sequentially continuous on a non first-countable Can you give me an example of a function which is sequentially continuous but not continuous? (I know that in first-countable spaces this is equivalent, but what about in spaces without this condition?)
Thank you :)
 A: Let $X$ be $\omega_1+1$ with the order topology. Define $f:X\to\Bbb R$ by $f(\alpha)=0$ if $\alpha<\omega_1$, and $f(\omega_1)=1$. Then $f$ is sequentially continuous but not continuous. The reason that $f$ is sequentially continuous is that if $\langle \alpha_n:n\in\omega\rangle$ is a convergent sequence in $X$ with limit $\alpha$, then either 


*

*$\alpha<\omega_1$, and there is an $m\in\omega$ such that $\alpha_n<\omega_1$ for all $n\ge m$, in which case $f(\alpha_n)=0=f(\alpha)$ for all $n\ge m$, or

*$\alpha=\omega_1$, in which case there is an $m\in\omega$ such that $\alpha_n=\omega_1$ for all $n\ge m$, in which case $f(\alpha_n)=1=f(\alpha)$ for all $n\ge m$.
And $f$ is discontinuous at $\omega_1$, because $f^{-1}\big[(0,2)\big]=\{\omega_1\}$, but $\omega_1$ is not an isolated point of $X$.
A: Take $\omega_1+1$ as an ordinal space, then $\omega_1$ does not have a countable neighborhood base.
Now consider the function which maps $\omega_1$ to $0$, and every other ordinal $\alpha\mapsto 2^\alpha$ (ordinal exponentiation).
By the definition of ordinal exponentiation, if $\alpha=\lim_n\alpha_n$ then $2^\alpha=\lim 2^{\alpha_n}$, so this is certainly sequentially continuous. But clearly this function is not continuous in $\omega_1$ (easily recognizable using long sequences).
A: Take a topological space $X$ of your choice, such there exists a subset $A$ whose closure is strictly larger than its sequential closure.
Take any $x$ which lies in the closure of $A$ but not in the sequential closure of $A$.
Now, we consider the topological space $A \cup \{x\}$ and we define
$$
f(y) = \begin{cases} 1 & y = x,\\ 0 & \text{else}.\end{cases}
$$
This function is sequentially continuous, but not continuous (note that only the trivial sequence $x_n \equiv x$ converges towards $x$).
A: Another example occurs in functional analysis. Let $X$ and $Y$ be Banach spaces such that $X$ is infinite-dimensional, and let $T \in B(X,Y)$ be compact and injective. Consider $T$ as a function $(X,\text{weak}) \to (Y,\text{norm})$, then:


*

*$T$ is sequentially continuous because compact operators map weakly convergent sequences to strongly convergent sequences.

*In order to see that $T$ is not continuous, we use that $T$ is injective. For any non-zero $x\in X$ we have $\|Tx\| > 0$, hence $\|T\lambda x\| > 1$ for $\lambda$ sufficiently large. It follows that the inverse image of the open unit ball (of $Y$) does not contain any non-zero linear subspace (of $X$). On the other hand, every weakly open neighbourhood of $0$ contains an infinite-dimensional subspace (of finite codimension), so we see that $T$ is not continuous.
