Let $(A_n)$ be a decreasing sequence of subsets of the topological space $(X,\mathcal T)$ such that
- $A_1$ is dense in $X$.
- $A_{n+1}$ is dense in $A_n$.
is $\bigcap_{n=1}^\infty A_n$ dense in $X$?
Let $(A_n)$ be a decreasing sequence of subsets of the topological space $(X,\mathcal T)$ such that
is $\bigcap_{n=1}^\infty A_n$ dense in $X$?
On
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 $\bigcap_{n\ge 1}A_n$?
On
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 $\mathbb{R}$.
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.