Sequence in Hausdorff Space can have at most $2^\mathfrak{c}$ many cluster points I need to show that in a Hausdorff space any sequence can have at most $2^\mathfrak{c}$ many cluster points. I've tried rigging some product space in such a way to make this work, but honestly I'm at a loss with how to proceed with this one. I'm also supposed to show that if I have the extra condition that this space is first countable, then I can have at most $2^\omega$ many cluster points with my sequence. 
I tried to think of an example of a Hausdorff space which has a sequence with $2^\mathfrak{c}$ many cluster points to get a feel for the this sort of space, but I couldn't even do that. I'd appreciate any guidance. Thanks.
 A: Notation. $|Y|$ is the cardinal of a set $Y$ and $P(Y)$ is the set of all subsets of $Y.$
Definition. A filter on a non-empty set $X$ is a non-empty $F\subset P(X)$ such that (i)  if $X\supset A\supset B\in F$ then $A\in F, $ and (ii) if $A,A'\in F$ then $A\cap A'\in F,  $ and (iii) $\emptyset \not \in F.$
Let $\Bbb G$ be the set of filters on $\omega.$
Let $S$ be a Hausdorff space and let $\emptyset \ne T=\{s_n:n\in \omega\}\subset S.$ 
Let $C$ be the set of cluster points of $T$.
(I)..... For  $p\in C$ let $U(p)$ be the family of open subsets of $X$ that contain $p.$ Let $V(p)=\{\{n\in \omega: s_n\in V\}: V\in U(p)\}$ and let $F(p)=\{A\cup B: a\in V(p)\land B\subset \omega\}.$
Now $F(p)$ is a filter on $\omega.$ So $F$ is a  function from $C$  into  $\Bbb G.$
If $p_1,p_2$ are unequal members of $C,$ there exist disjoint $U_1\in U(p_1), U_2\in U(p_2),$ so $\{n\in \omega:s_n\in U(p_1)\}$  and $\{n\in \omega:n\in U(p_2)\}$ are disjoint members of $F(p_1),F(p_2),$ respectively. By (ii) and (iii) of the def'n of a filter, we must have $F(p_1)\ne F(p_2).$
So $F$ is injective. So $|C|\le |\Bbb G|.$ And since $\Bbb G \subset P(P(\omega))$ we have $|\Bbb F|\le |P(P(\omega))|=2^{2^{\omega}}=2^c. $
(II)....If $S$ is also $1$st-countable, then for each $p\in C$ let $V(p)$ be a countable local base at  $p$  and let $W(p)=\{\{n\in \omega: s_n\in V \}:V\in V(p)\}.$
Now if $p_1,p_2$ are distinct members of $C$, then, as before, let $U_1,U_2$ be disjoint members of $U(p_1),U(p_2)$ respectively. Now let $U_1\supset V_1\in V(p_1)$ and $ U_2\supset V_2\in V(p_2).$ Then we see that $\{n\in \omega:s_n\in V_1\}\in W(p_1)\setminus W(p_2), $ so $W(p_1)\ne W(p_2).$ 
Now each $W(p)$ is a countable subset of $P(\omega),$ and the family of all countable subsets of $P(\omega)$ has cardinal $2^{\omega}.$ So, since $W$ is injective, therefore $|C|\le 2^{\omega}.$
Remark. If $X$ is any infinite set and $G$ is the set of all filters on $X,$ then $G\subset P(P(X))$ so $|G|\le 2^{2^{|X|}}.$ In fact we have $|G|=2^{2^{|X|}}$ but this is not obvious.
Remark. I do not know of a simple example with $|C|=2^c,$  but there probably is one.  An advanced example: With the discrete topology on $\Bbb N,$ the maximal (Cech-Stone) compactification $\beta \Bbb N$ is a compact Hausdorff space of cardinal $2^c$ in which the closure of any infinite subset also has cardinal $2^c.$ The space $\beta \Bbb N$ can also be described as the Wallman extension $w\Bbb N$, whose points are the $\subset$-maximal filters (the ultra-filters) on $\Bbb N.$ 
