Definition of “Normal topological space”
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I found this definition but i don't know if it is correct or not ?
"A topological space X is normal if for every pair of closed disjoint sets A and B of X there exists $f: Xto [0; 1]$ continuous, such that $f(A) = {0}$ and $f (B) = {1}$"
I need a reference please if it is correct .
general-topology definition
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
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add a comment |
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I found this definition but i don't know if it is correct or not ?
"A topological space X is normal if for every pair of closed disjoint sets A and B of X there exists $f: Xto [0; 1]$ continuous, such that $f(A) = {0}$ and $f (B) = {1}$"
I need a reference please if it is correct .
general-topology definition
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
$endgroup$
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Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
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– Justin Benfield
Jan 17 '18 at 9:21
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@JustinBenfield yes $f$ is supposed continuous
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– Vrouvrou
Jan 17 '18 at 9:22
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Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
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– Justin Benfield
Jan 17 '18 at 9:25
1
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Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
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– bof
Jan 17 '18 at 10:17
1
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@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
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– Paul Sinclair
Jan 17 '18 at 17:20
add a comment |
$begingroup$
I found this definition but i don't know if it is correct or not ?
"A topological space X is normal if for every pair of closed disjoint sets A and B of X there exists $f: Xto [0; 1]$ continuous, such that $f(A) = {0}$ and $f (B) = {1}$"
I need a reference please if it is correct .
general-topology definition
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
$endgroup$
I found this definition but i don't know if it is correct or not ?
"A topological space X is normal if for every pair of closed disjoint sets A and B of X there exists $f: Xto [0; 1]$ continuous, such that $f(A) = {0}$ and $f (B) = {1}$"
I need a reference please if it is correct .
general-topology definition
general-topology definition
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
edited Jan 17 '18 at 9:22
Vrouvrou
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
asked Jan 17 '18 at 9:15
VrouvrouVrouvrou
1,9261822
1,9261822
$begingroup$
Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:21
$begingroup$
@JustinBenfield yes $f$ is supposed continuous
$endgroup$
– Vrouvrou
Jan 17 '18 at 9:22
$begingroup$
Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:25
1
$begingroup$
Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
$endgroup$
– bof
Jan 17 '18 at 10:17
1
$begingroup$
@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
$endgroup$
– Paul Sinclair
Jan 17 '18 at 17:20
add a comment |
$begingroup$
Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:21
$begingroup$
@JustinBenfield yes $f$ is supposed continuous
$endgroup$
– Vrouvrou
Jan 17 '18 at 9:22
$begingroup$
Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:25
1
$begingroup$
Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
$endgroup$
– bof
Jan 17 '18 at 10:17
1
$begingroup$
@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
$endgroup$
– Paul Sinclair
Jan 17 '18 at 17:20
$begingroup$
Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:21
$begingroup$
Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:21
$begingroup$
@JustinBenfield yes $f$ is supposed continuous
$endgroup$
– Vrouvrou
Jan 17 '18 at 9:22
$begingroup$
@JustinBenfield yes $f$ is supposed continuous
$endgroup$
– Vrouvrou
Jan 17 '18 at 9:22
$begingroup$
Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:25
$begingroup$
Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:25
1
1
$begingroup$
Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
$endgroup$
– bof
Jan 17 '18 at 10:17
$begingroup$
Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
$endgroup$
– bof
Jan 17 '18 at 10:17
1
1
$begingroup$
@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
$endgroup$
– Paul Sinclair
Jan 17 '18 at 17:20
$begingroup$
@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
$endgroup$
– Paul Sinclair
Jan 17 '18 at 17:20
add a comment |
1 Answer
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You may take this as the definition of "normal", but is unusual. The standard definition is this:
A topological space $X$ is normal if for every pair of disjoint closed sets $A$ and $B$ of $X$ there exist disjoint open subsets $U$ and $V$ of $X$ such $A subset U$ and $B subset V$ (in other words, if any two disjoint closed subsets $A$ and $B$ of $X$ have disjoint open neighborhoods).
Urysohn's lemma (see bof's comment) says that your definition is equivalent to the standard definition. It was proved by Pavel Samuilovich Urysohn in 1925.
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1 Answer
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$begingroup$
You may take this as the definition of "normal", but is unusual. The standard definition is this:
A topological space $X$ is normal if for every pair of disjoint closed sets $A$ and $B$ of $X$ there exist disjoint open subsets $U$ and $V$ of $X$ such $A subset U$ and $B subset V$ (in other words, if any two disjoint closed subsets $A$ and $B$ of $X$ have disjoint open neighborhoods).
Urysohn's lemma (see bof's comment) says that your definition is equivalent to the standard definition. It was proved by Pavel Samuilovich Urysohn in 1925.
$endgroup$
add a comment |
$begingroup$
You may take this as the definition of "normal", but is unusual. The standard definition is this:
A topological space $X$ is normal if for every pair of disjoint closed sets $A$ and $B$ of $X$ there exist disjoint open subsets $U$ and $V$ of $X$ such $A subset U$ and $B subset V$ (in other words, if any two disjoint closed subsets $A$ and $B$ of $X$ have disjoint open neighborhoods).
Urysohn's lemma (see bof's comment) says that your definition is equivalent to the standard definition. It was proved by Pavel Samuilovich Urysohn in 1925.
$endgroup$
add a comment |
$begingroup$
You may take this as the definition of "normal", but is unusual. The standard definition is this:
A topological space $X$ is normal if for every pair of disjoint closed sets $A$ and $B$ of $X$ there exist disjoint open subsets $U$ and $V$ of $X$ such $A subset U$ and $B subset V$ (in other words, if any two disjoint closed subsets $A$ and $B$ of $X$ have disjoint open neighborhoods).
Urysohn's lemma (see bof's comment) says that your definition is equivalent to the standard definition. It was proved by Pavel Samuilovich Urysohn in 1925.
$endgroup$
You may take this as the definition of "normal", but is unusual. The standard definition is this:
A topological space $X$ is normal if for every pair of disjoint closed sets $A$ and $B$ of $X$ there exist disjoint open subsets $U$ and $V$ of $X$ such $A subset U$ and $B subset V$ (in other words, if any two disjoint closed subsets $A$ and $B$ of $X$ have disjoint open neighborhoods).
Urysohn's lemma (see bof's comment) says that your definition is equivalent to the standard definition. It was proved by Pavel Samuilovich Urysohn in 1925.
answered Dec 7 '18 at 13:44
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Paul Frost
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$begingroup$
Isn't $f$ supposed to be continuous here? Otherwise I don't see how this says anything about the topology of $X$.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:21
$begingroup$
@JustinBenfield yes $f$ is supposed continuous
$endgroup$
– Vrouvrou
Jan 17 '18 at 9:22
$begingroup$
Based on en.wikipedia.org/wiki/Normal_space, it seems that your definition is for a perfectly normal space.
$endgroup$
– Justin Benfield
Jan 17 '18 at 9:25
1
$begingroup$
Yes, the definition you found is correct; see Urysohn's Lemma. But where did you find it? PLEASE SOURCE YOUR QUOTATION.
$endgroup$
– bof
Jan 17 '18 at 10:17
1
$begingroup$
@JustinBenfield - Perfectly normal requires that $f^{-1}(0) = A$ and $f^{-1}(1) = B$. The conditions here are weaker, and by Urysohn's lemma are equivalent to the usual (and IMO, preferable) definition of normal.
$endgroup$
– Paul Sinclair
Jan 17 '18 at 17:20