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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 $q\in Y$.

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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
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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 $q \in Y$."

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    $\begingroup$ Amended to give an explicit neighborhood. $\endgroup$ – Randall Jan 20 at 4:58
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ $r$ could be 0.5 as in my example. You need to focus on $q \in Y$. $\endgroup$ – Randall Jan 20 at 5:30
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    $\begingroup$ Ah I get it now, thank you for your patience (: $\endgroup$ – Sean Lee Jan 20 at 5:31

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