I know that $[0,1]^{\Bbb R}$ with the product topology is not 1st countable. What I want now is to find a subset of $[0,1]^{\Bbb R}$ which is not closed but has all limit points. Does such a set exist? Then, what is it?
a set containing every limit points but not closed
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2You should define what you mean by "limit point" here; the usual definition is that $x$ is a limit point of $A$ if every neighborhood of $x$ contains a point of $A$ (other than $x$ itself). With this definition, in *any* topological space, a set is closed iff it contains all its limit points. So I suppose you want "limit point" to mean something else. – 2012-10-06
2 Answers
In general topology limit points are not defined solely by sequences, but rather by a much stronger construction called nets.
In turns it is equivalent to say "closed" and "contains all limit points of nets". It seems, to me, that you are asking for a sequentially-closed set which is not closed. Where "sequentially-" prefix means that we only require convergent sequences to have limit points.
Such sets exist, mainly because you already know that the space is compact and not first-countable and therefore sequentially-closed is not equivalent to closed anymore.
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0Of course, of course. The _mainly_ in the last sentence has the possibility of throwing people off. – 2012-10-06
You can consider the set of $(x_i)$ such that $x_i=0$, for all i except possibly a countable family.
This is sequentially closed (because a countable union of countable set is countable)
and this is dense (hence not closed) : any open of the basis is of the form : "the set of $(x_i)$ such that $x_{i_1} \in U_1 ... x_{i_n} \in U_n $ " so you can always find an element which is nul everywhere except on a finite set in it.
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0@Marc: In the original version of this answer it was quite unclear what was happening. – 2012-10-06