When is a subspace of a Scott space itself a Scott space?

0

$begingroup$

Suppose $P$ and $Q subseteq P$ are posets, and let $tau$ and $rho$ be their respective Scott topologies. Now $Q$ is also equipped with the subspace topology $tauvert_Q$ inherited from $P$. It is easy to see that:

$$tauvert_Q subseteq rho$$

I have not found an example when the reverse inclusion is not also true. So my question is:

When do the two topologies on $Q$ coincide?

general-topology order-theory

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

$endgroup$

add a comment | 

0

$begingroup$

Suppose $P$ and $Q subseteq P$ are posets, and let $tau$ and $rho$ be their respective Scott topologies. Now $Q$ is also equipped with the subspace topology $tauvert_Q$ inherited from $P$. It is easy to see that:

$$tauvert_Q subseteq rho$$

I have not found an example when the reverse inclusion is not also true. So my question is:

When do the two topologies on $Q$ coincide?

general-topology order-theory

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

$endgroup$

add a comment | 

0

0

0

$begingroup$

Suppose $P$ and $Q subseteq P$ are posets, and let $tau$ and $rho$ be their respective Scott topologies. Now $Q$ is also equipped with the subspace topology $tauvert_Q$ inherited from $P$. It is easy to see that:

$$tauvert_Q subseteq rho$$

I have not found an example when the reverse inclusion is not also true. So my question is:

When do the two topologies on $Q$ coincide?

general-topology order-theory

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

$endgroup$

Suppose $P$ and $Q subseteq P$ are posets, and let $tau$ and $rho$ be their respective Scott topologies. Now $Q$ is also equipped with the subspace topology $tauvert_Q$ inherited from $P$. It is easy to see that:

$$tauvert_Q subseteq rho$$

I have not found an example when the reverse inclusion is not also true. So my question is:

When do the two topologies on $Q$ coincide?

general-topology order-theory

general-topology order-theory

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

share|cite|improve this question

share|cite|improve this question

edited Jan 22 at 9:55

Bernard Hurley

edited Jan 22 at 9:55

Bernard Hurley

edited Jan 22 at 9:55

Bernard Hurley

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

asked Jan 22 at 9:13

Bernard HurleyBernard Hurley

1787

1787

add a comment | 

add a comment | 

1 Answer
1

active

oldest

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0

$begingroup$

I have found an example of posets $P$ and $Q$, as above where:

$$tauvert_Q neqrho$$

Let $P=mathcal{C}(mathbb{N})$, the power set of the natural numbers ordered by set inclusion. Let $Q$ be the subposet consisting of cofinite elements of $P$. That is:

$$Q={xin P : complement x mbox{is finite} }$$

where $complement x$ denotes the complement of $x$ in $mathbb{N}$.

Now $Q$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $S$ in $Q$ contains its supremum $bigcup S$. To see this pick $x_0in S$. If $x_0 = bigcup S$, we are done. If not, pick $yin Ssetminusdownarrow{x_0}$. Since $S$ is directed there is $x_1in S$ such that $x_0subsetneq x_1$ and $ysubseteq x_1$. Repeating the process yields an ascending chain $x_0, x_1ldots$ of elements that must be finite by ACC. So, for some $n$, $bigcup S = x_n in S$, as claimed.

Since every directed set in $Q$ contains its supremum, it follows that every up-set is open in its Scott topology $rho$. So setting

$$B={complement {0}, mathbb{N}}$$

we see that $Binrho$. I.e. it is open in the Scott topology on $Q$. Now $P$ is algebraic and its compact elements are precisely the finite subsets of $mathbb{N}$. Thus:

$$mathcal{B} = {uparrow{x} : xin P mbox{is finite}}$$

is a basis for $tau$, the Scott topology on $P$. Clearly each element of the basis contains an element of the form $complement {n}in Q$ where $nge 1in mathbb{N}$, and the same must be true for any $Cintau$. Since $B$ does not contain such an element it follows that for all $Cintau$, $Ccap Qneq B$. We conclude that the two topologies $rho$ and $tauvert_Q$ do not coincide.

share|cite|improve this answer

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

$endgroup$

add a comment | 

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1 Answer
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1 Answer
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0

$begingroup$

I have found an example of posets $P$ and $Q$, as above where:

$$tauvert_Q neqrho$$

Let $P=mathcal{C}(mathbb{N})$, the power set of the natural numbers ordered by set inclusion. Let $Q$ be the subposet consisting of cofinite elements of $P$. That is:

$$Q={xin P : complement x mbox{is finite} }$$

where $complement x$ denotes the complement of $x$ in $mathbb{N}$.

Now $Q$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $S$ in $Q$ contains its supremum $bigcup S$. To see this pick $x_0in S$. If $x_0 = bigcup S$, we are done. If not, pick $yin Ssetminusdownarrow{x_0}$. Since $S$ is directed there is $x_1in S$ such that $x_0subsetneq x_1$ and $ysubseteq x_1$. Repeating the process yields an ascending chain $x_0, x_1ldots$ of elements that must be finite by ACC. So, for some $n$, $bigcup S = x_n in S$, as claimed.

Since every directed set in $Q$ contains its supremum, it follows that every up-set is open in its Scott topology $rho$. So setting

$$B={complement {0}, mathbb{N}}$$

we see that $Binrho$. I.e. it is open in the Scott topology on $Q$. Now $P$ is algebraic and its compact elements are precisely the finite subsets of $mathbb{N}$. Thus:

$$mathcal{B} = {uparrow{x} : xin P mbox{is finite}}$$

is a basis for $tau$, the Scott topology on $P$. Clearly each element of the basis contains an element of the form $complement {n}in Q$ where $nge 1in mathbb{N}$, and the same must be true for any $Cintau$. Since $B$ does not contain such an element it follows that for all $Cintau$, $Ccap Qneq B$. We conclude that the two topologies $rho$ and $tauvert_Q$ do not coincide.

share|cite|improve this answer

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

$endgroup$

add a comment | 

0

$begingroup$

I have found an example of posets $P$ and $Q$, as above where:

$$tauvert_Q neqrho$$

Let $P=mathcal{C}(mathbb{N})$, the power set of the natural numbers ordered by set inclusion. Let $Q$ be the subposet consisting of cofinite elements of $P$. That is:

$$Q={xin P : complement x mbox{is finite} }$$

where $complement x$ denotes the complement of $x$ in $mathbb{N}$.

Now $Q$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $S$ in $Q$ contains its supremum $bigcup S$. To see this pick $x_0in S$. If $x_0 = bigcup S$, we are done. If not, pick $yin Ssetminusdownarrow{x_0}$. Since $S$ is directed there is $x_1in S$ such that $x_0subsetneq x_1$ and $ysubseteq x_1$. Repeating the process yields an ascending chain $x_0, x_1ldots$ of elements that must be finite by ACC. So, for some $n$, $bigcup S = x_n in S$, as claimed.

Since every directed set in $Q$ contains its supremum, it follows that every up-set is open in its Scott topology $rho$. So setting

$$B={complement {0}, mathbb{N}}$$

we see that $Binrho$. I.e. it is open in the Scott topology on $Q$. Now $P$ is algebraic and its compact elements are precisely the finite subsets of $mathbb{N}$. Thus:

$$mathcal{B} = {uparrow{x} : xin P mbox{is finite}}$$

is a basis for $tau$, the Scott topology on $P$. Clearly each element of the basis contains an element of the form $complement {n}in Q$ where $nge 1in mathbb{N}$, and the same must be true for any $Cintau$. Since $B$ does not contain such an element it follows that for all $Cintau$, $Ccap Qneq B$. We conclude that the two topologies $rho$ and $tauvert_Q$ do not coincide.

share|cite|improve this answer

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

$endgroup$

add a comment | 

0

0

0

$begingroup$

I have found an example of posets $P$ and $Q$, as above where:

$$tauvert_Q neqrho$$

Let $P=mathcal{C}(mathbb{N})$, the power set of the natural numbers ordered by set inclusion. Let $Q$ be the subposet consisting of cofinite elements of $P$. That is:

$$Q={xin P : complement x mbox{is finite} }$$

where $complement x$ denotes the complement of $x$ in $mathbb{N}$.

Now $Q$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $S$ in $Q$ contains its supremum $bigcup S$. To see this pick $x_0in S$. If $x_0 = bigcup S$, we are done. If not, pick $yin Ssetminusdownarrow{x_0}$. Since $S$ is directed there is $x_1in S$ such that $x_0subsetneq x_1$ and $ysubseteq x_1$. Repeating the process yields an ascending chain $x_0, x_1ldots$ of elements that must be finite by ACC. So, for some $n$, $bigcup S = x_n in S$, as claimed.

Since every directed set in $Q$ contains its supremum, it follows that every up-set is open in its Scott topology $rho$. So setting

$$B={complement {0}, mathbb{N}}$$

we see that $Binrho$. I.e. it is open in the Scott topology on $Q$. Now $P$ is algebraic and its compact elements are precisely the finite subsets of $mathbb{N}$. Thus:

$$mathcal{B} = {uparrow{x} : xin P mbox{is finite}}$$

is a basis for $tau$, the Scott topology on $P$. Clearly each element of the basis contains an element of the form $complement {n}in Q$ where $nge 1in mathbb{N}$, and the same must be true for any $Cintau$. Since $B$ does not contain such an element it follows that for all $Cintau$, $Ccap Qneq B$. We conclude that the two topologies $rho$ and $tauvert_Q$ do not coincide.

share|cite|improve this answer

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

$endgroup$

I have found an example of posets $P$ and $Q$, as above where:

$$tauvert_Q neqrho$$

Let $P=mathcal{C}(mathbb{N})$, the power set of the natural numbers ordered by set inclusion. Let $Q$ be the subposet consisting of cofinite elements of $P$. That is:

$$Q={xin P : complement x mbox{is finite} }$$

where $complement x$ denotes the complement of $x$ in $mathbb{N}$.

Now $Q$ clearly satisfies the Ascending Chain Condition. It follows that every directed set $S$ in $Q$ contains its supremum $bigcup S$. To see this pick $x_0in S$. If $x_0 = bigcup S$, we are done. If not, pick $yin Ssetminusdownarrow{x_0}$. Since $S$ is directed there is $x_1in S$ such that $x_0subsetneq x_1$ and $ysubseteq x_1$. Repeating the process yields an ascending chain $x_0, x_1ldots$ of elements that must be finite by ACC. So, for some $n$, $bigcup S = x_n in S$, as claimed.

Since every directed set in $Q$ contains its supremum, it follows that every up-set is open in its Scott topology $rho$. So setting

$$B={complement {0}, mathbb{N}}$$

we see that $Binrho$. I.e. it is open in the Scott topology on $Q$. Now $P$ is algebraic and its compact elements are precisely the finite subsets of $mathbb{N}$. Thus:

$$mathcal{B} = {uparrow{x} : xin P mbox{is finite}}$$

is a basis for $tau$, the Scott topology on $P$. Clearly each element of the basis contains an element of the form $complement {n}in Q$ where $nge 1in mathbb{N}$, and the same must be true for any $Cintau$. Since $B$ does not contain such an element it follows that for all $Cintau$, $Ccap Qneq B$. We conclude that the two topologies $rho$ and $tauvert_Q$ do not coincide.

share|cite|improve this answer

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

share|cite|improve this answer

share|cite|improve this answer

edited Jan 23 at 11:57

edited Jan 23 at 11:57

edited Jan 23 at 11:57

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

answered Jan 23 at 5:06

Bernard HurleyBernard Hurley

1787

1787

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

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