( X , T ) is a topological space & D is a subset of X
If D is Dense then d is closed set .
(( rem ,, closure of D = X ))
Oh, no, no, no....you better check your definitions and facts alright: the rationals $\displaystyle \mathbb{Q}$ are dense too in the reals $\displaystyle \mathbb{R}$ under the usual, euclidean topology.
Definition: if $\displaystyle \mathbb{X}$ is a topological space, a subset $\displaystyle S\subset\mathbb{X}$ is dense in $\displaystyle \mathbb{X}$ if $\displaystyle \overline{S}=\mathbb{X}\,\Longleftrightarrow\,S\ca p A \neq\emptyset\,\,\,\,\forall$open set $\displaystyle A\subset\mathbb{X}$.
So either you don't know the basic facts about dense subsets in topological spaces or else you're not actually asking what you think you are...
Tonio
I know that, and no: it isn't wrong. That's why I pointed out that the rationals are dense in the reals with the usual, euclidean topology. If you don't specify topology, by default most mathematicians will surely assume you're talking of the euclidean topology...don't you think?
Tonio
I think people isn't really sure what you're actually asking, partly because of the poor language and partly because of the even poorer redaction.
Here is your original question :
"( X , T ) is a topological space & D is a subset of X
If D is Dense then d is closed set .
(( rem ,, closure of D = X ))"
What are you asking here? To prove that if a set is dense in a topological space then it is closed? Well, you were already told: this is blatantly false, and as an easy counter-example take (guess what!)...yes, the rational numbers set $\displaystyle \mathbb{Q}$ in the real numbers set $\displaystyle \mathbb{R}$ with the euclidean topology.
If you're asking something else then I can't say what it is.
Tonio
@amro05:
The statement that In a topological space a dense subset is closed. is simply false.
To say that there are examples of spaces in which it is true does not change that fact.
Do you understand the difference in those two concepts?
What you have posted in this thread makes me doubt that you understand this question.