Direct limit of directed system of countable subset Let $X$ be a arbitrary topological space.
 Consider the directed system of all countable subspaces of X (with their inclusion maps).
Can we conclude the direct limit of this system is $X$?
This question relate to my previous question.
Let $X$ be a compact space, in especially X=$S^1$ with subspace topology of $R^2$, Is above proposition true for $S^1 ?$
 A: No, that's not true: Consider $\aleph_1$ or any other uncountable set equipped with the cocountable topology, i.e. the topology in which the non-empty open sets are the complements of countable sets. Any countable subset of this topological space is discrete, hence so is their direct limit (by the universal property, any map originating from the direct limit is continuous); however, the cocountable topology itself is not discrete.
Addendum: Of course, there is always a canonical bijective continuous map $(\lim\limits_{A\subseteq X\text{ countable}} A)\rightarrow X$, so the topology coming from the countable subsets is at least as fine as the original topology on $X$. The example above, however, shows that the former may be discrete even if $X$ was not.
Addendum 2: Any first countable topological space has the property in question. Considering the bijection from the first addendum, it suffices to show that the image of a closed set $Z$ in $\lim\limits_{A\subseteq X\text{ countable}} A$ is also closed in $X$. In other words, we have to show that if $Z\cap A$ is closed in $A$ for any countable $A\subseteq X$, then $Z$ is closed in $X$. Now, if $X$ is first countable, it suffices to show that $Z$ is closed under taking limits of (${\mathbb N}$-indexed) sequences in $X$ (as opposed to taking limits of the more general notion of nets). So if $(x_n)\to x$ in $X$ with $\{x_n\}\subseteq Z$, then the same holds in the countable subspace $A:=\{x_n\}_n\cup\{x\}$, and since $Z\cap A$ is closed in $A$, it follows that $x\in Z\cap A\subseteq Z$ as desired. 
