A noncontinuous function which preserves limits I know that if a given function $f$ between two topological spaces is continuous then the image of a convergent sequence is a convergent sequence, and $f$ preserves the limit in the sense that $x_n \to x \implies f(x_n) \to f(x)$.
The converse is true if the domain is first countable: If $X$ satisfies the first axiom of countability and $x_n \to x \implies f(x_n) \to f(x)$ for any convergent sequence in $X$ then $f$ is continuous. 
Thus this still holds if the condition on the domain is dropped? I've been trying to find an counterexample using non first countable spaces like the Sorgenfrey's Line and real numbers with the cofinite topology as the domain but every time I try to break the continuity of the function I end up with convergent sequences whose images don't converge. I was wondering if the statement isn't actually true and can be proved without the first axiom using a different technique maybe.
 A: Let $\tau$ be the co-countable topology in $\mathbb R$. Then a sequence $x_n$ in $(\mathbb R, \tau)$ converges to $x$ if and only if $x_n$ is eventually constant. In fact, suppose the sequence is often different from $x$, that is, given any $n \in \mathbb N$ there is a $m \geq n$ such that $x_m \neq x$. Then the set $\mathbb R \setminus \{x_m; x_m \neq x\}$ is an open neighborhood of $x$ which doens't contain any element of sequence $x_n$, so $x_n$ can not converge to $x$.
This implies every map $f$ on $(\mathbb R, \tau)$ has the property $x_n \to x \implies f(x_n) \to f(x)$, even the ones which are discontinuous. In particular, $f$ from $(\mathbb R, \tau)$ onto the real numbers with the usual topology maping $x$ to $x$ has this property and is not continuous, since $f^{-1}((0,1))$ is not an open of $(\mathbb R, \tau)$.
A: Let $X=\beta \mathbb N$ \ $\mathbb N$ where $\beta \mathbb N$ is the Cech-Stone (maximal) compactification  of the discrete space  $\mathbb N.$  Then $X$ is an infinite   compact Hausdorff space with no isolated points, and with the property that the only convergent sequences in $X$ are "eventually-constant" sequences. In fact,  if $S$ is an infinite subset of $X$ then the cardinal of $\overline S$ is $2^{2^{\aleph_0}}.$ So any $f:X\to X$ preserves convergent sequences, whether $f$ is continuous or not. Of course $X$ is not $1$st-countable.
