Dense sets - different intuitions are equivalent? Given a topological space $X$ and a subset $A$ that is dense in $X$, is it always true that $\forall x \in X$ there exists a sequence $\{a_n \in A\}$ of points such that the sequence converges to $x$?
It seems like this might not be true for all spaces; in my particular case I'm trying to prove it's true if $X$ is compact Hausdorff and $A$ is countable; but I can't seem to make any headway on my proof. It seems 'intuitively obvious', but that might be because I'm thinking of it in terms of metric spaces too much (the theorem is true for metric spaces $X$, since you can just restrict the size of a ball around a point $x$ and pick points $\{a_n\}$ from your dense subset to form a convergent sequence)
 A: Example 1. Let $X$ be an uncountable set with the cocountable topology, and let $A$ be any uncountable proper subset of $X.$ Then $A$ is dense in $X.$ There are no nontrivial convergent sequences in $X,$ i.e., every convergent sequence is eventually constant. This example is neither compact nor Hausdorff.
Example 2. Let $X=[0,\Omega],$ the set of all ordinals less than or equal to the first uncountable ordinal $\Omega=\omega_1,$ with the order topology, and let $A=[0,\Omega)$ be the set of all countable ordinals. Then $X$ is a compact Hausdorff space and $A$ is dense in $X,$ but no sequence in $A$ converges to the point $\Omega.$ This example is not separable.
Example 3. Let $X$ be a separable compact Hausdorff space of cardinality $2^{2^{\aleph_0}},$ for example the product space $[0,1]^{2^{\aleph_0}}$ (recall that the product of $2^{\aleph_0}$ separable spaces is separable), and let $A$ be a countable dense subset of $X.$ Inasmuch as there are only $2^{\aleph_0}$ sequences in $A,$ and each of those sequences converges to at most one point, it follows that for the majority of points $x$ in $X$ there is no sequence in $A$ converging to $x.$
A: Let $X$ be $\beta{\mathbb N}$, the Cech-Stone compactication of $\mathbb{N}$ (see wikipedia) has $\mathbb{N}$ as a dense subset and is compact Hausdorff. But for any $x \in \beta \mathbb{N} \setminus \mathbb{N}$. Then $x \in \overline{\mathbb{N}}$ but no sequence $(x_n)_n$ from $\mathbb{N}$ can converge to $x$. Suppose it did, then we can assume that $M = \{x_n : n \in \mathbb{N}\}$ is infinite (or the sequence could only converge to a point of $\mathbb{N}$ as it would have a constant subsequence). Split $M =M_1 \cup M_2$ in two infinite disjoint subsets
As $x$ can be seen as an ultrafilter $\mathscr{F}$ on $\mathbb{N}$) one of the $M_i$, say $M_{i_0}$, is not in $\mathscr{F}$. But then $(\mathbb{N}\setminus M_{i_0}) \cup \{x\}$ is a neighbourhood of $x$ that misses a infinitely many terms of the sequence, so $x_n \nrightarrow x$.
