Let $S\subset \mathbb{R}^n$ be a set. I work with the following definition of the relative interior
\begin{equation}
\text{relint}(S):=\{s\in S\,\,|\,\,\exists\,\delta>0: B(s,\delta)\cap \text{aff}(S) \subset S\}.
\end{equation}
My questions:
- Am I correct that this is nothing else than the interior of $S$ with respect to the subspace topology on $\text{aff}(S)$? I do not immediately see from the above definition why this is the case, but it seems to me like this is true.
- Is it correct that we have $\text{relint}(S)\neq \text{int}(S)\cap\text{aff}(S)$? I thought that we could choose $S$ to be a closed segment, where $\text{int}(S)$ is empty and $\text{relint}(S)$ equals the open segment.
- Is it true that we always have $\text{int}(S) \subset \text{relint}(S)$? I think this holds as $B(s,\delta)\subset S$ immediately implies $B(s,\delta)\cap\text{aff}(S)\subset S$.
Thank you in advance.
Best Answer
Yes this is true: from the definition of subspace topology interior, if $x \in \text{int}_{\text{aff}(C)}( C)$ then there exists a set of the form $U \cap \text{aff}(C)$ contained in $C$ that contains $x$ for some $U$ open in $\mathbb{R}^k$. Then choosing an open ball $B$ containing $x$ and contained in $U$, we obtain $x \in B\cap \text{aff}(C) \subseteq C$ so $x \in \text{ri}(C)$. See if you can prove the other inclusion yourself.
In general this is false, as you said. Take $[0,1]\times\{0\}$ in $\mathbb{R}^2$. The relative interior is $(0,1)$ while the interior is empty, and the affine hull is the $x$-axis.
True, for the reasons you stated.