A normed linear space $X$ is separable if and only if it has a denumerable
dense subset.
Proof: Suppose $X$ is separable. Then $X$ has a countable dense subset, say $D$. If $D$ is denumerable, there is nothing to prove. If $D$ is finite, define $D':=D\cup\{q_iy \mid q_i\in\mathbb Q\}$, where $y$ is some fixed element in $D$. Then $D'$ is a denumerable dense subset.
Conversely, if $X$ has a denumerable dense subset, then $X$ is separable by definition. $\blacksquare$
Is the above proof correct? It seems so simple. Is there anything that I'm overlooking?
Thank you.
Best Answer
The only mistake is you forgot to say $y \neq 0$. (If $X=\{0\}$ then the result is false).