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

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??

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.

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.

• How can we use sequential compactness though? we only know that its compact, not sequentially compact. Apr 26 at 0:49