[proofplan] The inclusion $K \subset (K^\circ)^\circ$ follows immediately from the definition of the polar set. For the reverse inclusion, we prove the contrapositive: if $x_0 \notin K$, then there is a vector $y \in K^\circ$ with $x_0 \cdot y > 1$. The separating vector is constructed directly from the closest point of $K$ to $x_0$, whose existence follows from closedness in finite dimension; convexity then gives the supporting inequality. [/proofplan] [step:Show every point of $K$ lies in the bipolar] Let $x_0 \in K$. To prove $x_0 \in (K^\circ)^\circ$, let $y \in K^\circ$. By the definition of the [polar set](/page/Polar%20Set), every point $x \in K$ satisfies $x \cdot y \leq 1$. Applying this with $x = x_0$ gives $x_0 \cdot y \leq 1$. Since this holds for every $y \in K^\circ$, the definition of the polar of $K^\circ$ gives $x_0 \in (K^\circ)^\circ$. Hence \begin{align*} K \subset (K^\circ)^\circ. \end{align*} [/step] [step:Choose a nearest point of $K$ to a point outside $K$] Let $x_0 \in \mathbb{R}^n \setminus K$. Define the distance from $x_0$ to $K$ by \begin{align*} d := \inf\{|x_0 - x| : x \in K\}. \end{align*} Since $K$ is nonempty, choose a sequence $(x_m)$ in $K$ such that $|x_0 - x_m| \to d$. For all sufficiently large $m$, we have $|x_0 - x_m| \leq d + 1$, and hence \begin{align*} |x_m| \leq |x_0| + |x_0 - x_m| \leq |x_0| + d + 1. \end{align*} Thus a tail of $(x_m)$ is bounded. By the [Bolzano-Weierstrass theorem](/page/Bolzano-Weierstrass%20Theorem), there is a subsequence $(x_{m_j})$ converging to some point $p \in \mathbb{R}^n$. Since $K$ is closed and every $x_{m_j}$ lies in $K$, we have $p \in K$. Continuity of the Euclidean norm gives \begin{align*} |x_0 - p| = \lim_{j \to \infty} |x_0 - x_{m_j}| = d. \end{align*} Because $x_0 \notin K$ and $p \in K$, we have $d = |x_0 - p| > 0$. [guided] We need an actual supporting vector, not only an infimum. The natural vector will point from the nearest point of $K$ to $x_0$, so we first prove that such a nearest point exists. Define \begin{align*} d := \inf\{|x_0 - x| : x \in K\}. \end{align*} Because $K$ is nonempty, this infimum is taken over a nonempty subset of $[0,\infty)$. Choose a sequence $(x_m)$ in $K$ such that $|x_0 - x_m| \to d$. This sequence is eventually bounded: for all sufficiently large $m$, \begin{align*} |x_0 - x_m| \leq d + 1, \end{align*} and therefore the triangle inequality gives \begin{align*} |x_m| \leq |x_0| + |x_0 - x_m| \leq |x_0| + d + 1. \end{align*} The [Bolzano-Weierstrass theorem](/page/Bolzano-Weierstrass%20Theorem) says that every bounded sequence in $\mathbb{R}^n$ has a convergent subsequence, so there is a subsequence $(x_{m_j})$ and a point $p \in \mathbb{R}^n$ such that $x_{m_j} \to p$. Closedness of $K$ is used here: since every $x_{m_j}$ belongs to $K$ and $K$ is closed, the limit $p$ also belongs to $K$. Finally, continuity of the Euclidean norm gives \begin{align*} |x_0 - p| = \lim_{j \to \infty} |x_0 - x_{m_j}| = d. \end{align*} Thus $p$ is a nearest point of $K$ to $x_0$. Since $x_0 \notin K$ but $p \in K$, the distance is positive: \begin{align*} d = |x_0 - p| > 0. \end{align*} [/guided] [/step] [step:Use convexity to obtain a supporting inequality] Define the vector $v \in \mathbb{R}^n$ by \begin{align*} v := x_0 - p. \end{align*} We claim that \begin{align*} x \cdot v \leq p \cdot v \end{align*} for every $x \in K$. Fix $x \in K$. Since $K$ is convex and $p,x \in K$, the point \begin{align*} p_t := (1-t)p + tx \end{align*} belongs to $K$ for every $t \in [0,1]$. Since $p$ minimizes the distance to $x_0$ over $K$, we have \begin{align*} |x_0 - p|^2 \leq |x_0 - p_t|^2 \end{align*} for every $t \in [0,1]$. Using $v = x_0 - p$ and $p_t = p + t(x-p)$, this becomes \begin{align*} |v|^2 \leq |v - t(x-p)|^2 = |v|^2 - 2t\,v \cdot (x-p) + t^2 |x-p|^2. \end{align*} Subtracting $|v|^2$ and dividing by $t > 0$ gives \begin{align*} 0 \leq -2\,v \cdot (x-p) + t |x-p|^2. \end{align*} Letting $t \downarrow 0$ yields $v \cdot (x-p) \leq 0$, equivalently \begin{align*} x \cdot v \leq p \cdot v. \end{align*} [guided] The nearest point $p$ gives a geometric normal vector. Define \begin{align*} v := x_0 - p. \end{align*} We will prove that the hyperplane through $p$ orthogonal to $v$ supports $K$. Fix $x \in K$. Convexity is used by moving from $p$ a small distance toward $x$. For each $t \in [0,1]$, define \begin{align*} p_t := (1-t)p + tx. \end{align*} Since $K$ is convex and both endpoints $p$ and $x$ lie in $K$, we have $p_t \in K$. Because $p$ is a nearest point of $K$ to $x_0$, the distance from $x_0$ to $p_t$ cannot be smaller than the distance from $x_0$ to $p$: \begin{align*} |x_0 - p|^2 \leq |x_0 - p_t|^2. \end{align*} Now rewrite this inequality using $v = x_0 - p$ and $p_t = p + t(x-p)$. Then \begin{align*} x_0 - p_t = x_0 - p - t(x-p) = v - t(x-p), \end{align*} so \begin{align*} |v|^2 \leq |v - t(x-p)|^2 = |v|^2 - 2t\,v \cdot (x-p) + t^2 |x-p|^2. \end{align*} Subtracting $|v|^2$ from both sides gives \begin{align*} 0 \leq -2t\,v \cdot (x-p) + t^2 |x-p|^2. \end{align*} For $t > 0$, divide by $t$: \begin{align*} 0 \leq -2\,v \cdot (x-p) + t |x-p|^2. \end{align*} Letting $t \downarrow 0$ removes the quadratic error term and gives \begin{align*} 0 \leq -2\,v \cdot (x-p), \end{align*} hence $v \cdot (x-p) \leq 0$. This is exactly \begin{align*} x \cdot v \leq p \cdot v. \end{align*} [/guided] [/step] [step:Normalize the supporting vector to separate $x_0$ from the bipolar] Since $0 \in K$, the supporting inequality with $x = 0$ gives \begin{align*} 0 \leq p \cdot v. \end{align*} Also, \begin{align*} x_0 \cdot v = (p+v)\cdot v = p \cdot v + |v|^2 > p \cdot v, \end{align*} because $|v| = d > 0$. Define $\lambda \in (0,\infty)$ by \begin{align*} \lambda := \begin{cases} \frac{1}{p \cdot v}, & \text{if } p \cdot v > 0,\\ \frac{2}{x_0 \cdot v}, & \text{if } p \cdot v = 0. \end{cases} \end{align*} Define $y \in \mathbb{R}^n$ by \begin{align*} y := \lambda v. \end{align*} If $p \cdot v > 0$, then for every $x \in K$, \begin{align*} x \cdot y = \lambda x \cdot v \leq \lambda p \cdot v = 1, \end{align*} while \begin{align*} x_0 \cdot y = \lambda x_0 \cdot v > \lambda p \cdot v = 1. \end{align*} If $p \cdot v = 0$, then for every $x \in K$, \begin{align*} x \cdot y = \lambda x \cdot v \leq 0 \leq 1, \end{align*} while \begin{align*} x_0 \cdot y = \frac{2}{x_0 \cdot v} x_0 \cdot v = 2 > 1. \end{align*} In both cases, $y \in K^\circ$ and $x_0 \cdot y > 1$. Hence $x_0 \notin (K^\circ)^\circ$. [guided] The supporting inequality gives $x \cdot v \leq p \cdot v$ for every $x \in K$. To turn this into membership in the polar set, we need the right-hand side to be at most $1$ after scaling. The assumption $0 \in K$ ensures that the supporting level is nonnegative: substituting $x = 0$ gives \begin{align*} 0 \cdot v \leq p \cdot v, \end{align*} so \begin{align*} 0 \leq p \cdot v. \end{align*} The outside point lies strictly beyond the supporting hyperplane because \begin{align*} x_0 \cdot v = (p+v)\cdot v = p \cdot v + |v|^2. \end{align*} Since $|v| = |x_0-p| > 0$, this gives \begin{align*} x_0 \cdot v > p \cdot v. \end{align*} There are two normalization cases. If $p \cdot v > 0$, define \begin{align*} \lambda := \frac{1}{p \cdot v}. \end{align*} If $p \cdot v = 0$, then $x_0 \cdot v = |v|^2 > 0$, and we define \begin{align*} \lambda := \frac{2}{x_0 \cdot v}. \end{align*} In either case $\lambda > 0$, so define \begin{align*} y := \lambda v. \end{align*} When $p \cdot v > 0$, the supporting inequality implies that for every $x \in K$, \begin{align*} x \cdot y = \lambda x \cdot v \leq \lambda p \cdot v = 1. \end{align*} Thus $y \in K^\circ$. At the same time, \begin{align*} x_0 \cdot y = \lambda x_0 \cdot v > \lambda p \cdot v = 1. \end{align*} So $y$ is a polar vector that excludes $x_0$ from the bipolar. When $p \cdot v = 0$, the supporting inequality gives $x \cdot v \leq 0$ for every $x \in K$. Therefore, for every $x \in K$, \begin{align*} x \cdot y = \lambda x \cdot v \leq 0 \leq 1, \end{align*} so again $y \in K^\circ$. The chosen scaling gives \begin{align*} x_0 \cdot y = \frac{2}{x_0 \cdot v} x_0 \cdot v = 2 > 1. \end{align*} Thus in both cases we have constructed $y \in K^\circ$ with $x_0 \cdot y > 1$, which means $x_0 \notin (K^\circ)^\circ$. [/guided] [/step] [step:Conclude equality of $K$ and its bipolar] We have shown that every $x_0 \in K$ belongs to $(K^\circ)^\circ$. We have also shown that every $x_0 \in \mathbb{R}^n \setminus K$ fails to belong to $(K^\circ)^\circ$. Therefore \begin{align*} (K^\circ)^\circ = K. \end{align*} This proves the theorem. [/step]