What is an example of a nonmetrizable topological space? Find topological spaces $X$ and $Y$ and a function $f:X\to Y$ which is not
continuous, but which has the property that $f(x_n)\to f(x)$ in $Y$ whenever $x_n\to x$ in $X$
I know this is true if $X$ is metrizable, so I want a counterexample when $X$ is not metrizable.
 A: You want an example of a function $f:X\to Y$ that is sequentially continuous but not continuous. Let $X=\omega_1+1$, the space of all ordinals up to and including the first uncountable ordinal, with the order topology, and let $Y=\{0,1\}$ with the discrete topology. Let
$$f:X\to Y:\alpha\mapsto\begin{cases}
0,&\text{if }\alpha<\omega_1\\
1,&\text{if }\alpha=\omega_1\;.
\end{cases}$$
Then $f$ is not continuous, because $f^{-1}\big[\{1\}\big]=\{\omega_1\}$ is not open. However, $f$ is sequentially continuous, because a sequence $\langle\alpha_n:n\in\omega\rangle$ in $X$ converges to $\omega_1$ if and only if there is an $m\in\omega$ such that $\alpha_n=\omega_1$ for all $n\ge m$ and hence such that $f(\alpha_n)=1=f(\omega_1)$ for all $n\ge m$.
Added: Here’s a simpler version of essentially the same idea, though in this case the space $X$ is no longer compact. Define a new topology $\tau$ on $\Bbb R$ as follows: 
$$\tau=\{U\subseteq\Bbb R:0\notin U\text{ of }\Bbb R\setminus U\text{ is countable}\}\;.$$
Let $X$ be $\Bbb R$ with the topology $\tau$. If $x\in\Bbb R\setminus\{0\}$, then $\{x\}$ is open: $x$ is an isolated point of $x$. However, the all nbhds of $0$ are big: if $x\in U\in\tau$, then $\Bbb R\setminus U$ is a countable set. (Note: countable includes finite.)
It’s a standard and pretty straightforward exercise to show that if $\sigma=\langle x_n:n\in\Bbb N\rangle$ is a convergent sequence in $X$, then $\sigma$ is eventually constant, i.e., there are an $m\in\Bbb N$ and an $x\in X$ such that $x_n=x$ for all $n\ge m$. In other words, the only sequences in $X$ that converge are the ones that absolutely have to converge, because from some point on they’re constant.
Now define
$$f:X\to Y:x\mapsto\begin{cases}
0,&\text{if }x\ne 0\\
1,&\text{if }x=0\;.
\end{cases}$$
Then $f$ is not continuous, because $f^{-1}\big[\{1\}\big]=\{0\}$, which is not open in $X$. However, $f$ is sequentially continuous: if $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $X$ converging to some $x\in X$, there is an $m\in\Bbb N$ such that $x_n=x$ for all $n\ge m$, so of course $f(x_n)=f(x)$ for all $n\ge m$, and $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(x)$.
