# Set $C$ contains set $B$ which is the largest subset of $C$ relatively open in $E$

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 $E$.

Proof:

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 $E$:

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

• 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 $F$. – DanielWainfleet Jan 27 '16 at 23:51
• I dk if my comment came thru.My laptop is acting funny. But you are mistaken.I'll be back. – 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.