Mixing
Why is this counterexample wrong? (Theorem about Open subsets and Open relative to)
$begingroup$
In Principles of Mathematical Analysis by Walter Rudin, Theorem 2.30 states that:
Suppose $Y subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y cap G $ for some open subset $G$ of $X$.
I've thought of the following "counterexample":
If $X = mathbb{R}^2$, $Y = ([0,2],0)$, $G = B_1(0)$ (i.e. the open ball of radius 1 centered at 0), then the theorem implies that $E = Y cap G = ([0,1),0)$ is open relative to $Y$, which it is clearly not (due to $(0,0) in E$).
Could someone point out why this "counterexample" is wrong?
Edit:
Here is the definition of "Open Relative to":
$E$ is open relative to $Y$ if for each $p in E$ there is an associated $r>0$ such that $q in E$ whenever $d(p,q) < r$ and $qin Y$.
real-analysis general-topology
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
$endgroup$
add a comment |
$begingroup$
In Principles of Mathematical Analysis by Walter Rudin, Theorem 2.30 states that:
Suppose $Y subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y cap G $ for some open subset $G$ of $X$.
I've thought of the following "counterexample":
If $X = mathbb{R}^2$, $Y = ([0,2],0)$, $G = B_1(0)$ (i.e. the open ball of radius 1 centered at 0), then the theorem implies that $E = Y cap G = ([0,1),0)$ is open relative to $Y$, which it is clearly not (due to $(0,0) in E$).
Could someone point out why this "counterexample" is wrong?
Edit:
Here is the definition of "Open Relative to":
$E$ is open relative to $Y$ if for each $p in E$ there is an associated $r>0$ such that $q in E$ whenever $d(p,q) < r$ and $qin Y$.
real-analysis general-topology
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
$endgroup$
3
$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39
add a comment |
$begingroup$
In Principles of Mathematical Analysis by Walter Rudin, Theorem 2.30 states that:
Suppose $Y subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y cap G $ for some open subset $G$ of $X$.
I've thought of the following "counterexample":
If $X = mathbb{R}^2$, $Y = ([0,2],0)$, $G = B_1(0)$ (i.e. the open ball of radius 1 centered at 0), then the theorem implies that $E = Y cap G = ([0,1),0)$ is open relative to $Y$, which it is clearly not (due to $(0,0) in E$).
Could someone point out why this "counterexample" is wrong?
Edit:
Here is the definition of "Open Relative to":
$E$ is open relative to $Y$ if for each $p in E$ there is an associated $r>0$ such that $q in E$ whenever $d(p,q) < r$ and $qin Y$.
real-analysis general-topology
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
$endgroup$
In Principles of Mathematical Analysis by Walter Rudin, Theorem 2.30 states that:
Suppose $Y subset X$. A subset $E$ of $Y$ is open relative to $Y$ if and only if $E = Y cap G $ for some open subset $G$ of $X$.
I've thought of the following "counterexample":
If $X = mathbb{R}^2$, $Y = ([0,2],0)$, $G = B_1(0)$ (i.e. the open ball of radius 1 centered at 0), then the theorem implies that $E = Y cap G = ([0,1),0)$ is open relative to $Y$, which it is clearly not (due to $(0,0) in E$).
Could someone point out why this "counterexample" is wrong?
Edit:
Here is the definition of "Open Relative to":
$E$ is open relative to $Y$ if for each $p in E$ there is an associated $r>0$ such that $q in E$ whenever $d(p,q) < r$ and $qin Y$.
real-analysis general-topology
real-analysis general-topology
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
edited Jan 20 at 5:03
Sean Lee
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
asked Jan 20 at 4:47
Sean LeeSean Lee
393111
393111
3
$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39
add a comment |
3
$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39
3
3
$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
That fact that $(0,0) in E$ doesn't make Rudin wrong. The set $[0,0.5)$ is an open neighborhood in $Y$ about $(0,0)$ that is contained in $E$, so $(0,0)$ is an interior point of $E$.
Note that this line of reasoning fails in $X$ because there an open set about $(0,0)$ will contain a ball that leaves the $x$-axis, and for sure, $E$ is not open in $X$.
Edit: seeing your updated question with the definition, it seems you've ignored the very important clause "and $q in Y$."
answered Jan 20 at 4:56
RandallRandall
10.2k11230
$endgroup$
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
|
show 4 more comments
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1 Answer
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$begingroup$
That fact that $(0,0) in E$ doesn't make Rudin wrong. The set $[0,0.5)$ is an open neighborhood in $Y$ about $(0,0)$ that is contained in $E$, so $(0,0)$ is an interior point of $E$.
Note that this line of reasoning fails in $X$ because there an open set about $(0,0)$ will contain a ball that leaves the $x$-axis, and for sure, $E$ is not open in $X$.
Edit: seeing your updated question with the definition, it seems you've ignored the very important clause "and $q in Y$."
answered Jan 20 at 4:56
RandallRandall
10.2k11230
$endgroup$
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
|
show 4 more comments
$begingroup$
That fact that $(0,0) in E$ doesn't make Rudin wrong. The set $[0,0.5)$ is an open neighborhood in $Y$ about $(0,0)$ that is contained in $E$, so $(0,0)$ is an interior point of $E$.
Note that this line of reasoning fails in $X$ because there an open set about $(0,0)$ will contain a ball that leaves the $x$-axis, and for sure, $E$ is not open in $X$.
Edit: seeing your updated question with the definition, it seems you've ignored the very important clause "and $q in Y$."
answered Jan 20 at 4:56
RandallRandall
10.2k11230
$endgroup$
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
|
show 4 more comments
$begingroup$
That fact that $(0,0) in E$ doesn't make Rudin wrong. The set $[0,0.5)$ is an open neighborhood in $Y$ about $(0,0)$ that is contained in $E$, so $(0,0)$ is an interior point of $E$.
Note that this line of reasoning fails in $X$ because there an open set about $(0,0)$ will contain a ball that leaves the $x$-axis, and for sure, $E$ is not open in $X$.
Edit: seeing your updated question with the definition, it seems you've ignored the very important clause "and $q in Y$."
answered Jan 20 at 4:56
RandallRandall
10.2k11230
$endgroup$
That fact that $(0,0) in E$ doesn't make Rudin wrong. The set $[0,0.5)$ is an open neighborhood in $Y$ about $(0,0)$ that is contained in $E$, so $(0,0)$ is an interior point of $E$.
Note that this line of reasoning fails in $X$ because there an open set about $(0,0)$ will contain a ball that leaves the $x$-axis, and for sure, $E$ is not open in $X$.
Edit: seeing your updated question with the definition, it seems you've ignored the very important clause "and $q in Y$."
answered Jan 20 at 4:56
RandallRandall
10.2k11230
edited Jan 20 at 5:05
answered Jan 20 at 4:56
RandallRandall
10.2k11230
answered Jan 20 at 4:56
RandallRandall
10.2k11230
answered Jan 20 at 4:56
RandallRandall
10.2k11230
10.2k11230
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
|
show 4 more comments
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
2
2
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
Jan 20 at 4:58
1
1
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
Jan 20 at 5:04
1
1
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
Jan 20 at 5:06
1
1
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
Jan 20 at 5:30
1
1
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
Jan 20 at 5:31
|
show 4 more comments
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$begingroup$
In many expositions, the statement you've given is actually the definition of relative open (especially in topology). If Rudin makes this a theorem, what is his definition of relative open? Answerers would need this in order to answer your question correctly.
$endgroup$
– Randall
Jan 20 at 5:00
$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
Jan 20 at 6:25
$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
Jan 20 at 9:36
$begingroup$
@fleablood Yes, it should still be open relative (mistake on my part).
$endgroup$
– Sean Lee
Jan 20 at 13:39