If F is a compact topological field, then F is finite.

Question

Im trying to see why every compact topological field must be finite. Assuming the topological space is not the trivial topology.

Also: Does compact imply limit point compact in a topological field??

Answer 1

Let's be more general:

For a topological space and an element x of it $\mathfrak{U}(x)$ denotes the set of all open neighborhoods of $x$.

Let $R$ be a topological ring. A subset $A$ of $R$ is called bounded, if for all $U \in \mathfrak{U}(0)$ there is a $V \in \mathfrak{U}(0)$ such that $ A \cdot V \subset U$.

Lemma If $A \subset R$ is compact, then $A$ is bounded.

Proof. Let $U \in \mathfrak{U}(0)$. For $a \in A$, by continuity of $\cdot$, pick $W_a \in \mathfrak{U}(a)$ and $V_a \in \mathfrak{U}(0)$ such that $W_a \cdot V_a \subset U$. Since $A$ is compact there are $a_1,... ,a_n \in A$ such that $A \subset \bigcup_{i=1}^n W_{a_i}$. Then $V := \bigcap_{i=1}^n V_{a_i} \in \mathfrak{U}(0)$ and $A \cdot V \subset U$.

Now, let $F$ be a compact, T2 topological field. Assume $F$ is not discrete. By T2 there is a $U \in \mathfrak{U}(0)$ such that $1 \notin U$. By the lemma there is a $V \in \mathfrak{U}(0)$ such that $F \cdot V \subset U$. Since $F$ is not discrete, there is a $0 \neq v \in V$. Hence $1 = 1/v \cdot v \in U$. Contradiction. Hence, $F$ is discrete and therefore finite.

Note that we didn't use continuity of inversion.

Answer 2

I'll assume sequential compactness,

If $F$ is a topological field with non-discrete topology then there is a sequence $a_n \in F^*, a_n\to 0$.

If $F$ is compact then there is a subsequence such that $1/a_{k_j}$ converges, to some $ b\in F$.

The multiplication is continuous so the sequence $1=a_{k_j} (1/a_{k_j})$ converges to $0 \times b = 0$, this is a contradiction.