Are segments in a Partially ordered space closed/open?
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I've been trying to figure out what I can say about the following definition:
https://en.wikipedia.org/wiki/Partially_ordered_space
A topological space $X$ with a partial order $leq$, is called a partially ordered space if it's graph $Gr(leq)subset Xtimes X$, is a closed subset of $Xtimes X$.
Are the following sets open/closed in $X$?:
$l(x)={y in X: xleq y }$ and $r(x)={y in X: yleq x }$
In general, for a topological space $X$ with a binary relation whose graph is closed, can I say that $l_R(x)={ yin X: (x,y)in R }$ or $r_R(x)={ yin X: (y,x)in R }$ are closed/ open?
general-topology order-theory
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
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add a comment |
$begingroup$
I've been trying to figure out what I can say about the following definition:
https://en.wikipedia.org/wiki/Partially_ordered_space
A topological space $X$ with a partial order $leq$, is called a partially ordered space if it's graph $Gr(leq)subset Xtimes X$, is a closed subset of $Xtimes X$.
Are the following sets open/closed in $X$?:
$l(x)={y in X: xleq y }$ and $r(x)={y in X: yleq x }$
In general, for a topological space $X$ with a binary relation whose graph is closed, can I say that $l_R(x)={ yin X: (x,y)in R }$ or $r_R(x)={ yin X: (y,x)in R }$ are closed/ open?
general-topology order-theory
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
$endgroup$
add a comment |
$begingroup$
I've been trying to figure out what I can say about the following definition:
https://en.wikipedia.org/wiki/Partially_ordered_space
A topological space $X$ with a partial order $leq$, is called a partially ordered space if it's graph $Gr(leq)subset Xtimes X$, is a closed subset of $Xtimes X$.
Are the following sets open/closed in $X$?:
$l(x)={y in X: xleq y }$ and $r(x)={y in X: yleq x }$
In general, for a topological space $X$ with a binary relation whose graph is closed, can I say that $l_R(x)={ yin X: (x,y)in R }$ or $r_R(x)={ yin X: (y,x)in R }$ are closed/ open?
general-topology order-theory
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
$endgroup$
I've been trying to figure out what I can say about the following definition:
https://en.wikipedia.org/wiki/Partially_ordered_space
A topological space $X$ with a partial order $leq$, is called a partially ordered space if it's graph $Gr(leq)subset Xtimes X$, is a closed subset of $Xtimes X$.
Are the following sets open/closed in $X$?:
$l(x)={y in X: xleq y }$ and $r(x)={y in X: yleq x }$
In general, for a topological space $X$ with a binary relation whose graph is closed, can I say that $l_R(x)={ yin X: (x,y)in R }$ or $r_R(x)={ yin X: (y,x)in R }$ are closed/ open?
general-topology order-theory
general-topology order-theory
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
asked Jan 21 at 13:04
Keen-ameteurKeen-ameteur
1,455316
1,455316
add a comment |
add a comment |
1 Answer
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Consider following embeddings
$$alpha(x):Xto Xtimes X$$
$$alpha(x)(y)=(x,y)$$
$$beta(x):Xto Xtimes X$$
$$beta(x)(y)=(y,x)$$
Both are continuous for any $x$. Therefore if $Gr(leq)$ is closed then so is the preimage. But $alpha(x)^{-1}(Gr(leq))=l(x)$ and $beta(x)^{-1}(Gr(leq))=r(x)$. Proving that both subsets are closed.
The choice of $Gr(leq)$ is clearly irrelevant in this context. It will work for any binary relation which is closed in $Xtimes X$. Also it will work if we switch "closed" to "open".
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
$endgroup$
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
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@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider following embeddings
$$alpha(x):Xto Xtimes X$$
$$alpha(x)(y)=(x,y)$$
$$beta(x):Xto Xtimes X$$
$$beta(x)(y)=(y,x)$$
Both are continuous for any $x$. Therefore if $Gr(leq)$ is closed then so is the preimage. But $alpha(x)^{-1}(Gr(leq))=l(x)$ and $beta(x)^{-1}(Gr(leq))=r(x)$. Proving that both subsets are closed.
The choice of $Gr(leq)$ is clearly irrelevant in this context. It will work for any binary relation which is closed in $Xtimes X$. Also it will work if we switch "closed" to "open".
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
$endgroup$
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
add a comment |
$begingroup$
Consider following embeddings
$$alpha(x):Xto Xtimes X$$
$$alpha(x)(y)=(x,y)$$
$$beta(x):Xto Xtimes X$$
$$beta(x)(y)=(y,x)$$
Both are continuous for any $x$. Therefore if $Gr(leq)$ is closed then so is the preimage. But $alpha(x)^{-1}(Gr(leq))=l(x)$ and $beta(x)^{-1}(Gr(leq))=r(x)$. Proving that both subsets are closed.
The choice of $Gr(leq)$ is clearly irrelevant in this context. It will work for any binary relation which is closed in $Xtimes X$. Also it will work if we switch "closed" to "open".
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
$endgroup$
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
add a comment |
$begingroup$
Consider following embeddings
$$alpha(x):Xto Xtimes X$$
$$alpha(x)(y)=(x,y)$$
$$beta(x):Xto Xtimes X$$
$$beta(x)(y)=(y,x)$$
Both are continuous for any $x$. Therefore if $Gr(leq)$ is closed then so is the preimage. But $alpha(x)^{-1}(Gr(leq))=l(x)$ and $beta(x)^{-1}(Gr(leq))=r(x)$. Proving that both subsets are closed.
The choice of $Gr(leq)$ is clearly irrelevant in this context. It will work for any binary relation which is closed in $Xtimes X$. Also it will work if we switch "closed" to "open".
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
$endgroup$
Consider following embeddings
$$alpha(x):Xto Xtimes X$$
$$alpha(x)(y)=(x,y)$$
$$beta(x):Xto Xtimes X$$
$$beta(x)(y)=(y,x)$$
Both are continuous for any $x$. Therefore if $Gr(leq)$ is closed then so is the preimage. But $alpha(x)^{-1}(Gr(leq))=l(x)$ and $beta(x)^{-1}(Gr(leq))=r(x)$. Proving that both subsets are closed.
The choice of $Gr(leq)$ is clearly irrelevant in this context. It will work for any binary relation which is closed in $Xtimes X$. Also it will work if we switch "closed" to "open".
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
answered Jan 21 at 13:14
freakishfreakish
12.5k1630
12.5k1630
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
add a comment |
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
I'm not sure why $alpha_x$ and $beta_x$ are continuous. Can you explain?
$endgroup$
– Keen-ameteur
Jan 21 at 13:58
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
@Keen-ameteur it follows from the universal property of the product topology. Their composition with every projection is continuous. Because it is either constant or the identity.
$endgroup$
– freakish
Jan 21 at 13:59
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
$begingroup$
Thank you for your answer.
$endgroup$
– Keen-ameteur
Jan 21 at 14:15
add a comment |
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