Question 1199314
**1. Key Idea**

* The crux of the argument lies in the uniqueness of limits in complete metric spaces. 

**2. Proof Outline**

* **Assume:** We have a complete subfield F of Q_p. This means F is a field itself, and it's complete with respect to the same p-adic absolute value |.|_p as Q_p.

* **Show Q_p ⊆ F:**
    * Take any x ∈ Q_p. 
    * Since Q_p is complete, there exists a Cauchy sequence (x_n) in Q ⊆ F such that x_n → x in Q_p. 
    * Since F is a subfield of Q_p, it contains all rational numbers (Q). Therefore, (x_n) is also a Cauchy sequence in F.
    * Since F is complete, this Cauchy sequence (x_n) must converge to a limit y in F.
    * **Uniqueness of Limits:** In any metric space (and hence in Q_p and F), limits are unique. Since x_n → x in Q_p and x_n → y in F, we must have x = y.
    * This shows that for any x ∈ Q_p, there exists a corresponding y ∈ F (namely, y = x). 
    * Therefore, Q_p ⊆ F.

* **Conclusion:**

    * We started with F ⊆ Q_p and proved Q_p ⊆ F. 
    * Combining these, we conclude that F = Q_p.

**In essence:**

* The completeness of both F and Q_p, along with the uniqueness of limits in complete metric spaces, forces any complete subfield of Q_p to be equal to Q_p itself.

**Note:** This proof relies heavily on the uniqueness of limits in complete metric spaces. 

Let me know if you have any other questions or would like to explore related concepts!