Proof of Sets Cut Out by Linear Inequalities in $\mathbb{R}^n$ Are Closed

Throughout, closedness in $\mathbb{R}^{n}$ refers to the metric topology $\mathcal{T}_{d_{E}}$, which exists by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii): a subset of $\mathbb{R}^{n}$ is closed when its complement is open in $(\mathbb{R}^{n},d_{E})$, by Closed Subset of a Topological Space (i) and Open Subset of a Metric Space (i), and $B(y,r)$ is the open ball of the base setting. We prove (ii) first; the other two clauses follow from it.

For (ii), let $P$ be the set in question and let $y\in\mathbb{R}^{n}\setminus P$, so that $w_{0}\cdot y>\beta$ for some $w_{0}\in W$. Put

$$r=\frac{w_{0}\cdot y-\beta}{\lVert w_{0}\rVert+1}>0.$$

Let $y'\in B(y,r)$, that is, $\lVert y'-y\rVert<r$ (Open Ball in a Metric Space (ii)). By the Cauchy--Schwarz inequality The Cauchy-Schwarz Inequality on $\mathbb{R}^n$ (i),

$$w_{0}\cdot y'=w_{0}\cdot y+w_{0}\cdot(y'-y)\ge w_{0}\cdot y-\lVert w_{0}\rVert\,\lVert y'-y\rVert>w_{0}\cdot y-(\lVert w_{0}\rVert+1)\,r=\beta,$$

so $y'\notin P$. Hence $B(y,r)\subseteq\mathbb{R}^{n}\setminus P$. As $y$ was an arbitrary point of the complement, the complement of $P$ is open, and $P$ is closed.

Clause (i) is the special case $W=\{w\}$ of (ii).

For (iii), Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (ii) and bilinearity give

$$H^{\le}(a,\alpha)=\{y:a\cdot y\le\alpha\},\qquad H^{\ge}(a,\alpha)=\{y:(-a)\cdot y\le-\alpha\},$$

and both are closed by (i), applied with $w=a$ and with $w=-a$ respectively. By Hyperplanes and Half-Spaces in $\mathbb{R}^n$ (i), $H(a,\alpha)=H^{\le}(a,\alpha)\cap H^{\ge}(a,\alpha)$. Its complement is the union of the two complements, which are open, so it is open by Topological Space (iii), and $H(a,\alpha)$ is closed.