Mixing
Properties of open subsets of the real numbers
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Let $G$ be an open subset of $mathbb{R}$ . Then:
a) Is the set $H={xy|x,yin G text{and} 0notin G}$ open in $mathbb{R}$?
b) Is the set $G=mathbb{R}$ if $0in G$ and $forall x,yin G, x+yin G$?
I think the answer to both the problems is yes. But, the question is, should we use the group theoretic properties or topological properties to prove the statements? And how should we exactly proceed. Any hints? Thanks beforehand.
general-topology real-numbers topological-groups
asked yesterday
vidyarthi
2,6881831
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up vote
0
down vote
favorite
Let $G$ be an open subset of $mathbb{R}$ . Then:
a) Is the set $H={xy|x,yin G text{and} 0notin G}$ open in $mathbb{R}$?
b) Is the set $G=mathbb{R}$ if $0in G$ and $forall x,yin G, x+yin G$?
I think the answer to both the problems is yes. But, the question is, should we use the group theoretic properties or topological properties to prove the statements? And how should we exactly proceed. Any hints? Thanks beforehand.
general-topology real-numbers topological-groups
asked yesterday
vidyarthi
2,6881831
It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $G$ be an open subset of $mathbb{R}$ . Then:
a) Is the set $H={xy|x,yin G text{and} 0notin G}$ open in $mathbb{R}$?
b) Is the set $G=mathbb{R}$ if $0in G$ and $forall x,yin G, x+yin G$?
I think the answer to both the problems is yes. But, the question is, should we use the group theoretic properties or topological properties to prove the statements? And how should we exactly proceed. Any hints? Thanks beforehand.
general-topology real-numbers topological-groups
asked yesterday
vidyarthi
2,6881831
Let $G$ be an open subset of $mathbb{R}$ . Then:
a) Is the set $H={xy|x,yin G text{and} 0notin G}$ open in $mathbb{R}$?
b) Is the set $G=mathbb{R}$ if $0in G$ and $forall x,yin G, x+yin G$?
I think the answer to both the problems is yes. But, the question is, should we use the group theoretic properties or topological properties to prove the statements? And how should we exactly proceed. Any hints? Thanks beforehand.
general-topology real-numbers topological-groups
general-topology real-numbers topological-groups
asked yesterday
vidyarthi
2,6881831
asked yesterday
vidyarthi
2,6881831
edited yesterday
asked yesterday
vidyarthi
2,6881831
asked yesterday
vidyarthi
2,6881831
asked yesterday
vidyarthi
2,6881831
2,6881831
It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday
add a comment |
It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday
It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday
add a comment |
2 Answers
2
active
oldest
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up vote
1
down vote
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $xneq0$ is a homeomorphism. So then our set is a union (over $xin G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-varepsilon,varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
answered yesterday
Y. Forman
11.3k423
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
add a comment |
up vote
0
down vote
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x in G$ there is an $epsilon_x>0$ so that for any $x': x-epsilon_x < x' < x + epsilon_x$ we know $x'in G$ and for any $y in G$ there is an $epsilon_y>0$ so that for any $y': y-epsilon_y < y' < y + epsilon_y$ we know $y'in G$.
Does it follow that for $xyin H$ that there is a $epsilon_{xy}> 0$ so that for any $z: xy-epsilon_{xy} < z < xy + epsilon_{xy}$ that $z in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x in G$ then by induction $n*x in G$ for all $ninmathbb N$. Use archimedian property and the fact that for all $epsilon > 0$ there is an $0 < x < epsilon; x in G$.
answered yesterday
fleablood
65.5k22682
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $xneq0$ is a homeomorphism. So then our set is a union (over $xin G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-varepsilon,varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
answered yesterday
Y. Forman
11.3k423
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
add a comment |
up vote
1
down vote
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $xneq0$ is a homeomorphism. So then our set is a union (over $xin G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-varepsilon,varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
answered yesterday
Y. Forman
11.3k423
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
add a comment |
up vote
1
down vote
up vote
1
down vote
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $xneq0$ is a homeomorphism. So then our set is a union (over $xin G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-varepsilon,varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
answered yesterday
Y. Forman
11.3k423
We're asking if a set is open, so we're going to have to use at least some topological properties. For a, I'd use the fact that multiplication by a fixed $xneq0$ is a homeomorphism. So then our set is a union (over $xin G$) of sets homeomorphic to $G$.
b I'd approach differently. Here we can use the Archimedean property. In particular, $G$ must contain some interval $[-varepsilon,varepsilon]$. Every real number is a finite positive integer multiple of something in that interval.
answered yesterday
Y. Forman
11.3k423
edited yesterday
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
answered yesterday
Y. Forman
11.3k423
11.3k423
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
add a comment |
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
1
1
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
edited the post. see now for part b. and, could you please elaborate your answers. I mean how does homeomorphism arise here?
– vidyarthi
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
@vidyarthi Edited and elaborated. See if you can work the full proofs out based on what I've sketched here, and let me know if you're still stuck.
– Y. Forman
yesterday
add a comment |
up vote
0
down vote
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x in G$ there is an $epsilon_x>0$ so that for any $x': x-epsilon_x < x' < x + epsilon_x$ we know $x'in G$ and for any $y in G$ there is an $epsilon_y>0$ so that for any $y': y-epsilon_y < y' < y + epsilon_y$ we know $y'in G$.
Does it follow that for $xyin H$ that there is a $epsilon_{xy}> 0$ so that for any $z: xy-epsilon_{xy} < z < xy + epsilon_{xy}$ that $z in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x in G$ then by induction $n*x in G$ for all $ninmathbb N$. Use archimedian property and the fact that for all $epsilon > 0$ there is an $0 < x < epsilon; x in G$.
answered yesterday
fleablood
65.5k22682
add a comment |
up vote
0
down vote
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x in G$ there is an $epsilon_x>0$ so that for any $x': x-epsilon_x < x' < x + epsilon_x$ we know $x'in G$ and for any $y in G$ there is an $epsilon_y>0$ so that for any $y': y-epsilon_y < y' < y + epsilon_y$ we know $y'in G$.
Does it follow that for $xyin H$ that there is a $epsilon_{xy}> 0$ so that for any $z: xy-epsilon_{xy} < z < xy + epsilon_{xy}$ that $z in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x in G$ then by induction $n*x in G$ for all $ninmathbb N$. Use archimedian property and the fact that for all $epsilon > 0$ there is an $0 < x < epsilon; x in G$.
answered yesterday
fleablood
65.5k22682
add a comment |
up vote
0
down vote
up vote
0
down vote
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x in G$ there is an $epsilon_x>0$ so that for any $x': x-epsilon_x < x' < x + epsilon_x$ we know $x'in G$ and for any $y in G$ there is an $epsilon_y>0$ so that for any $y': y-epsilon_y < y' < y + epsilon_y$ we know $y'in G$.
Does it follow that for $xyin H$ that there is a $epsilon_{xy}> 0$ so that for any $z: xy-epsilon_{xy} < z < xy + epsilon_{xy}$ that $z in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x in G$ then by induction $n*x in G$ for all $ninmathbb N$. Use archimedian property and the fact that for all $epsilon > 0$ there is an $0 < x < epsilon; x in G$.
answered yesterday
fleablood
65.5k22682
What "group properties" could possibly be of any use? Especially as no-one claimed $G$ was a sub group of $mathbb R$.
1) I'd use that that standard that because $G$ is open, then for any $x in G$ there is an $epsilon_x>0$ so that for any $x': x-epsilon_x < x' < x + epsilon_x$ we know $x'in G$ and for any $y in G$ there is an $epsilon_y>0$ so that for any $y': y-epsilon_y < y' < y + epsilon_y$ we know $y'in G$.
Does it follow that for $xyin H$ that there is a $epsilon_{xy}> 0$ so that for any $z: xy-epsilon_{xy} < z < xy + epsilon_{xy}$ that $z in H$?
That should be straightforward (albeit tedious) to prove.
2) is different. Hint: If $x in G$ then by induction $n*x in G$ for all $ninmathbb N$. Use archimedian property and the fact that for all $epsilon > 0$ there is an $0 < x < epsilon; x in G$.
answered yesterday
fleablood
65.5k22682
answered yesterday
fleablood
65.5k22682
answered yesterday
fleablood
65.5k22682
answered yesterday
fleablood
65.5k22682
65.5k22682
add a comment |
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It's a topology question so use topological properties. Namely if $x,yin G$ then there are neighorhoods around $x,y$ entirely in $G$. Does that mean there are neighborhoods around the product?
– fleablood
yesterday
Just because a set is called $G$ doesn't mean it is a group. Other than the fact that multiplication and addition are binary operations it's hard to so what "group theoretic properities" (of $mathbb R$ because $G$ is not a group) could be useful.
– fleablood
yesterday