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Let $(A_n)$ be a decreasing sequence of subsets of the topological space $(X,\mathcal T)$ such that

  1. $A_1$ is dense in $X$.
  2. $A_{n+1}$ is dense in $A_n$.

is $\bigcap_{n=1}^\infty A_n$ dense in $X$?

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1. and 2. can be equivalently replaced by: $A_n$ is dense in $X$. – 1015 Mar 2 '13 at 17:54
@julien: I think 1 & 2 imply $A_n$ is dense in $X$. But if the $A_n$ are dense in $X$, are 1 & 2 true? – user59671 Mar 2 '13 at 18:22
If $A_{n+1}$ is dense in $X$, then $A_{n+1}$ is a fortiori dense in any set that contains it, eg $A_n$. – 1015 Mar 2 '13 at 18:29
up vote 10 down vote accepted

Hint: Take $\{r_n,n\in\Bbb N\}$ an enumeration of rationals and $X_n:=\{r_k,k\geqslant n\}$. Each $X_n$ is dense in $X:=\Bbb R$ for the usual topology, but the intersection is empty.

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HINT: Let $\{q_n:n\in\Bbb Z^+\}$ be an enumeration of $\Bbb Q$. For $n\in\Bbb Z^+$ let $A_n=\{q_k:k\ge n\}$. What is $\bigcap_{n\ge 1}A_n$?

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The best counterexample has already been given.

Let me simply point out that if your property holds for every such sequence $A_n$, then in particular $X$ would have to be a Baire space.

And there exist topological spaces which are not Baire spaces.

As you can see here, a typical counterexample is $\mathbb{Q}$ with the usual topology induced by the metric on $\mathbb{R}$.

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thanks. also mentioned in – user59671 Mar 2 '13 at 18:27
@CutieKrait not quite, you are looking for a Baire space, not a set with the property of Baire. See – Henno Brandsma Mar 2 '13 at 21:36
@HennoBrandsma Thanks for this useful precision. – 1015 Mar 2 '13 at 21:40

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