[proofplan]
We verify each property of $\mathcal{O}_K$ separately. Openness follows from the fact that $\mathcal{O}_K = B(0, 1)$ is a closed ball, which is open by the [Closed Balls Are Open](/theorems/???) theorem. The subring property is checked directly using the ultrametric inequality and multiplicativity. The ideal property of the sub-level sets follows from the ultrametric inequality and multiplicativity of $|\cdot|$. The unit group characterisation follows from $|x^{-1}| = |x|^{-1}$.
[/proofplan]
[step:Show $\mathcal{O}_K$ is an open subset of $K$]
By definition, $\mathcal{O}_K = \{x \in K : |x| \leq 1\} = B(0, 1)$, the closed ball of radius $1$ centred at $0$. By the [Closed Balls Are Open](/theorems/???) theorem (which applies since $(K, |\cdot|)$ is non-archimedean), $B(0, 1)$ is an open subset of $K$.
Similarly, for each $r \in (0, 1]$, the sets $\{x : |x| \leq r\} = B(0, r)$ and $\{x : |x| < r\} = B^\circ(0, r)$ are both open (the former by the same theorem, and the latter as an open ball).
[/step]
[step:Show $\mathcal{O}_K$ is a subring of $K$]
We verify the subring axioms.
**Contains $1$ and $-1$:** Since $|\cdot|$ is an absolute value, $|1| = 1 \leq 1$ and $|-1| = |{-1}| = 1 \leq 1$ (because $|{-1}|^2 = |(-1)^2| = |1| = 1$ and $|{-1}| > 0$, so $|{-1}| = 1$). Thus $1, -1 \in \mathcal{O}_K$.
**Closed under addition:** Let $x, y \in \mathcal{O}_K$, so $|x| \leq 1$ and $|y| \leq 1$. By the ultrametric inequality:
\begin{align*}
|x + y| \leq \max(|x|, |y|) \leq 1.
\end{align*}
Therefore $x + y \in \mathcal{O}_K$.
**Closed under multiplication:** Let $x, y \in \mathcal{O}_K$. By multiplicativity of $|\cdot|$:
\begin{align*}
|xy| = |x| \cdot |y| \leq 1 \cdot 1 = 1.
\end{align*}
Therefore $xy \in \mathcal{O}_K$.
[/step]
[step:Show $\{x : |x| < r\}$ and $\{x : |x| \leq r\}$ are ideals of $\mathcal{O}_K$ for $r \in (0,1]$]
Let $\mathfrak{a}_r = \{x \in K : |x| \leq r\}$ (the argument for $\{x : |x| < r\}$ is identical with strict inequalities).
**Additive subgroup:** If $x, y \in \mathfrak{a}_r$, then $|x + y| \leq \max(|x|, |y|) \leq r$ by the ultrametric inequality, so $x + y \in \mathfrak{a}_r$. Since $|-x| = |x| \leq r$, we have $-x \in \mathfrak{a}_r$. Also $|0| = 0 \leq r$, so $0 \in \mathfrak{a}_r$.
**Absorption:** Let $a \in \mathcal{O}_K$ and $y \in \mathfrak{a}_r$. Then
\begin{align*}
|ay| = |a| \cdot |y| \leq 1 \cdot r = r,
\end{align*}
so $ay \in \mathfrak{a}_r$. This shows $\mathfrak{a}_r$ is closed under multiplication by elements of $\mathcal{O}_K$, i.e., $\mathfrak{a}_r$ is an ideal of $\mathcal{O}_K$.
[guided]
An ideal $\mathfrak{a}$ of a ring $R$ must satisfy two properties: (i) $\mathfrak{a}$ is an additive subgroup of $R$, and (ii) $R \cdot \mathfrak{a} \subseteq \mathfrak{a}$ (absorption by ring elements).
For property (i): given $x, y \in \mathfrak{a}_r$ with $|x|, |y| \leq r$, the ultrametric inequality gives $|x + y| \leq \max(|x|, |y|) \leq r$, so $x + y \in \mathfrak{a}_r$. Since $|-x| = |{-1}| \cdot |x| = |x| \leq r$, we have $-x \in \mathfrak{a}_r$. And $0 \in \mathfrak{a}_r$ since $|0| = 0 \leq r$.
For property (ii): given $a \in \mathcal{O}_K$ (so $|a| \leq 1$) and $y \in \mathfrak{a}_r$ (so $|y| \leq r$), multiplicativity gives $|ay| = |a| \cdot |y| \leq 1 \cdot r = r$, so $ay \in \mathfrak{a}_r$.
The condition $r \leq 1$ is essential for the absorption property: we need $|a| \leq 1$ (i.e., $a \in \mathcal{O}_K$), not $|a| \leq r$. If $r > 1$, the set $\{x : |x| \leq r\}$ is still an additive subgroup but need not be an ideal of $\mathcal{O}_K$ (though this is moot, since $\{x : |x| \leq r\}$ for $r > 1$ properly contains $\mathcal{O}_K$).
[/guided]
[/step]
[step:Characterise the unit group as $\mathcal{O}_K^\times = \{x : |x| = 1\}$]
**($\supseteq$):** Suppose $|x| = 1$. Then $x \neq 0$, so $x^{-1}$ exists in $K$, and $|x^{-1}| = |x|^{-1} = 1$. Therefore both $x \in \mathcal{O}_K$ and $x^{-1} \in \mathcal{O}_K$, so $x \in \mathcal{O}_K^\times$.
**($\subseteq$):** Suppose $x \in \mathcal{O}_K^\times$, so both $x, x^{-1} \in \mathcal{O}_K$. Then $|x| \leq 1$ and $|x^{-1}| = |x|^{-1} \leq 1$. The second inequality gives $|x| \geq 1$. Combining: $|x| = 1$.
[/step]
