Sequential Continuity does not imply Continuity We say that a function, $f: X \to Y$ ($X, Y$ are topological spaces) is sequentially continuous if $f(x_n)$ converges to $f(x)$ whenever $x_n$ converges $x$. Give an example of a function that is sequentially continuous but not continuous.
I tried letting $X$ be $\mathbb R$ with cofinite topology and $Y$ be $\mathbb R$ with discrete topology where $f$ is identity map but $f$ is neither sequential continuous or continuous. There are no simple solutions online since most use ordinal sets and we have not yet covered that in our topology class.

Note that in the following, by countable I mean not uncountable, that is, countable means"finite or countably infinite". 
Consider a space $X$ in which every non-empty $V\subset X$ is open if and only if $X\backslash V$ is countable .This is called the co-countable topology on $X$. A sequence $(p_n)_{n \in \mathbb N}$ of points in $X$ cannot converge, in any sense, to a point $p \in X$ if $\{n \in\mathbb N : p_n=p\}$ is finite. Hence any convergent sequence in $X$ is eventually constant, and therefore any function $f:X\to Y$ to any space $Y$ is sequentially continuous. Now if $X$ is uncountable then it is not a discrete space so there exist discontinuous functions on $X$. For an example, let $X=\mathbb R$ (the reals) and let $Y$ be the reals with the usual topology, and let $f=\text{id}_{\mathbb R}$. The inverse $f^{-1}(0,1)$ of the open interval $(0,1)$ is not open in $X$. For another example, let $X$ be any uncountable set, with the co-countable topology, let $Y$ be the set $X$ with the discrete topology, and let $f=\text{id}_X$. Then $\{p\}$ is open in $Y$ for any $p \in Y$ but $f^{-1}\{p\}=\{p\}$ is not open in $X$. 
