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.
 A: 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 $X$. 
