# a set containing every limit points but not closed

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?

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You 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. – Nate Eldredge Oct 6 '12 at 15:09

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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Note that there are sequential spaces which are not first-countable. – Arthur Fischer Oct 6 '12 at 11:15
@Arthur: Good point, but those are not compact, are they now? – Asaf Karagila Oct 6 '12 at 11:15
Of course, of course. The mainly in the last sentence has the possibility of throwing people off. – Arthur Fischer Oct 6 '12 at 11:35

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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jwchoi wants a not closed set.. – Berci Oct 6 '12 at 11:24
@Berci: This set is not closed (read the answer: being dense, if it were also closed it would have to be everything, but it isn't). – Marc van Leeuwen Oct 6 '12 at 14:31
@Marc: In the original version of this answer it was quite unclear what was happening. – Arthur Fischer Oct 6 '12 at 14:36