$X={1,2,3},U={∅,{1},{1,2},{1,2,3}}$ Is $(X,U)$ Hausdorff space?
$begingroup$
Consider $X={1,2,3}$ and a topology
$$U={∅,{1},{1,2},{1,2,3}}$$
Is $(X,U)$ a Hausdorff space?
My solution:
Points $x$ and $y$ in a topological space $X$ can be separated by neighbourhoods if there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $y$ such that $U$ and $V$ are disjoint ($U ∩ V = ∅)$.
$X$ is a Hausdorff space if all distinct points in $X$ are pairwise neighborhood-separable.
So by the defintion we pick element 1 and 2. The "smallest" neighbourhood of 1 is ${1}$, and of 2 is ${1,2}$. And intersection isn't the empty set. So this isn't true.
My question is this approach ok? Can we say that neighbourhood of 1 is ${1}$?
general-topology metric-spaces
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
$endgroup$
add a comment |
$begingroup$
Consider $X={1,2,3}$ and a topology
$$U={∅,{1},{1,2},{1,2,3}}$$
Is $(X,U)$ a Hausdorff space?
My solution:
Points $x$ and $y$ in a topological space $X$ can be separated by neighbourhoods if there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $y$ such that $U$ and $V$ are disjoint ($U ∩ V = ∅)$.
$X$ is a Hausdorff space if all distinct points in $X$ are pairwise neighborhood-separable.
So by the defintion we pick element 1 and 2. The "smallest" neighbourhood of 1 is ${1}$, and of 2 is ${1,2}$. And intersection isn't the empty set. So this isn't true.
My question is this approach ok? Can we say that neighbourhood of 1 is ${1}$?
general-topology metric-spaces
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
$endgroup$
add a comment |
$begingroup$
Consider $X={1,2,3}$ and a topology
$$U={∅,{1},{1,2},{1,2,3}}$$
Is $(X,U)$ a Hausdorff space?
My solution:
Points $x$ and $y$ in a topological space $X$ can be separated by neighbourhoods if there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $y$ such that $U$ and $V$ are disjoint ($U ∩ V = ∅)$.
$X$ is a Hausdorff space if all distinct points in $X$ are pairwise neighborhood-separable.
So by the defintion we pick element 1 and 2. The "smallest" neighbourhood of 1 is ${1}$, and of 2 is ${1,2}$. And intersection isn't the empty set. So this isn't true.
My question is this approach ok? Can we say that neighbourhood of 1 is ${1}$?
general-topology metric-spaces
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
$endgroup$
Consider $X={1,2,3}$ and a topology
$$U={∅,{1},{1,2},{1,2,3}}$$
Is $(X,U)$ a Hausdorff space?
My solution:
Points $x$ and $y$ in a topological space $X$ can be separated by neighbourhoods if there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $y$ such that $U$ and $V$ are disjoint ($U ∩ V = ∅)$.
$X$ is a Hausdorff space if all distinct points in $X$ are pairwise neighborhood-separable.
So by the defintion we pick element 1 and 2. The "smallest" neighbourhood of 1 is ${1}$, and of 2 is ${1,2}$. And intersection isn't the empty set. So this isn't true.
My question is this approach ok? Can we say that neighbourhood of 1 is ${1}$?
general-topology metric-spaces
general-topology metric-spaces
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
edited Dec 4 '18 at 13:30
Xander Henderson
14.3k103554
14.3k103554
asked Dec 4 '18 at 9:22
josfjosf
266216
asked Dec 4 '18 at 9:22
josfjosf
266216
asked Dec 4 '18 at 9:22
josfjosf
266216
266216
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your approach is fine. ${1}$ is definitely a neighborhood of $1$ (that's what it means when ${1}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Your approach is fine. ${1}$ is definitely a neighborhood of $1$ (that's what it means when ${1}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
$endgroup$
add a comment |
$begingroup$
Your approach is fine. ${1}$ is definitely a neighborhood of $1$ (that's what it means when ${1}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
$endgroup$
add a comment |
$begingroup$
Your approach is fine. ${1}$ is definitely a neighborhood of $1$ (that's what it means when ${1}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
$endgroup$
Your approach is fine. ${1}$ is definitely a neighborhood of $1$ (that's what it means when ${1}$ appears in the definition of $U$ like that), and in this case each element actually has a smallest neighborhood, so it suffices to use those to prove / disprove Hausdorff-ness.
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
answered Dec 4 '18 at 9:24
ArthurArthur
113k7110193
113k7110193
add a comment |
add a comment |
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