This is a proof that a compact field is finite, so you might want to read that first. To help you compare this proof to the NSA proof that a compact field is finite (which uses non-standard analysis), I have endeavored to use similar terms in both.

Let F be a compact field, and suppose to the contrary that F is infinite. Take an infinite sequence of different elements of F. Applying compactness, we see that that sequence must have a subsequence yn which converges to a limit. Denote x = limn→yn. Define zn = yn-x. Then (subtraction is continuous in a topological field) limn→∞zn = 0.

Now define wn=zn-1 (the sequence of multiplicative inverses of the z's). F is compact, so there exists a convergent subsequence of wn: say v = limk→∞wnk exists, for some increasing sequence of indices nk. By continuity (recall that znk→0),

0*v = limk→∞ znk*wnk = limk→∞ 1 = 1
so 0*v=1. But this cannot be! (For instance, we deduce that 1=0*v=(0+0)*v=2*(0*v)=2, which is false).

So our assumption that F is infinite cannot hold; a compact field F must be finite.

The other proof is probably "simpler". Except that you need to know more logic, to work with non-standard analysis...

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