Why is this counterexample wrong? (Theorem about Open subsets and Open relative to)












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    5 hours ago












  • $begingroup$
    Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
    $endgroup$
    – DanielWainfleet
    29 mins ago
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
    5 hours ago












  • $begingroup$
    Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
    $endgroup$
    – DanielWainfleet
    29 mins ago














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago







Sean Lee

















asked 5 hours ago









Sean LeeSean Lee

1748




1748








  • 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
    5 hours ago












  • $begingroup$
    Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
    $endgroup$
    – DanielWainfleet
    29 mins ago














  • 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
    5 hours ago












  • $begingroup$
    Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
    $endgroup$
    – fleablood
    3 hours ago












  • $begingroup$
    On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
    $endgroup$
    – DanielWainfleet
    29 mins ago








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
5 hours ago






$begingroup$
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
5 hours ago














$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
3 hours ago






$begingroup$
Why does $(0,0) in E$ meany $E$ is not open relatively to $Y$?????
$endgroup$
– fleablood
3 hours ago














$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
29 mins ago




$begingroup$
On notation: You should write $Y=[0,1]times {0}$ and $E=[0,1)times {0}.$
$endgroup$
– DanielWainfleet
29 mins ago










1 Answer
1






active

oldest

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4












$begingroup$

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 $q in Y$."






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Amended to give an explicit neighborhood.
    $endgroup$
    – Randall
    5 hours ago






  • 1




    $begingroup$
    @Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
    $endgroup$
    – Sean Lee
    5 hours ago






  • 1




    $begingroup$
    @SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    $r$ could be 0.5 as in my example. You need to focus on $q in Y$.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    Ah I get it now, thank you for your patience (:
    $endgroup$
    – Sean Lee
    4 hours ago











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

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 $q in Y$."






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Amended to give an explicit neighborhood.
    $endgroup$
    – Randall
    5 hours ago






  • 1




    $begingroup$
    @Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
    $endgroup$
    – Sean Lee
    5 hours ago






  • 1




    $begingroup$
    @SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    $r$ could be 0.5 as in my example. You need to focus on $q in Y$.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    Ah I get it now, thank you for your patience (:
    $endgroup$
    – Sean Lee
    4 hours ago
















4












$begingroup$

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 $q in Y$."






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Amended to give an explicit neighborhood.
    $endgroup$
    – Randall
    5 hours ago






  • 1




    $begingroup$
    @Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
    $endgroup$
    – Sean Lee
    5 hours ago






  • 1




    $begingroup$
    @SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    $r$ could be 0.5 as in my example. You need to focus on $q in Y$.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    Ah I get it now, thank you for your patience (:
    $endgroup$
    – Sean Lee
    4 hours ago














4












4








4





$begingroup$

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 $q in Y$."






share|cite|improve this answer











$endgroup$



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 $q in Y$."







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 5 hours ago









RandallRandall

9,63611230




9,63611230








  • 2




    $begingroup$
    Amended to give an explicit neighborhood.
    $endgroup$
    – Randall
    5 hours ago






  • 1




    $begingroup$
    @Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
    $endgroup$
    – Sean Lee
    5 hours ago






  • 1




    $begingroup$
    @SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    $r$ could be 0.5 as in my example. You need to focus on $q in Y$.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    Ah I get it now, thank you for your patience (:
    $endgroup$
    – Sean Lee
    4 hours ago














  • 2




    $begingroup$
    Amended to give an explicit neighborhood.
    $endgroup$
    – Randall
    5 hours ago






  • 1




    $begingroup$
    @Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
    $endgroup$
    – Sean Lee
    5 hours ago






  • 1




    $begingroup$
    @SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    $r$ could be 0.5 as in my example. You need to focus on $q in Y$.
    $endgroup$
    – Randall
    4 hours ago








  • 1




    $begingroup$
    Ah I get it now, thank you for your patience (:
    $endgroup$
    – Sean Lee
    4 hours ago








2




2




$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
5 hours ago




$begingroup$
Amended to give an explicit neighborhood.
$endgroup$
– Randall
5 hours ago




1




1




$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
5 hours ago




$begingroup$
@Randall Edited to include the definition of "relative open". Thank you; I understand why the "counterexample" was wrong.
$endgroup$
– Sean Lee
5 hours ago




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
4 hours ago






$begingroup$
@SeanLee Excellent. Yours is a common misconception that nearly everyone has when first learning. Good to clear it up now.
$endgroup$
– Randall
4 hours ago






1




1




$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
4 hours ago






$begingroup$
$r$ could be 0.5 as in my example. You need to focus on $q in Y$.
$endgroup$
– Randall
4 hours ago






1




1




$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
4 hours ago




$begingroup$
Ah I get it now, thank you for your patience (:
$endgroup$
– Sean Lee
4 hours ago


















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