Boyd & Vandenberghe, example 2.2 — interior and relative interior points

Question

I am reading Boyd & Vandenberghe's Convex Optimization. As I have not yet studied topology, I am having some trouble fully understanding example 2.2 on page 23:

Example 2.2 $ \ $ Consider a square in the $(x_1,x_2)$-plane in $\mathbf{R}^3$, defined as $$C=\{x\in\mathbf{R}^3\mid-1\leq x_1\leq1,\,-1\leq x_2\leq1,\,x_3=0\}.$$ Its affine hull is the $(x_1,x_2)$-plane, i.e., $\mathbf{af}\!\mathbf{f}\,C=\{x\in\mathbf{R}^3\mid x_3=0\}$. The interior of $C$ is empty, but the relative interior is $$\mathbf{relint}\,C=\{x\in\mathbf{R}^3\mid-1<x_1<1,\,-1<x_2<1,\,x_3=0\}.$$ Its boundary (in $\mathbf{R}^3$) is itself; its relative boundary is the wire-frame outline, $$\{x\in\mathbf{R}^3\mid\max\{|x_1|,|x_2|\}=1,\,x_3=0\}.$$

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Specifically, I was not fully understanding the concepts of interior and relative interior points. So I began conducting research, which resulted in me posting an answer. To ensure that I understood the concepts, especially within the context of example 2.2, I decided to "explain" the idea to myself, which resulted in the following explanation:

We have a square in the $(x_1, x_2)$-plane in $\mathbb{R}^3$, and, since $x_3 = 0$, the affine hull of $C$, $\mathbf{af}\!\mathbf{f}\, C$, is all of the $(x_1, x_2)$-plane, $\mathbf{af}\!\mathbf{f}\, C = \{ x \in \mathbb{R}^3 | x_3 = 0 \}$, as was stated.

The issue is that we had $B_{\delta}(x) \not \subseteq S$ (where $B_{\delta}(x)$ is the ball of radius $\delta$ entered at the point $x$), for any $\delta > 0$, where the subset of points $S$ is considered as a subset of $\mathbb{R}^3$. This can be thought of as the interior relative to $\mathbb{R}^3$. Instead, we now consider the subset of points $S$ as a subset of $\mathbf{af}\!\mathbf{f}\,C = \{ x \in \mathbb{R}^3 | x_3 = 0 \}$. This can be thought of as the interior relative to the affine hull.

Note that, when $S$ was considered as a subset of $\mathbb{R}^3$, we would have the ball $B_{\delta}(x) \subseteq S$ for any $\delta > 0$, where $S \subseteq \mathbb{R}^3$ is a set of points. Assuming our distance function $d$ for the associated metric space $(\mathbb{R}^3, d)$ is just the Euclidean distance, the ball $B_{\delta}(x)$ would literally just be a ball. However, when $S$ is considered as a subset of the affine hull $\mathbf{af}\!\mathbf{f}\,C$, we would have the ball $B_{\delta}(x) \subseteq S$ for any $\delta > 0$, where $S \subseteq \mathbf{af}\!\mathbf{f}\,C = \{ x \in \mathbb{R}^3 | x_3 = 0 \}$ is a set of points. Assuming our distance function $d$ for the associated metric space $(\mathbf{af}\!\mathbf{f}\,C, d)$ is just the Euclidean distance, the ball $B_{\delta}(x)$ would be the open disc. This is why the relative interior of $C$, $\mathbf{relint}\,C$, is $\underline{not}$ empty, unlike the interior of $C$ -- because the ball $B_{\delta}(x)$ is now the open disc.

I would greatly appreciate it if people could please take the time to review my understanding by ensuring that everything I wrote above is correct. Are there any errors? Can I improve my explanation?

Answer 1

It is all clear if you remember that in topology interior is always relative. We have a topological space $X$ and a set $A\subset X$, and then we define the interior of $A$ using open balls in $X$. So in fact to be precise it should be called interior of $A$ in $X$. The same applies to boundary.

Now the "relative interior" and "interior" in the sense of Boyd are simply "interior in $R^2$" and "interior in $R^3$".