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Let $X$ a space; and $Y$ is a dense subset of $X$. However there may be some subset $Z$ of $Y$ such that $Z$ is also dense in $X$. Now I want to delete some elements of $Y$ which seem needless, so that I can make the new dense subset $Z$ of $Y$ be the smallest. How can I get it?

My idea is this: If there exists $y \in \overline{Y\setminus \{y\}}$, then delete $y$, because $Y\setminus \{y\}$ is also dense in $X$. However what should I do next step?

Thanks ahead:)

Paul
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    In general, there isn't any smallest. E.g., given any set of rationals dense in the reals, you can always remove one more rational and still have a dense subset. – Gerry Myerson Mar 25 '13 at 12:29

2 Answers2

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Note that for $X=\Bbb R$, removing any finite subset of $\Bbb Q$ is still a dense subset. Enumerate $\Bbb Q$ as $q_n$, and then consider $Q_n=\{q_k\mid k>n\}$. This is a decreasing chain of dense subsets, and the intersection is empty.

Asaf Karagila
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As Gerry has pointed out, this construction is not possible.

Let me compare this to, e.g., the construction of the closure of a set $A$ as the smallest closed set containing $A$. In this case you have the property that any intersection of closed sets containing $A$ is again closed and contains $A$. Therefore, you can write $$\bar A = \bigcap_{B \supset A, \; B\text{ closed}} B.$$

In the case of dense subsets, this strategy does not work.

gerw
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