Topology on $X$ that is $T_0$ but not $T_1$
Show that if $text{card}(X) geq 2$, there is a topology on $X$ that is $T_0$ but not $T_1$
Consider the point set ${a,b}$ where $b>a$. Then, the topological space of this set is $T_0$ as there is an open neighborhood of one that does not contain the other. But this is not $T_1$ as $b$ won't be closed. By induction, the following holds for $text{card}(X) geq 2$. Is this okay?
general-topology
asked Nov 25 at 13:23
Dom
1627
add a comment |
Show that if $text{card}(X) geq 2$, there is a topology on $X$ that is $T_0$ but not $T_1$
Consider the point set ${a,b}$ where $b>a$. Then, the topological space of this set is $T_0$ as there is an open neighborhood of one that does not contain the other. But this is not $T_1$ as $b$ won't be closed. By induction, the following holds for $text{card}(X) geq 2$. Is this okay?
general-topology
asked Nov 25 at 13:23
Dom
1627
Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11
add a comment |
Show that if $text{card}(X) geq 2$, there is a topology on $X$ that is $T_0$ but not $T_1$
Consider the point set ${a,b}$ where $b>a$. Then, the topological space of this set is $T_0$ as there is an open neighborhood of one that does not contain the other. But this is not $T_1$ as $b$ won't be closed. By induction, the following holds for $text{card}(X) geq 2$. Is this okay?
general-topology
asked Nov 25 at 13:23
Dom
1627
Show that if $text{card}(X) geq 2$, there is a topology on $X$ that is $T_0$ but not $T_1$
Consider the point set ${a,b}$ where $b>a$. Then, the topological space of this set is $T_0$ as there is an open neighborhood of one that does not contain the other. But this is not $T_1$ as $b$ won't be closed. By induction, the following holds for $text{card}(X) geq 2$. Is this okay?
general-topology
general-topology
asked Nov 25 at 13:23
Dom
1627
asked Nov 25 at 13:23
Dom
1627
asked Nov 25 at 13:23
Dom
1627
asked Nov 25 at 13:23
Dom
1627
asked Nov 25 at 13:23
Dom
1627
1627
Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11
add a comment |
Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11
Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11
add a comment |
1 Answer
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No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $mathcal{T} = {A subseteq X: p in A} cup {emptyset}$, which is a topology (the included point topology).
Then $(X,mathcal{T})$ is $T_0$ (if $x neq y$ then one of them is not $p$ (say $x$) and then $Xsetminus{x}$ is open and contains $y$ and not $x$), but not $T_1$ as ${p}$ is not closed, as $Xsetminus {p}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $mathcal{T}_u={emptyset,X} cup{(x,infty): x in X}$, where $(x,infty) = {p in X: x < p}$.
This is $T_0$ because if $x neq y$, then $x < y$ (WLOG) and then $y in (x,infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of ${q}$ when $p < q$ (as chosen above).
answered Nov 25 at 13:36
Henno Brandsma
104k346113
add a comment |
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No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $mathcal{T} = {A subseteq X: p in A} cup {emptyset}$, which is a topology (the included point topology).
Then $(X,mathcal{T})$ is $T_0$ (if $x neq y$ then one of them is not $p$ (say $x$) and then $Xsetminus{x}$ is open and contains $y$ and not $x$), but not $T_1$ as ${p}$ is not closed, as $Xsetminus {p}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $mathcal{T}_u={emptyset,X} cup{(x,infty): x in X}$, where $(x,infty) = {p in X: x < p}$.
This is $T_0$ because if $x neq y$, then $x < y$ (WLOG) and then $y in (x,infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of ${q}$ when $p < q$ (as chosen above).
answered Nov 25 at 13:36
Henno Brandsma
104k346113
add a comment |
No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $mathcal{T} = {A subseteq X: p in A} cup {emptyset}$, which is a topology (the included point topology).
Then $(X,mathcal{T})$ is $T_0$ (if $x neq y$ then one of them is not $p$ (say $x$) and then $Xsetminus{x}$ is open and contains $y$ and not $x$), but not $T_1$ as ${p}$ is not closed, as $Xsetminus {p}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $mathcal{T}_u={emptyset,X} cup{(x,infty): x in X}$, where $(x,infty) = {p in X: x < p}$.
This is $T_0$ because if $x neq y$, then $x < y$ (WLOG) and then $y in (x,infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of ${q}$ when $p < q$ (as chosen above).
answered Nov 25 at 13:36
Henno Brandsma
104k346113
add a comment |
No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $mathcal{T} = {A subseteq X: p in A} cup {emptyset}$, which is a topology (the included point topology).
Then $(X,mathcal{T})$ is $T_0$ (if $x neq y$ then one of them is not $p$ (say $x$) and then $Xsetminus{x}$ is open and contains $y$ and not $x$), but not $T_1$ as ${p}$ is not closed, as $Xsetminus {p}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $mathcal{T}_u={emptyset,X} cup{(x,infty): x in X}$, where $(x,infty) = {p in X: x < p}$.
This is $T_0$ because if $x neq y$, then $x < y$ (WLOG) and then $y in (x,infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of ${q}$ when $p < q$ (as chosen above).
answered Nov 25 at 13:36
Henno Brandsma
104k346113
No induction needed. Let $p$ and $q$ be distinct points of $X$.
Define $mathcal{T} = {A subseteq X: p in A} cup {emptyset}$, which is a topology (the included point topology).
Then $(X,mathcal{T})$ is $T_0$ (if $x neq y$ then one of them is not $p$ (say $x$) and then $Xsetminus{x}$ is open and contains $y$ and not $x$), but not $T_1$ as ${p}$ is not closed, as $Xsetminus {p}$ is not open (it's non-empty because of $q$ and does not contain $p$, so it's not in $mathcal{T}$).
As an alternative, put a linear order on $X$ (uses axiom of choice) and we have two points $p < q$ WLOG.
Define the topology (upper topology) by $mathcal{T}_u={emptyset,X} cup{(x,infty): x in X}$, where $(x,infty) = {p in X: x < p}$.
This is $T_0$ because if $x neq y$, then $x < y$ (WLOG) and then $y in (x,infty)$ and $x$ is not in that set. It is not $T_1$, e.g. because $p$ will always be in the closure of ${q}$ when $p < q$ (as chosen above).
answered Nov 25 at 13:36
Henno Brandsma
104k346113
edited Nov 25 at 14:04
answered Nov 25 at 13:36
Henno Brandsma
104k346113
answered Nov 25 at 13:36
Henno Brandsma
104k346113
answered Nov 25 at 13:36
Henno Brandsma
104k346113
104k346113
add a comment |
add a comment |
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Is the mathematical induction really required?
– Aniruddha Deshmukh
Nov 25 at 13:28
First of all, why can we speak of $b>a$? Secondly, what do you mean by "the topological space"?
– Jonas Lenz
Nov 25 at 13:31
You have to define a topology on $X$ with the properties that are asked for. On any set we can define the discrete, the indiscrete (trivial) topology, included and excluded point topologies etc. On infinite $X$ the cofinite topology and on uncountable ones the co-countable ones. Linear orders on a set induce at least three topologies (upper, lower and their combination, the "linear topology") etc.
– Henno Brandsma
Nov 25 at 14:11