Please let me know if you think this proof is OK.

Given a set $E\subset \mathbb{R}^n$ and $C\subset E$, prove that $C$ contains a set $B$ which is the largest subset of $C$ relatively open in $E$.


Consider the set $H\subset \mathbb{R}^n$ defined as $H:=\bigcup\limits_{\vec{x}\in C^\circ}B_\varepsilon(\vec{x})$ for all $\vec{x}\in C^\circ$ and $\varepsilon > 0$, such that each open ball $B_\varepsilon(\vec{x})$ is contained in $C$. [$C^\circ$ is the interior of $C$]. Then $H$ is open in $\mathbb{R}^n$ and $B:=H \cap C$ is relatively open in $C$ by definition. Moreover, $H=C^\circ$, the interior of $C$, is the largest open subset of $C$, by definition. Since $C\subset E$, and $B$ is relatively open in $C$, $B$ must be relatively open in $E$:

$H\cap C = H\cap E=B$.

  • $\begingroup$ If by $C^o$ you mean the interior of $C$ in $R^n$ then this is wrong. Let $E$ be the set of members of $R^n$ ith rational co-ordinates, and let $C=E$. Then $H$ is empty, but $C$ isrelatively open in the subspace $E$ because $C=E.$ Consider that $E$ is a topological space, and every subset $C$ of a top. space $E$ has an interior $int_E(C)$, defined as the union of all $E$-open subsets of $C$, and it is the largest $E$-open subset of $C$. If $E$ is a subspace of $F$ then $int_E(C)=\cup \{t\cap C :t\in T_F\}$ where $T_F$ is the family of open sets in $F$. $\endgroup$ – DanielWainfleet Jan 27 '16 at 23:51
  • 1
    $\begingroup$ I dk if my comment came thru.My laptop is acting funny. But you are mistaken.I'll be back. $\endgroup$ – DanielWainfleet Jan 27 '16 at 23:53

Let $\{U_\alpha\}$ be the collection of all subsets of $C$ that are relatively open in $E$. The union $\cup U_\alpha$ is relatively open from the fact arbitrary unions of open sets are open.

The collection $\{U_\alpha\}$ may contain the empty set exclusively.


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.