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?


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 $q\in Y$.

  • 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 '19 at 5:00
  • $\begingroup$ Why does $(0,0) \in E$ meany $E$ is not open relatively to $Y$????? $\endgroup$ – fleablood Jan 20 '19 at 6:25
  • $\begingroup$ On notation: You should write $Y=[0,1]\times \{0\}$ and $E=[0,1)\times \{0\}.$ $\endgroup$ – DanielWainfleet Jan 20 '19 at 9:36
  • $\begingroup$ @fleablood Yes, it should still be open relative (mistake on my part). $\endgroup$ – Sean Lee Jan 20 '19 at 13:39

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 $q \in Y$."

  • 2
    $\begingroup$ Amended to give an explicit neighborhood. $\endgroup$ – Randall Jan 20 '19 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 '19 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 '19 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 '19 at 5:30
  • 1
    $\begingroup$ Ah I get it now, thank you for your patience (: $\endgroup$ – Sean Lee Jan 20 '19 at 5:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.