Identity for the boundary of an open set with boundary in a Hilbert space Let $H$ be a $\mathbb R$-Hilbert space, $x\in H$ with $\left\|x\right\|_H=1$, $\alpha\in\mathbb R$ and \begin{align}\mathbb H_{x,\:\alpha}&:=\left\{y\in H:\langle x,y\rangle_H\ge\alpha\right\},\\\mathbb H_{x,\:\alpha}^\circ&:=\left\{y\in H:\langle x,y\rangle_H>\alpha\right\}.\end{align} If $U\subseteq\mathbb H_{x,\:\alpha}$, then $$\partial U:=U\setminus\mathbb H_{x,\:\alpha}^\circ.$$

Let $U$ be an open subset (in the subspace topology) of $\mathbb H_{x,\:\alpha}$ and $\operatorname{Int}U$ denote the interior of $U$ in $H$. How can we show that $\partial U=U\setminus\operatorname{Int}U$ and hence $\partial U$ is well-defined, i.e. independent of the choice of $x$ and $\alpha$?

Since $U$ is $\mathbb H_{x,\:\alpha}$-open, $$U=V\cap\mathbb H_{x,\:\alpha}\tag1$$ for some open subset $V$ of $H$. Now, $V\cap\mathbb H_{x,\:\alpha}^\circ$ is $H$-open and hence $$V\cap\mathbb H_{x,\:\alpha}^\circ\subseteq\operatorname{Int}U\tag2.$$ Let $y\in\operatorname{Int}U$. Then the closed ball $\overline B_\varepsilon(y)$ around $y$ with radius $\varepsilon$ is contained in $\operatorname{Int}U$, $$\overline B_\varepsilon(y)\subseteq\operatorname{Int}U\tag3,$$ for some $\varepsilon>0$. In particular, $$z:=y-\varepsilon x\in\operatorname{Int}U\tag4.$$ Since $\operatorname{Int}U\subseteq U\subseteq\mathbb H_{x,\:\alpha}$, $(4)$ implies $$\alpha\le\langle x,z\rangle_H=\langle x,y\rangle_H-\varepsilon<\langle x,y\rangle_H\tag5$$ and hence $$y\in\mathbb H_{x,\:\alpha}^\circ\tag6.$$

How can we conclude?

 A: The question consists of two parts:

*

*Let $Y:=\mathbb{H}_{x,\alpha}$, then its interior is $$Y^\circ=\{y:\langle x,y\rangle>\alpha\}$$
Proof: Call the RHS set $W$. If $\langle x,y\rangle>\alpha$ then the ball $B_\epsilon(y)$, $\epsilon<\langle x,y\rangle-\alpha$ is entirely within $Y^\circ$, $$\|y-z\|<\epsilon\implies \langle x,z\rangle=\langle x,y\rangle+\langle x,z-y\rangle\ge\langle x,y\rangle-\epsilon>\alpha$$
If $y\in Y$ is such that $\langle x,y\rangle=\alpha$, then any ball $B_\epsilon(y)$ contains points outside $Y$, for example, $z=y-\frac{\epsilon}{2} x$, $$\langle x,z\rangle=\alpha-\frac{\epsilon}{2}<\alpha$$ As $W$ is the largest open set in $Y$, $W=Y^\circ$.


*$U\setminus Y^\circ=U\setminus U^\circ$, given that $U$ is open in $Y$ (meaning $U=V\cap Y$ for some $V$ open in $H$). This part is entirely topological in character.

Since $U\subseteq Y$, then $U^{\circ c}\supseteq Y^{\circ c}$, so $U\setminus Y^{\circ}\subseteq U\setminus U^{\circ}$.
Conversely, $U\cap U^{\circ c}\subseteq Y^{\circ c}$ since $U^\circ=(V\cap Y)^\circ=V\cap Y^\circ$.
