Mixing
Open dense subset of $mathbb{R}$.
$begingroup$
I found this question in one of the posts
Let $Asubseteqmathbb{R}$ be open and dense. Show that
$$mathbb{R}={x+y:x,yin A}$$
The poster stated that it is not too hard to prove but I can't seem to figure out the proof. Can someone
help me out.
general-topology topological-vector-spaces
$endgroup$
|
show 13 more comments
$begingroup$
I found this question in one of the posts
Let $Asubseteqmathbb{R}$ be open and dense. Show that
$$mathbb{R}={x+y:x,yin A}$$
The poster stated that it is not too hard to prove but I can't seem to figure out the proof. Can someone
help me out.
general-topology topological-vector-spaces
$endgroup$
1
$begingroup$
I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
1
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
1
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
2
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
1
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50
|
show 13 more comments
$begingroup$
I found this question in one of the posts
Let $Asubseteqmathbb{R}$ be open and dense. Show that
$$mathbb{R}={x+y:x,yin A}$$
The poster stated that it is not too hard to prove but I can't seem to figure out the proof. Can someone
help me out.
general-topology topological-vector-spaces
$endgroup$
I found this question in one of the posts
Let $Asubseteqmathbb{R}$ be open and dense. Show that
$$mathbb{R}={x+y:x,yin A}$$
The poster stated that it is not too hard to prove but I can't seem to figure out the proof. Can someone
help me out.
general-topology topological-vector-spaces
general-topology topological-vector-spaces
asked Jan 16 at 19:04
user397197
1
$begingroup$
I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
1
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
1
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
2
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
1
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50
|
show 13 more comments
1
$begingroup$
I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
1
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
1
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
2
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
1
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50
1
1
$begingroup$
I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
$begingroup$
I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
1
1
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
1
1
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
2
2
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
1
1
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50
|
show 13 more comments
1 Answer
1
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$begingroup$
Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$.
Consider any $z in mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x in A$ and $y in A$.
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
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$begingroup$
Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$.
Consider any $z in mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x in A$ and $y in A$.
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$.
Consider any $z in mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x in A$ and $y in A$.
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$.
Consider any $z in mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x in A$ and $y in A$.
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
$endgroup$
Since $A$ is open and nonempty, it contains some open interval $J = (a,b)$ with $a < b$.
Consider any $z in mathbb R$. Since $A$ is dense, $z - J = (z-b, z-a)$ contains some member $x$ of $A$. Thus $x = z - y$ where $a < y < b$, so $z = x+y$ where $x in A$ and $y in A$.
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
answered Jan 16 at 19:15
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
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I don't think the statement is correct. Take for example rationals. Set of rationals is dense and it is open in some sense(non-topological). However sum of any two rationals will again produce rational number.
$endgroup$
– kolobokish
Jan 16 at 19:09
1
$begingroup$
@kolobokish What in the world do you mean by "open in some sense (non-topological)"? Open (in mathematics) is a topological property.
$endgroup$
– Robert Israel
Jan 16 at 19:10
1
$begingroup$
@stackuser rest assured the rationals are definitely not open.
$endgroup$
– Randall
Jan 16 at 19:18
2
$begingroup$
@kolobokish Perhaps you are trying to mention that we can choose different topologies on $Bbb R$, i.e. "open" might describe different sets in different (topological) contexts.
$endgroup$
– Omnomnomnom
Jan 16 at 19:19
1
$begingroup$
In any reasonable interpretation of such a problem, in the absence of any additional assumptions, one has to interpret this in the euclidean topology, not in any topology you feel like in order to make it wrong.
$endgroup$
– Randall
Jan 16 at 19:50