2
$\begingroup$

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??

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • 7
    $\begingroup$ How can we use sequential compactness though? we only know that its compact, not sequentially compact. $\endgroup$ Apr 26 at 0:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.