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A subspace $X$ of a convergence space $Y$ is strictly dense if for every $y \in Y$ and every filter $\mathcal F$ converging to $y$, there is a filter $\mathcal G$ containing $X$ that converges to $y$ such that $\bar{\mathcal G} \subseteq \mathcal F$. Here, $\bar{\mathcal G}$ is just the filter generated by the closure of all the elements of $\mathcal G$.

I want to prove that if $Y$ is a topological space and $X$ is dense is $Y$, then $X$ is strictly dense. This is my current argument: Suppose $\mathcal F$ converges to $y$. Then $\mathcal U_y \subseteq \mathcal F$, where $\mathcal U_y$ is the neighborhood filter of $y$. Since $X$ is dense in $Y$, it intersects every open set containing $y$, hence there is a filter $\mathcal G$ that contains $X$ and all the elements of $\mathcal U_y$. Since $\mathcal U_y \subseteq \mathcal G$, we have that $\mathcal G$ converges to $y$. Now somehow I need to show that $\bar{\mathcal G} \subseteq \mathcal F$, but I have no idea how to go about it. Any tips?

Addendum: The filter $\mathcal G$ in my argument above is the one generated by the sets of the form $U \cap X$, where $U$ is an open set containing $y$. The filter $\bar{\mathcal G}$ is generated by the closure of such sets.

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I agree with your choice of $\mathcal{G}$, as the filter generated by the sets $U \cap X$, where $U$ runs over the open neighbourhoods of $y$.

Now, for a dense subset $X$ of $Y$ and any open set $O$, $\overline{O \cap X} = \overline{O}$.

Proof: the left to right inclusion is obvious from $O \cap X \subset O$. To see the right to left inclusion, let $p$ be any point of $\overline{O}$ and let $U_p$ be any open neighbourhood of $p$. Then $O \cap U_p$ is non-empty (as $p \in \overline{O}$) and open, and so must intersect $X$, as $X$ is dense, and so $(O \cap U_p) \cap X = U_p \cap (O \cap X)$ is non-empty. Note that we have showed that every open neighbourhood $U_p$ intersects $O \cap X$, so $p \in \overline{O \cap X}$, which shows the reverse inclusion.

So if we have a generating set for $\overline{\mathcal{G}}$, which (you say) is of the form $\overline{U \cap X}$ for $U \in \mathcal{U}_y$, it is in fact also of the form $\overline{U}$, which is in $\mathcal{F}$ as an enlargement of $U \in \mathcal{F}$.

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I was thinking about proving that $\overline{O \cap X} = \bar O$. Looks like my thoughts were in the right direction. Thanks. – echoone May 23 '13 at 19:50

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