is the least integer greater than or equal to $-1$ for which we have not yet concluded that the proposition holds.
On the one hand, suppose $ind(X)le n,$ and let $$mathcal B:={Vsubseteq X:Vtext{ open and }ind(partial V)le n-1}.$$ By inductive hypothesis, $ind'(partial V)le n-1$ for each $Vinmathcal B,$ and it is readily shown from $ind(X)le n$ that $mathcal B$ is a basis for the topology on $X$. Also, taking any $xin X$ and any open $U$ with $xin U,$ there is an open $V$ with $xin V$ and $overline Vsubseteq U$, and so $X$ is regular. [Poster's note: I can't believe I missed that.]
On the other hand, suppose that $X$ is regular, and $ind'(X)le n.$ Take any $xin X$ and any open $U$ with $xin U$. Since $X$ is regular and $x$ lies in the open set $U,$ then there is some open $W$ such that $xin W$ and $overline Wsubseteq U$. Since $x$ lies in the open set $W$ and $mathcal B$ is a basis for the topology on $X,$ then there is some $Vinmathcal B$ such that $xin Vsubseteq W$. Then $overline Vsubseteqoverline Wsubseteq U$, and since $Vinmathcal B,$ then $ind'(partial V)le n-1.$ By inductive hypothesis, $ind(partial V)le n-1.$ Thus, $ind(X)le n$. $Box$
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
$endgroup$
$DeclareMathOperator{ind}{ind}$Of course, if I'd simply thought about it a bit more before posting, I'd not have asked the question in the first place. Ah, well. Hopefully, this will help other users in the future.
To distinguish between the two versions of small inductive dimension, I will let $ind(cdot)$ represent the initially presented version, and $ind'(cdot)$ represent the alternate version.
Proposition: Given an integer $nge-1$, the following are equivalent.
$ind(X)le n$--meaning that for every point $xin X$ and for every open set $U$ such that $xin U,$ there exists an open $V$ with $xin V$ such that $overline Vsubseteq U$ and $ind(partial V)le n-1.$
- The space $X$ is regular, and $ind'(X)le n$--meaning that $X$ has a basis $mathcal B$ such that for every $Uinmathcal B$ we have $ind'(partial U)le n-1$.
Proof: We proceed inductively on $n$, with the $n=-1$ case immediate. Suppose that $n$ is the least integer greater than or equal to $-1$ for which we have not yet concluded that the proposition holds.
On the one hand, suppose $ind(X)le n,$ and let $$mathcal B:={Vsubseteq X:Vtext{ open and }ind(partial V)le n-1}.$$ By inductive hypothesis, $ind'(partial V)le n-1$ for each $Vinmathcal B,$ and it is readily shown from $ind(X)le n$ that $mathcal B$ is a basis for the topology on $X$. Also, taking any $xin X$ and any open $U$ with $xin U,$ there is an open $V$ with $xin V$ and $overline Vsubseteq U$, and so $X$ is regular. [Poster's note: I can't believe I missed that.]
On the other hand, suppose that $X$ is regular, and $ind'(X)le n.$ Take any $xin X$ and any open $U$ with $xin U$. Since $X$ is regular and $x$ lies in the open set $U,$ then there is some open $W$ such that $xin W$ and $overline Wsubseteq U$. Since $x$ lies in the open set $W$ and $mathcal B$ is a basis for the topology on $X,$ then there is some $Vinmathcal B$ such that $xin Vsubseteq W$. Then $overline Vsubseteqoverline Wsubseteq U$, and since $Vinmathcal B,$ then $ind'(partial V)le n-1.$ By inductive hypothesis, $ind(partial V)le n-1.$ Thus, $ind(X)le n$. $Box$
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
edited Jan 22 at 0:39
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
answered Sep 6 '13 at 4:20
Cameron BuieCameron Buie
85.6k772160
85.6k772160
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
add a comment |
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
1
1
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
$begingroup$
"Hopefully, this will help other users in the future." More than 4 years later but it did help me!
$endgroup$
– Alessandro Codenotti
Oct 29 '17 at 22:05
add a comment |
$begingroup$
Yes, you are right. As I know small inductive dimension is usually defined only for regular spaces. Similary large inductive dimension is usually defined only for normal spaces and covering dimension only for completely regular spaces.
Also note that for being zero-dimensional the conditions are always equivalent and it implies regularity.
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
$endgroup$
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
add a comment |
$begingroup$
Yes, you are right. As I know small inductive dimension is usually defined only for regular spaces. Similary large inductive dimension is usually defined only for normal spaces and covering dimension only for completely regular spaces.
Also note that for being zero-dimensional the conditions are always equivalent and it implies regularity.
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
$endgroup$
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
add a comment |
$begingroup$
Yes, you are right. As I know small inductive dimension is usually defined only for regular spaces. Similary large inductive dimension is usually defined only for normal spaces and covering dimension only for completely regular spaces.
Also note that for being zero-dimensional the conditions are always equivalent and it implies regularity.
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
$endgroup$
Yes, you are right. As I know small inductive dimension is usually defined only for regular spaces. Similary large inductive dimension is usually defined only for normal spaces and covering dimension only for completely regular spaces.
Also note that for being zero-dimensional the conditions are always equivalent and it implies regularity.
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
answered Sep 5 '13 at 8:00
user87690user87690
6,5761825
6,5761825
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
add a comment |
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
$begingroup$
(+1) True, zero-dimensionality makes the conditions equivalent (and in fact implies complete regularity), but I was wondering if regularity was both necessary and sufficient for the definitions to agree. It turns out that it is, and I feel silly for not realizing it sooner. Proof to follow. Also, I've seen small inductive dimension and covering dimension defined for arbitrary spaces (though spaces with finite large inductive dimension are immediately normal), so I don't think those conventions are necessarily universal.
$endgroup$
– Cameron Buie
Sep 6 '13 at 3:30
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