Continuous function at a point characterized by sequences Suppose that $X$ is a sequential space. If $Y$ is a topological space, it is proven in Engelking that $f:X\rightarrow Y$ is continuous (on the whole space) if and only if for each sequence $\{x_n\}_{n\in\mathbb N}$ in $X$ converging to a point $x\in X$ the sequence $\{f(x_n)\}_{n\in\mathbb N}$ in $Y$ converges to $f(x)$.
But I cannot imagine the condition in terms of sequences for a function $f:X\rightarrow Y$ ($X$ and $Y$ as before) to be continuous only at a point $a\in X$. I started supposing that the necessary and sufficient condition for that is that for any sequence $\{x_n\}_{n\in\mathbb N}$ in $X$ converging to $a$, the sequences $\{f(x_n)\}_{n\in\mathbb N}$ in $Y$ converges to $f(a)$. If $f$ is continuous it is obvious that the condition holds. But I don't know how to prove that such a condition is enough (if it is, that is not clear).
By definition, I would have to prove that for each neighbourhood $U'$ of $f(a)$, there is a neoghbourhood $U$ of $a$ such that $U\subseteq f^{-1}(U')$. I think, this condition is equivalent to prove that for any open set $V'$ containing $f(a)$, the set $f^{-1}(V')$ contains $a$ and is open too. Since $X$ is sequential, I will be done if I'm able to prove that $f^{-1}(U')$ is sequentially open. Or that the complement of $f^{-1}(U')$ in $X$, $f^{-1}(Y\setminus U')$ is sequentially closed. But I don't know how. The problem is that those definitions imply sequences with different limits, but I know that $f(x_n)$ converges to $f(a)$ only when $x_n$ converges to $a$.
So that:
1.- What is the characterization in terms of sequences for a function $f:X\rightarrow Y$ to be continuous when $X$ is sequential?
2.- In case that my condition was right, how can I prove the result?
Thanks.
Remark. Notice that I'm not supposing $X$ to be Hausdorff. 
EDIT
Following hamamAbdallah's advice, I think I can give a proof. Suppose that $f$ is sequentially continuous but not continuous. By definition, there exists a neighbourhood $U'$ of $f(a)$ such that there is no $U$ neighborhood of $a$ satisfying $f(U)\subseteq U'$. The idea now is define s sequence that converges to $a$ but that its image does not converge to $f(a)$.  For each neighbourhood of $a$, call $A_U=f(U)\setminus U'$. Consider a net $(x_U)$, where $x_U\in f^{-1}(A_U)$. This net converges to $a$. I claim that since $X$ is sequential, I can pick a subnet of $(x_U)$ that is a sequence and that also converges to $a$. Call it $(x_n)_n$. But by definition of $A_U$, the sequence $(f(x_n))_n$ should not be able to converge to $f(a)$, which is a contradiction. 
Is this proof valid? How can I justify that the subnet I claimed actually exists? 
 A: Here I present a counterexample to my question. I think I have find a function that is sequentially continuous at $a$, i.e. for any sequence $(x_n)_n$ on $X$ converging to $a$, the sequence $(f(x_n))_n$ converges to $f(a)$, but that $f$ is not continuous at $a$, i.e. there exists a neighbourhood of $f(a)$ such that there doesn't exist a neighbourhood of $a$ which image is contained into the first.
Let $X=\mathbb Z^+\cup\{\ast\}\cup(\mathbb Z^+\times\mathbb Z^+)$ be the Arens' space (see here) and let $Y$ be the real line. Consider a function $f:X\rightarrow Y$ defined as follows:
$$
f(x)=\begin{cases}
1/x, & \mbox{if } x\in\mathbb Z^+ \\
0, & \mbox{if } x=\ast \\
m^n, &\mbox{if } x=(m,n)\in\mathbb Z^+\times\mathbb Z^+ .
\end{cases}
$$
I claim that $f$ is sequentially continuous at $a=\ast$. Indeed, the unique sequences that converge to $\ast$ are $x_n=n$ for $n\geq n_0$ or $x_n=\ast$ for $n\geq n_0'$. In the firts case we have that $f(x_n)=1/n$ while in the second and $f(x_n)=0$ (we are assuming $n_0=n_0'=0$). In both cases the sequences $(f(x_n))$ converge to $0$ in $Y$. However $f$ is not continuous at $a$: A neighbourhood of $\ast$ contains infinitely many isolated points $(m,n)$ which $f$ maps to big numbers $m^n$. Hence, for a neighbourhood say $V=(-1,1)$ of $0$, there is no $U$ neighbourhood of $\ast$ such that $f(U)\subseteq V$. 
Remark. Notice that this $f$ is not unique. We could set other functions $f$ with different values on $Z=\mathbb Z^+\times \mathbb Z^+$; for example that which is constant (different from $0$) on this subset. The important think here is that the set $f(Z)$ must be unbounded or have a positie infimum.

Let me give some details about the topology of $X$. In terms of neighbourhoods, the topology of $X$ is described as follows:


*

*Points belonging to the plane $\mathbb Z^+\times \mathbb Z^+$ are isolated.

*For positive integers $n$, a fundamental system of neighbourhoods is given by the set $\{B_m(n)\}_{m\in\mathbb N^*}$, where $B_m(n)$ consists in the point $n$ together with the $n$-th column, except the firts $m$ points. For example, $B_3(2)=\{2\}\cup\{(2,4),(2,5),(2,6),\dots\}$.

*And finally the point $\ast$. Neighbourhoods for it consists of the whole space $X$ minus finitely many $B_1(n)$ (i.e. some columns of the plane $\mathbb Z^+\times\mathbb Z^+$ together with the naturals that label such columns). And, in the resulting set, you are allow to remove also finite many points of each of the remaining columns (so can remove infinitely many in total).
Now, my first claim:
The unique sequences that converge to $\ast$ are of the form $x_n=\ast$ for  $n\geq n_0$ or $x_n= n$ for $n\geq n_0'$.
Well, it is clear that the above sequences converge to $\ast$ (the first is obvious and for the second, recall that neighbourhoods of $\ast$ have all but a finite number of $n\in\mathbb Z^+$). Now suppose that $(x_n)$ is a sequences on $X$ but is not of the above form. We can suppose that all the terms are isolated points. Now, if the sequence is eventually constant we can remove that point from a neighbourhood of $\ast$. And the result follows. If it is non-constant but all the points all contained in a column, we can remove that, and $\ast$ is not a limit neither. On the other hand, if the sequences is a row, since we were allowed to remove a finite quantity of points of each row, we can define a neighbourhood $V$ of $\ast$ such that $x_n\not\in V$ for all $n\in\mathbb N$. And we are done, because there is no more convergent sequences on $X$.
