Prove that the limit of a real sequence (if it exists) can't be a complex number 
I was wondering whether a sequence, all of whose terms are real numbers can converge (if convergence is at all possible) to a complex number, or more generally to any other number field.  

Actually I think that answer to this this question should be answered at the outset before trying to prove any of the limit theorems, but unfortunately neither our analysis class professor and neither the tutor cared to prove this. 
I have tried to prove this but each time I find that my argument becomes circular. Is there any way to prove this? 
 A: A sequence $\{z_i\}$ of complex numbers converges if and only if each of the sequences $\{Re(z_i)\}$ and $\{Im(z_i)\}$ converge, where $Re(z)$ and $Im(z)$ are the real and imaginary parts of $z$ respectively.
If for each $i \in \mathbb{N}$ we have $Im(z_i) = 0$ and $\{z_i\}$ converges, then $\{Im(z_i)\} \rightarrow 0$ and $\lim_{i \rightarrow \infty}z_i$ is real.
In other words you can think of a sequence of reals as a sequence of complex numbers with imaginary part 0; then the convergence of the sequence implies that the imaginary part of the limit is zero.
A: No. A sequence of real values can not converge to any other field under the normal axioms. This is called the completeness axiom of $\mathbb{R}$. It states that the supremum or infimum of a subset of the real numbers must be in the reals. In this sense every sequence of real numbers converges to a real number. I do think that this can be extended into the next smallest ordinal (i.e. the one above the continuum ordinal). However, it is not typical to do so.
