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Math Help - Q. in Dense -----> @ Topology

  1. #1
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    Q. in Dense -----> @ Topology

    ( 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 ))

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    Quote Originally Posted by amro05 View Post
    ( 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 ))

    I don't understand the problem. This statement is not true. Take the rationals as a dense subset of the reals with the usual topology.
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    Quote Originally Posted by Focus View Post
    I don't understand the problem. This statement is not true. Take the rationals as a dense subset of the reals with the usual topology.
    first of all >>>> thnx alout
    now ,, I`m very shore that this Q is true
    and by the way ,, the rationals is not a dense in R (( there is the irrationals in R who didn`t intersect the rationals ))
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    Quote Originally Posted by amro05 View Post
    first of all >>>> thnx alout
    now ,, I`m very shore that this Q is true
    and by the way ,, the rationals is not a dense in R (( there is the irrationals in R who didn`t intersect the rationals ))

    Oh, no, no, no....you better check your definitions and facts alright: the rationals \mathbb{Q} are dense too in the reals \mathbb{R} under the usual, euclidean topology.

    Definition: if \mathbb{X} is a topological space, a subset S\subset\mathbb{X} is dense in \mathbb{X} if \overline{S}=\mathbb{X}\,\Longleftrightarrow\,S\ca  p A \neq\emptyset\,\,\,\,\forallopen set 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
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    Quote Originally Posted by tonio View Post
    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...
    Well quite simply put, the only closed dense subset is the whole space, so the question seems odd.
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    Quote Originally Posted by Focus View Post
    Well quite simply put, the only closed dense subset is the whole space, so the question seems odd.

    I don't understand: if you pay attention I was addressing the OP who told YOU that the rationals aren't dense in the reals, which of course is wrong.
    I was not addressing you.

    Tonio
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    Quote Originally Posted by tonio View Post

    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
    THANX ALOUT TONIO
    SO YOU THINK THAT , I DIDN`T KNOW ANY THINK ABOUT DENSE ( AND WAY NOT , SO AS TOPLOGY )
    THAT IS A BUTIFUL WAY TO SOLVE ANY Q.
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    Quote Originally Posted by tonio View Post
    I don't understand: if you pay attention I was addressing the OP who told YOU that the rationals aren't dense in the reals, which of course is wrong.
    I was not addressing you.

    Tonio
    If X = R then
    T={ {} , R , Q , Q' } it is a toplolgy (( if u don`t know ))
    hear Q ( the rationals ) is not a dense

    what can u say about it ,, is it wrong ?????
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    Quote Originally Posted by amro05 View Post
    THANX ALOUT TONIO
    SO YOU THINK THAT , I DIDN`T KNOW ANY THINK ABOUT DENSE ( AND WAY NOT , SO AS TOPLOGY )
    THAT IS A BUTIFUL WAY TO SOLVE ANY Q.

    I didn't say you don't know topology. Read again my post...and don't write only with capital letters, please.

    Tonio
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    Quote Originally Posted by amro05 View Post
    If X = R then
    T={ {} , R , Q , Q' } it is a toplolgy (( if u don`t know ))
    hear Q ( the rationals ) is not a dense

    what can u say about it ,, is it wrong ?????

    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
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    Quote Originally Posted by tonio View Post
    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
    two or three days tolking about Q ( dense or not ) ...
    what about the real qustion ????
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    Quote Originally Posted by amro05 View Post
    two or three days tolking about Q ( dense or not ) ...
    what about the real qustion ????

    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 \mathbb{Q} in the real numbers set \mathbb{R} with the euclidean topology.
    If you're asking something else then I can't say what it is.

    Tonio
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    @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.
    Last edited by mr fantastic; January 8th 2011 at 12:52 PM.
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    Quote Originally Posted by Plato View Post
    @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.
    I agree.

    This thread has started to create unnecessary and undesirable frustration. Thread closed.
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