Mixing
If $tau$ is strictly finer than the standard topology $tau_{st}$ on $mathbb{R}$, prove their difference $tau...
$begingroup$
I think I have a proof, but it's a little convoluted.
Since $tau$ is strictly finer than than $tau_{st}$, there is an open set $U in tau$ such that $U notin tau_{st}$. For every $x in mathbb{R}$, the sets $U cup (-infty, x+1) $ and $U cup (x, +infty)$ are both open in $tau$. However, at most one of the two can be in $tau_{st}$. For the sake of contradiction, assume both are in $tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x in mathbb{R}$ we have constructed at least one set open in $tau$ which is not in $tau_{st}$, proving their difference is not countable.
Is this correct? Can it be done easier?
general-topology proof-verification
asked Jan 24 at 22:43
kmmkmm
1637
$endgroup$
add a comment |
$begingroup$
I think I have a proof, but it's a little convoluted.
Since $tau$ is strictly finer than than $tau_{st}$, there is an open set $U in tau$ such that $U notin tau_{st}$. For every $x in mathbb{R}$, the sets $U cup (-infty, x+1) $ and $U cup (x, +infty)$ are both open in $tau$. However, at most one of the two can be in $tau_{st}$. For the sake of contradiction, assume both are in $tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x in mathbb{R}$ we have constructed at least one set open in $tau$ which is not in $tau_{st}$, proving their difference is not countable.
Is this correct? Can it be done easier?
general-topology proof-verification
asked Jan 24 at 22:43
kmmkmm
1637
$endgroup$
$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33
add a comment |
$begingroup$
I think I have a proof, but it's a little convoluted.
Since $tau$ is strictly finer than than $tau_{st}$, there is an open set $U in tau$ such that $U notin tau_{st}$. For every $x in mathbb{R}$, the sets $U cup (-infty, x+1) $ and $U cup (x, +infty)$ are both open in $tau$. However, at most one of the two can be in $tau_{st}$. For the sake of contradiction, assume both are in $tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x in mathbb{R}$ we have constructed at least one set open in $tau$ which is not in $tau_{st}$, proving their difference is not countable.
Is this correct? Can it be done easier?
general-topology proof-verification
asked Jan 24 at 22:43
kmmkmm
1637
$endgroup$
I think I have a proof, but it's a little convoluted.
Since $tau$ is strictly finer than than $tau_{st}$, there is an open set $U in tau$ such that $U notin tau_{st}$. For every $x in mathbb{R}$, the sets $U cup (-infty, x+1) $ and $U cup (x, +infty)$ are both open in $tau$. However, at most one of the two can be in $tau_{st}$. For the sake of contradiction, assume both are in $tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x in mathbb{R}$ we have constructed at least one set open in $tau$ which is not in $tau_{st}$, proving their difference is not countable.
Is this correct? Can it be done easier?
general-topology proof-verification
general-topology proof-verification
asked Jan 24 at 22:43
kmmkmm
1637
asked Jan 24 at 22:43
kmmkmm
1637
asked Jan 24 at 22:43
kmmkmm
1637
asked Jan 24 at 22:43
kmmkmm
1637
asked Jan 24 at 22:43
kmmkmm
1637
1637
$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33
add a comment |
$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33
$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33
$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33
add a comment |
1 Answer
1
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Instead of the sets you are considering look at $Usetminus {x}$ for $x in mathbb R$. You can see that $Usetminus {x}neq Usetminus {y}$ whenever $x,y in U$ and $x neq y$ ; also $Usetminus {x} in tau_{st} $ for at most one $x$ (because $U =Usetminus {x})cup Usetminus {y}))$ for $x neq y$ and $Usetminus {x} in tau $ for all $x$. When $U$ is countable consideration of the intervals $U cup (x,x+1)$ can be used. I leave the details to you.
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
|
show 2 more comments
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1 Answer
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$begingroup$
Instead of the sets you are considering look at $Usetminus {x}$ for $x in mathbb R$. You can see that $Usetminus {x}neq Usetminus {y}$ whenever $x,y in U$ and $x neq y$ ; also $Usetminus {x} in tau_{st} $ for at most one $x$ (because $U =Usetminus {x})cup Usetminus {y}))$ for $x neq y$ and $Usetminus {x} in tau $ for all $x$. When $U$ is countable consideration of the intervals $U cup (x,x+1)$ can be used. I leave the details to you.
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
|
show 2 more comments
$begingroup$
Instead of the sets you are considering look at $Usetminus {x}$ for $x in mathbb R$. You can see that $Usetminus {x}neq Usetminus {y}$ whenever $x,y in U$ and $x neq y$ ; also $Usetminus {x} in tau_{st} $ for at most one $x$ (because $U =Usetminus {x})cup Usetminus {y}))$ for $x neq y$ and $Usetminus {x} in tau $ for all $x$. When $U$ is countable consideration of the intervals $U cup (x,x+1)$ can be used. I leave the details to you.
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
|
show 2 more comments
$begingroup$
Instead of the sets you are considering look at $Usetminus {x}$ for $x in mathbb R$. You can see that $Usetminus {x}neq Usetminus {y}$ whenever $x,y in U$ and $x neq y$ ; also $Usetminus {x} in tau_{st} $ for at most one $x$ (because $U =Usetminus {x})cup Usetminus {y}))$ for $x neq y$ and $Usetminus {x} in tau $ for all $x$. When $U$ is countable consideration of the intervals $U cup (x,x+1)$ can be used. I leave the details to you.
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
$endgroup$
Instead of the sets you are considering look at $Usetminus {x}$ for $x in mathbb R$. You can see that $Usetminus {x}neq Usetminus {y}$ whenever $x,y in U$ and $x neq y$ ; also $Usetminus {x} in tau_{st} $ for at most one $x$ (because $U =Usetminus {x})cup Usetminus {y}))$ for $x neq y$ and $Usetminus {x} in tau $ for all $x$. When $U$ is countable consideration of the intervals $U cup (x,x+1)$ can be used. I leave the details to you.
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
edited Jan 25 at 5:21
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
answered Jan 24 at 23:38
Kavi Rama MurthyKavi Rama Murthy
68k53068
68k53068
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
|
show 2 more comments
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
Won't $U cup {x}^c$ equal $U cup {y}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$?
$endgroup$
– kmm
Jan 24 at 23:45
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
@kmm You are right. I have corrected the proof.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:49
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
I am sorry if I am missing something obvious, but if $x in U^c$, won't $U cup {x}^c$ just be $mathbb{R} setminus {x}$, which is in $tau_{st}$, so it's not in $tau setminus tau_{st}$
$endgroup$
– kmm
Jan 24 at 23:53
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
@kmm Please check if the proof is OK now.
$endgroup$
– Kavi Rama Murthy
Jan 25 at 0:19
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
$begingroup$
It looks good to me, thank you very much.
$endgroup$
– kmm
Jan 25 at 0:24
|
show 2 more comments
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$begingroup$
The argument is incomplete. You have to consider the possibility that $U cup (x,infty) =U cup (y,infty)$ may hold for lots of pairs $x,y$.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:33