If G is open and dense subset in $\mathbb R$ then show that $G\setminus\{x\}$ is also open and dense in R. is it true in general metric space? If $G$ is open and dense subset in $\mathbb R$ then show that $G\setminus\{x\}$ is also open and dense in $\mathbb R$.  is it true in general metric space? 
I know as $G$ is open and singleton set $\{x\}$ is closed so that $G\setminus\{x\}$ has to be open, but how to show it is dense in $\mathbb R$. next the same will true in general Metric space.
 A: Take as your universe an interval and an isolated point. 
As universe it is open and dense. Remove the isolated point, what remains is open
(as its complement is the closed point) but sadly not dense anymore.
For your question you would need the property that every point is a density point of the universe (or the dense set).
A: It's not true in a general metric space.  Let $X$ be a metric space consisting of a single point $x$, with define $d(x,x) = 0$.  Then $G = \{x\}$ is open and dense in $X$, but $G \setminus \{x\} = \emptyset$ is no longer dense.
A: First, we prove that $G-\{x\}$ is open and dense in general topological space $X$ of $T_2$ (Hausdorff space) and above. 
Since $G-\{x\}=G\cap \{x\}^c$, and $\{x\}^c$ is open ($\{x\}$ is closed), $G-\{x\}$ is open.
Next we prove that $(G-\{x\})^c$ is nowhere dense in $X$. Since 
$$
(G-\{x\})^c=(G\cap \{x\}^c)^c=G^c\cup \{x\}
$$
if $\varnothing\ne O\subset(G-\{x\})^c$, $O$ is open, then $O\subset G^c$ for $\{x\}$ is closed, and so $O\subset (G^c)^o$. Since $G$ is dense in $X$, $G^c$ is nowhere dense in $X$, i.e. $(G^c)^o=\varnothing$. But this means that $O=\varnothing$, contradiction. Thus $(G-\{x\})^c$ is nowhere dense in $X$, and $(G-\{x\})$ is dense in $X$.
Edit:
Since $\Bbb{R}$ is Hausdorff space, above hold in $\Bbb{R}$. 
For general metric space that $X$ has isolated point, above doesn't hold.
A: $G\setminus\{x\}$ will be dense whenever $\{x\}$ isn't open, which, as stated in other answers, is true as long as $x$ isn't an isolated point.
An example that I like where this happens (and isn't just discrete where every subset is open) is $X=\{\frac{1}{n}:n\in\mathbb{N}^+\}\cup\{0\}$ viewed as a subspace of $\mathbb{R}$. So the metric on it is the usual metric on the real numbers. In this space, every point except 0 is isolated, while 0 isn't isolated because it has points arbitrarily close to it. 
And to see this is a counterexample to the property you are asking about, $X\setminus\{1\}$ is not dense, since it doesn't intersect $\{1\}$ which must be open because it equals $\{x\in X:|x-1|<\frac{1}{4}\}$.
