Show that $overline{mathbb{Q}cap (0,1)}=[0,1]$.
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Show that $overline{mathbb{Q}cap (0,1)}=[0,1]$
My attemp: If $E=mathbb{Q}cap (0,1)$, then $$overline{mathbb{Q}cap (0,1)}=overline{mathbb{Q}}cap overline{(0,1)}=mathbb{R}cap [0,1]=[0,1]$$
This is right? Thanks!
calculus real-analysis analysis proof-verification
asked Nov 12 at 15:24
pepo kirk
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up vote
1
down vote
favorite
Show that $overline{mathbb{Q}cap (0,1)}=[0,1]$
My attemp: If $E=mathbb{Q}cap (0,1)$, then $$overline{mathbb{Q}cap (0,1)}=overline{mathbb{Q}}cap overline{(0,1)}=mathbb{R}cap [0,1]=[0,1]$$
This is right? Thanks!
calculus real-analysis analysis proof-verification
asked Nov 12 at 15:24
pepo kirk
62
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
1
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that $overline{mathbb{Q}cap (0,1)}=[0,1]$
My attemp: If $E=mathbb{Q}cap (0,1)$, then $$overline{mathbb{Q}cap (0,1)}=overline{mathbb{Q}}cap overline{(0,1)}=mathbb{R}cap [0,1]=[0,1]$$
This is right? Thanks!
calculus real-analysis analysis proof-verification
asked Nov 12 at 15:24
pepo kirk
62
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Show that $overline{mathbb{Q}cap (0,1)}=[0,1]$
My attemp: If $E=mathbb{Q}cap (0,1)$, then $$overline{mathbb{Q}cap (0,1)}=overline{mathbb{Q}}cap overline{(0,1)}=mathbb{R}cap [0,1]=[0,1]$$
This is right? Thanks!
calculus real-analysis analysis proof-verification
calculus real-analysis analysis proof-verification
asked Nov 12 at 15:24
pepo kirk
62
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 15:24
pepo kirk
62
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 12 at 15:34
asked Nov 12 at 15:24
pepo kirk
62
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pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 12 at 15:24
pepo kirk
62
asked Nov 12 at 15:24
pepo kirk
62
62
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
pepo kirk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
1
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30
add a comment |
1
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
1
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30
1
1
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
1
1
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30
add a comment |
1 Answer
1
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oldest
votes
up vote
4
down vote
Your attempt doesn't work. $overline{Xcap Y}=overline{X}cap overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $Xcap Y={}$, so $overline{Xcap Y}={}$, but $overline{X}cap overline{Y}=[0,1]cap[1,2]={1}$.
I would do it like this. Since $X=Bbb Qcap (0,1)$ is contained in $Y=(0,1)$, we have $overline{X}subseteq overline{Y}=[0,1]$. To do the other way around, we note that for any $ain [0,1]$, there exists a sequence $(x_1,x_2,ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1in Bbb Qcap (a,1)$. For $k=2,3,ldots$, pick $x_kin Bbb{Q}$ in $left(a,frac{a+x_{k-1}}{2}right)$. So, $|x_k-a|leq frac{|x_1-a|}{2^{k-1}}$, so $x_kto a$.
answered Nov 12 at 15:42
Snookie
2308
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Your attempt doesn't work. $overline{Xcap Y}=overline{X}cap overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $Xcap Y={}$, so $overline{Xcap Y}={}$, but $overline{X}cap overline{Y}=[0,1]cap[1,2]={1}$.
I would do it like this. Since $X=Bbb Qcap (0,1)$ is contained in $Y=(0,1)$, we have $overline{X}subseteq overline{Y}=[0,1]$. To do the other way around, we note that for any $ain [0,1]$, there exists a sequence $(x_1,x_2,ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1in Bbb Qcap (a,1)$. For $k=2,3,ldots$, pick $x_kin Bbb{Q}$ in $left(a,frac{a+x_{k-1}}{2}right)$. So, $|x_k-a|leq frac{|x_1-a|}{2^{k-1}}$, so $x_kto a$.
answered Nov 12 at 15:42
Snookie
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
4
down vote
Your attempt doesn't work. $overline{Xcap Y}=overline{X}cap overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $Xcap Y={}$, so $overline{Xcap Y}={}$, but $overline{X}cap overline{Y}=[0,1]cap[1,2]={1}$.
I would do it like this. Since $X=Bbb Qcap (0,1)$ is contained in $Y=(0,1)$, we have $overline{X}subseteq overline{Y}=[0,1]$. To do the other way around, we note that for any $ain [0,1]$, there exists a sequence $(x_1,x_2,ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1in Bbb Qcap (a,1)$. For $k=2,3,ldots$, pick $x_kin Bbb{Q}$ in $left(a,frac{a+x_{k-1}}{2}right)$. So, $|x_k-a|leq frac{|x_1-a|}{2^{k-1}}$, so $x_kto a$.
answered Nov 12 at 15:42
Snookie
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
4
down vote
up vote
4
down vote
Your attempt doesn't work. $overline{Xcap Y}=overline{X}cap overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $Xcap Y={}$, so $overline{Xcap Y}={}$, but $overline{X}cap overline{Y}=[0,1]cap[1,2]={1}$.
I would do it like this. Since $X=Bbb Qcap (0,1)$ is contained in $Y=(0,1)$, we have $overline{X}subseteq overline{Y}=[0,1]$. To do the other way around, we note that for any $ain [0,1]$, there exists a sequence $(x_1,x_2,ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1in Bbb Qcap (a,1)$. For $k=2,3,ldots$, pick $x_kin Bbb{Q}$ in $left(a,frac{a+x_{k-1}}{2}right)$. So, $|x_k-a|leq frac{|x_1-a|}{2^{k-1}}$, so $x_kto a$.
answered Nov 12 at 15:42
Snookie
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your attempt doesn't work. $overline{Xcap Y}=overline{X}cap overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $Xcap Y={}$, so $overline{Xcap Y}={}$, but $overline{X}cap overline{Y}=[0,1]cap[1,2]={1}$.
I would do it like this. Since $X=Bbb Qcap (0,1)$ is contained in $Y=(0,1)$, we have $overline{X}subseteq overline{Y}=[0,1]$. To do the other way around, we note that for any $ain [0,1]$, there exists a sequence $(x_1,x_2,ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1in Bbb Qcap (a,1)$. For $k=2,3,ldots$, pick $x_kin Bbb{Q}$ in $left(a,frac{a+x_{k-1}}{2}right)$. So, $|x_k-a|leq frac{|x_1-a|}{2^{k-1}}$, so $x_kto a$.
answered Nov 12 at 15:42
Snookie
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 12 at 15:48
answered Nov 12 at 15:42
Snookie
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 12 at 15:42
Snookie
2308
answered Nov 12 at 15:42
Snookie
2308
2308
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Snookie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
pepo kirk is a new contributor. Be nice, and check out our Code of Conduct.
pepo kirk is a new contributor. Be nice, and check out our Code of Conduct.
pepo kirk is a new contributor. Be nice, and check out our Code of Conduct.
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1
Can you put what overbar means into the question? I assume you mean closure but it means different things in different contexts...
– Mason
Nov 12 at 15:28
No, this is wrong as you show something else than you want to show. In particular, you use the result you want to show in the first "=" ...
– Stockfish
Nov 12 at 15:29
1
Also: It seems a little silly that your first equality is what you are trying to prove: Maybe $overline{mathbb{Q}cap (0,1)}=mathbb{R}cap [0,1]=[0,1]=overline{mathbb{Q}}cap overline{(0,1)}$ makes more sense. Though, I still would argue that you haven't quite done due diligence.
– Mason
Nov 12 at 15:30