Re: a subset in X A is dense iff X intersection A is not empty


"W. Dale Hall" <mailtodhall@xxxxxxxxx> wrote in message
news:yahxf.128$PL5.112@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
>
> John Smith wrote:
>> I am reading a book about manifolds and this is something I should know
>> but I don't know how to prove it.
>> Show that a subset A in X is dense if and only if every nonempty open set
>> in X contains a point of A.
>
> This is true for any topological space X and dense subset A.
>
> The proof is a simple application of the definition of a
> dense subset. You do need to use both of the following facts:
>
> 1. The complement of a closed subset is open,
> 2. The only closed set containing a dense subset
> is the whole space X.
>
> If you'll assure me that this isn't a homework problem, I'll post
> a simple proof (I've written one up for the occasion, but just
> now decided I should check on the homework issue; no offense is
> intended)
>
> Dale
no offense taken. I don't know how to assure you that it's not hw. but the
university I go to don't give any hw. you study the whole thing on your
own. read the book and understand it mostly by yourself. here is a version
of my proof:
(<=) cl(A) is per definition a subset of X.
let q be in X. if q is in A then q is in cl(A) since A is a subset
of cl(A). so asumme that q is not in A. let U be an arbitrary neighborhood
of q.
Since U\{q} intersection A = U intersection A is not empty is q a limit
point of A. that is q is in cl(A).

is this right? I just need to prove (=>).


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