Showing $mathbb{Q} cap [a,b]$ is an open set in $mathbb{Q}$ for irrational $a$, $b$.
$begingroup$
I came up with this lemma (although not confident enough about it) while solving Baby Rudin. In the chapter "Basic Topology", I attempted to solve question 16, in which $mathbb{Q}$ is regarded as the metric space with the usual metric, whose subset $E$ contains rationals $p$, such that $2<p^2<3$. The question asks to show that $E$ is not compact.
I now take the family of sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ where $n$ runs through $1$ to $infty$. Clearly, for no finite $n=m$, it can cover $E$. Hence, we are done.
But, in this process, I wanted to make sure, that the sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ are all open in $mathbb{Q}$.
I just want you to kindly verify the validity of my approach to the problem and the lemma which follows.
LEMMA:
$mathbb{Q} cap [a,b]$ is an open set in $mathbb{Q}$, where $a$, $b$ are irrational.
Proof:
We know, the ordered set of rational numbers doesn't have the supremum property.
Hence the set $P = { t in mathbb{Q}: a < t < b}$ , although bounded, doesn't have the supremum ( and infimum) in $mathbb {Q}$. We now take a very small rational $epsilon >0$, such that $a< t -epsilon < t < t + epsilon < b$ , hence $N_{epsilon}(t) subset P$. This is the case for every $t in P$. Hence $P$ is open.
Additional Question: Can $mathbb{Q}$ alone be regarded as a discrete metric space, since any subset of it is both open and closed? [If false, please provide counterexample or arguments].
Thank you.
real-analysis proof-verification compactness real-numbers rational-numbers
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
$endgroup$
|
show 3 more comments
$begingroup$
I came up with this lemma (although not confident enough about it) while solving Baby Rudin. In the chapter "Basic Topology", I attempted to solve question 16, in which $mathbb{Q}$ is regarded as the metric space with the usual metric, whose subset $E$ contains rationals $p$, such that $2<p^2<3$. The question asks to show that $E$ is not compact.
I now take the family of sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ where $n$ runs through $1$ to $infty$. Clearly, for no finite $n=m$, it can cover $E$. Hence, we are done.
But, in this process, I wanted to make sure, that the sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ are all open in $mathbb{Q}$.
I just want you to kindly verify the validity of my approach to the problem and the lemma which follows.
LEMMA:
$mathbb{Q} cap [a,b]$ is an open set in $mathbb{Q}$, where $a$, $b$ are irrational.
Proof:
We know, the ordered set of rational numbers doesn't have the supremum property.
Hence the set $P = { t in mathbb{Q}: a < t < b}$ , although bounded, doesn't have the supremum ( and infimum) in $mathbb {Q}$. We now take a very small rational $epsilon >0$, such that $a< t -epsilon < t < t + epsilon < b$ , hence $N_{epsilon}(t) subset P$. This is the case for every $t in P$. Hence $P$ is open.
Additional Question: Can $mathbb{Q}$ alone be regarded as a discrete metric space, since any subset of it is both open and closed? [If false, please provide counterexample or arguments].
Thank you.
real-analysis proof-verification compactness real-numbers rational-numbers
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
$endgroup$
4
$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
1
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
2
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
1
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03
|
show 3 more comments
$begingroup$
I came up with this lemma (although not confident enough about it) while solving Baby Rudin. In the chapter "Basic Topology", I attempted to solve question 16, in which $mathbb{Q}$ is regarded as the metric space with the usual metric, whose subset $E$ contains rationals $p$, such that $2<p^2<3$. The question asks to show that $E$ is not compact.
I now take the family of sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ where $n$ runs through $1$ to $infty$. Clearly, for no finite $n=m$, it can cover $E$. Hence, we are done.
But, in this process, I wanted to make sure, that the sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ are all open in $mathbb{Q}$.
I just want you to kindly verify the validity of my approach to the problem and the lemma which follows.
LEMMA:
$mathbb{Q} cap [a,b]$ is an open set in $mathbb{Q}$, where $a$, $b$ are irrational.
Proof:
We know, the ordered set of rational numbers doesn't have the supremum property.
Hence the set $P = { t in mathbb{Q}: a < t < b}$ , although bounded, doesn't have the supremum ( and infimum) in $mathbb {Q}$. We now take a very small rational $epsilon >0$, such that $a< t -epsilon < t < t + epsilon < b$ , hence $N_{epsilon}(t) subset P$. This is the case for every $t in P$. Hence $P$ is open.
Additional Question: Can $mathbb{Q}$ alone be regarded as a discrete metric space, since any subset of it is both open and closed? [If false, please provide counterexample or arguments].
Thank you.
real-analysis proof-verification compactness real-numbers rational-numbers
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
$endgroup$
I came up with this lemma (although not confident enough about it) while solving Baby Rudin. In the chapter "Basic Topology", I attempted to solve question 16, in which $mathbb{Q}$ is regarded as the metric space with the usual metric, whose subset $E$ contains rationals $p$, such that $2<p^2<3$. The question asks to show that $E$ is not compact.
I now take the family of sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ where $n$ runs through $1$ to $infty$. Clearly, for no finite $n=m$, it can cover $E$. Hence, we are done.
But, in this process, I wanted to make sure, that the sets $mathbb{Q} cap [- sqrt{3} +1/(n+4), sqrt{3} -1/(n+4)]$ are all open in $mathbb{Q}$.
I just want you to kindly verify the validity of my approach to the problem and the lemma which follows.
LEMMA:
$mathbb{Q} cap [a,b]$ is an open set in $mathbb{Q}$, where $a$, $b$ are irrational.
Proof:
We know, the ordered set of rational numbers doesn't have the supremum property.
Hence the set $P = { t in mathbb{Q}: a < t < b}$ , although bounded, doesn't have the supremum ( and infimum) in $mathbb {Q}$. We now take a very small rational $epsilon >0$, such that $a< t -epsilon < t < t + epsilon < b$ , hence $N_{epsilon}(t) subset P$. This is the case for every $t in P$. Hence $P$ is open.
Additional Question: Can $mathbb{Q}$ alone be regarded as a discrete metric space, since any subset of it is both open and closed? [If false, please provide counterexample or arguments].
Thank you.
real-analysis proof-verification compactness real-numbers rational-numbers
real-analysis proof-verification compactness real-numbers rational-numbers
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
asked Dec 20 '18 at 15:49
Subhasis BiswasSubhasis Biswas
508311
508311
4
$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
1
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
2
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
1
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03
|
show 3 more comments
4
$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
1
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
2
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
1
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03
4
4
$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
1
1
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
2
2
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
1
1
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03
|
show 3 more comments
0
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$begingroup$
An alternate approach: Observe that when $a$ and $b$ are irrational, $mathbb{Q}cap[a,b]=mathbb{Q}cap(a,b)$.
$endgroup$
– Michael Burr
Dec 20 '18 at 15:53
$begingroup$
Can you please check the proof :(
$endgroup$
– Subhasis Biswas
Dec 20 '18 at 15:54
1
$begingroup$
The converse also holds: for any $a,binBbb R$ such that $a<b$, $Bbb{Q}cap[a,b]$ is an open subset of $Bbb Q$ if and only if both $a$ and $b$ are irrational numbers.
$endgroup$
– user593746
Dec 20 '18 at 15:54
2
$begingroup$
Why is every subset of $mathbb{Q}$ both open and closed? You haven't shown that. Is ${0}$ open?
$endgroup$
– Michael Burr
Dec 20 '18 at 15:54
1
$begingroup$
Your formula for $epsilon$ should depend on $a$ and $b$. Something like $min{frac{t-a}2,...}$.
$endgroup$
– Michael Burr
Dec 20 '18 at 16:03