Let $K$ be a subfield of $\mathbb C$ not contained in $\mathbb R$. Is $K$ always dense in $\mathbb C$?
I have studied to show a set $A$ is dense in $B$ we will have to show that for any element $x$ in $B$ there exists a sequence in $A$ converging to $x$.But dont know what should be my approach here?
It is true; here is an explicit proof.
We know that as $K$ is a subfield of $\mathbb{C}$, we must have $\mathbb{Q} \subset K$. As $K$ is not a subfield of $\mathbb{R}$, there is some element $\alpha \in \mathbb{C}\backslash\mathbb{R}$ with $\alpha \in K$. So, in particular, $\mathbb{Q}(\alpha) \subset K$, and it suffices to prove that $\mathbb{Q}(\alpha)$ is dense in $\mathbb{C}$.
As a real vector space, $\mathbb{C}$ has basis $\{1,\alpha\}$. Thus any $\beta \in \mathbb{C}$ can be written uniquely as $\lambda + \mu \alpha$ for some $\lambda, \mu \in \mathbb{R}$.
Now, an equivalent definition of 'dense' to the one you allude to above is that for every open set $U \subset \mathbb{C}$, we have $U \cap K \neq \emptyset$. By passing to a subset of $U$, we may assume that $U$ takes the form $U = \{\lambda + \mu \alpha: \lambda \in (a_1,b_2), \mu \in (b_1,b_2)\},$ with $a_i, b_i \in \mathbb{R}$. Then as $\mathbb{Q}$ is dense in $\mathbb{R}$, we can pick $a,b \in \mathbb{Q}$ such that $a \in (a_1,a_2), b \in (b_1,b_2)$. Then $a + b\alpha \in \mathbb{Q}(\alpha)\cap U$, so the intersection is non-empty and hence $K$ is dense in $\mathbb{C}$.