Boyd & Vandenberghe, question 2.31(d) — stuck on simple problem regarding interior of a dual cone In Boyd & Vandenberghe's Convex Optimization, question 2.31(d) asks to prove that the interior of the dual cone $K^*$ is equal to
$$\text{int } K^* = \{ z \mid z^\top x > 0, \,\forall x \in K \} \tag{1}$$
Recall that the dual cone of a cone $K$ is the set:
$$K^* = \{ y \mid y^\top x \ge 0 , \forall x \in K \}.$$
I've spent a solid chunk of time trying to prove this simple and seemingly evident statement about the interior of the dual cone but keep getting stuck on the problem of bounding the inner product of $x$ with a perturbed version of $z$ (details provided below).  Frustratingly, the proof in the book's answer key (available online) takes as given this very fact that that I am stuck on proving.
My work in proving statement (1) is given below.  I hope someone can show me the piece of mathematical technology I'm missing.  Thanks!

Question 2.31(d):
Let $K$ be a convex cone and $K^* = \{ y \mid y^\top x \ge 0 $ for all $ x \in K \}$ be its dual cone.
Let $S = \{ z \mid z^\top x > 0 $ for all $ x \in K \}.$   Show that $S = \text{int } K^*.$
Certainly $S \subseteq K^*.$  Now consider some arbitrary point $z_0 \in S$.  For all $x \in K$ we have $z_0^\top x > 0$.  It's clear that we need to find an $\epsilon$ such that for all $z' \in D(z_0, \epsilon)$,
$$z'^\top x > 0 ,\,\forall x \in K$$
Unfortunately, I don't know how to show $z'^\top x > 0$ for $z' \in D(z_0, \epsilon)$ when $\epsilon$ is chosen sufficiently small. I do know we can write $z'$ as $z_0 + \gamma u$ where $\|u\| = 1$ and $\gamma \in (0,\epsilon)$.  And I do know that
$$z_0^\top x - \gamma \|x\| ~\le~ z_0^\top x + \gamma u^\top x ~\le~ z_0^\top x + \gamma \|x\|$$
where $z_0^\top x > 0$ and $\gamma \|x\| \ge 0$. However, I don't know how to show the critical piece, that
$$z_0^\top x - \gamma \|x\| > 0$$
when $\epsilon$ is chosen sufficiently small since $x$ can range over $K$ and therefore $\|x\|$ can be arbitrarily large. Frustratingly, the solution in B&V just takes it for granted that for sufficiently small $\epsilon$,
$$z_0^\top x + \gamma u^\top x > 0$$
I've looked online for some matrix perturbation theory results to apply but nothing I've found has been useful.
Any help is greatly appreciated.
 A: Okay, the critical piece I was missing is that for any $z$, $z^\top x > 0 \iff z^\top x / \|x\| > 0.$
So let $x_0 = \text{arg inf } \{ z^\top x \mid x \in \text{cl}(K), \|x\|=1 \}$
and let $c_{x_0} = z^\top x_0 > 0$.  We know $x_0$ exists because the set $\text{cl}(K)\cap \{x \mid \|x\| = 1 \}$ is closed and bounded.
Now let $\varepsilon = c_{x_0} / 2$ so that for any $z' = z + \gamma u$ where  $\|u\|=1$ and  $\gamma \in (0,\varepsilon)$, we have
$$z'^\top x_0 \ge c_{x_0} - \gamma > c_{x_0} / 2 > 0.$$
Furthermore, for any other $x \in \text{cl}(K)\cap \{x \mid \|x\| = 1 \}$ we know
$$z'^\top x \ge z'^\top x_0 > 0.$$
Therefore, for any $x \in \text{cl}(K) \backslash \{0\}$ we have
$$z'^\top x / \|x\| > 0$$
which proves $z'^\top x > 0$ whenever $z' \in D(z,\varepsilon)$.
The other piece of the proof which is to show that $z^\top x = 0$ for $x \in \text{cl}(K) \backslash \{0\}$ only when $z$ is on the boundary of $K^*$.  This isn't very difficult.
A: The question has an additional bug: $K$ should be assumed to be closed. For instance, take $K$ to be the positive orthant together with $0$, i.e., $K=\{x\in\mathbb{R}^{n}:x_{i}>0,i=1,\ldots n\} \cup \{0\}$. Then $K^{\ast}=\mathbb{R}_{+}^{n}=\{x\in \mathbb{R}^{n}:x_{i}\geq0,i=1,\ldots n\}$. Every element $z\in K^{\ast}\backslash \{0\}$ satisfies $z^{T}x>0$ for all $x\in K\backslash \{0\}$, even those at the boundary of $K^{\ast}$.
