Problem regarding closed set and dense set In a metric space $X$ if $S$ is a proper subset of $X$ and $S$ is closed . Does it imply that it is not dense. But not closed does not imply dense. Is this correct ?
Additionally, are irrational numbers dense in $R$ ? How to prove it ?
 A: I think you're referring to the fact that the closure of a dense subset is the whole space itself. And no; non-closed sets are not necessarily dense; take, e.g. $B(0; 1)$ in $\mathbb R$ ; it is open, non-closed , and it is not dense; its closure is the closed ball $B(0;1)$ , i.e., the set {$x: ||x||\leq 1$}. You have both the irrationals and the rationals both dense and neither is closed and neither is open.
And to show irrationals are dense, for every irrational x , the constant sequence {$x,x,x...,x,...$} converges to $x$ , and for q rational, the sequence:
$\frac {2^{1/2}}{n}+q$ is a sequence of irrationals that converges to $q$
A: To prove the irrationals are dense in $\mathbb{R}$, it suffices to show that every basis element has a non-empty intersection with the set of irrationals. 
A: If a set is closed and dense, then it is the whole space, which is permissible.
The set $(0,1)$ is not closed in $\mathbb{R}$ and is not dense.
The set $\mathbb{Q}$ is not closed in $\mathbb{R}$ but is dense.
Let $C$ be a closed set (in $\mathbb{R}$) containing the irrationals. Let $x \in \mathbb{R}$. If $x$ is irrational, then $x \in C$, otherwise $x$ is rational. Let $\alpha$ be irrational, then  $x+\frac{\alpha}{n} \in C$ and $x+\frac{\alpha}{n} \to x$. Since $C$ is closed, we have $x \in C$. Hence $C=\mathbb{R}$, and it follows that the irrationals are dense in $\mathbb{R}$.
