MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I asked this question at but I didn't get any answer there.

In David Marker's Model Theory, there is a exercise said " we can view a countable Henkin construction as a forcing construction ". But in the book the Henkin construction is used to prove the compactness theorem. He didn't prove the completeness theorem.

So my question is can we use forcing method to prove the completeness of first-order logic?

Thank you in advance.

share|cite|improve this question
Five hours is not even a full night sleep for many people in the world. So you have waited that long? Oh dear. Also, you may want to add [forcing] tag to the MSE question. For completeness of the post, here is a link to the MSE post:… – Asaf Karagila Dec 18 '12 at 8:36
I added [forcing] tag to the MSE question. Sorry for my impatience, I just forget the time difference. – Chao Chen Dec 18 '12 at 9:10
The Henkin construction can be used to prove the completeness theorem as well as compactness, in fact it was invented for that very purpose. You should find this in any textbook on logic which is not specifically targeted on model theory, and so does not try to avoid syntax at all costs. – Emil Jeřábek Dec 18 '12 at 13:04
I have voted to close this question; it is on scope for MSE but I do not think it is the sort of "research level" question that this site is intended for. Even if the question is closed here it can be answered in detail on MSE. – Carl Mummert Dec 18 '12 at 13:51
@Chao: If you know how to prove completeness using Henkin completion, and you know how to recast Henkin completion for countable theories as a forcing construction in the proof of compactness, you can do exactly the same thing with Henkin completion in the proof of completeness. What is your question then? – Emil Jeřábek Dec 19 '12 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.