A function on an LCH space that is sequentially continuous but nowhere continuous

Question

This question is an extension of Example of topological spaces where sequential continuity does not imply continuity.

In my answer to that question, I gave an example of a topological space $X$ and a function $f : X \to \{0,1\}$ which is sequentially continuous but nowhere continuous. The space $X$ is completely regular but not locally compact.

Is there an example of a locally compact Hausdorff space $X$, another topological space $Y$, and a function $f : X \to Y$ which is sequentially continuous but nowhere continuous?

It will be even better if $X$ is compact Hausdorff and/or $Y$ is some nice space like $\{0,1\}$ or $[0,1]$.

If we step outside ZFC, we can get an affirmative answer. Suppose $\kappa$ is a measurable cardinal, so that there is a countably additive measure $\mu : 2^{\kappa} \to \{0,1\}$ such that all finite sets have measure 0. Then take $X = 2^{\kappa}$ with the product topology (think of the power set of $\kappa$ as the product of $\kappa$ many copies of the discrete space $\{0,1\}$) which is compact Hausdorff, $Y = \{0,1\}$, and $f = \mu$. The countable additivity of $\mu$ guarantees sequential continuity. But the finite sets are dense in $X$, as are the cofinite sets. So every nonempty open set in $X$ contains a finite set and a cofinite sets, whose measures are 0 and 1 respectively. Thus $\mu$ is nowhere continuous.

But I would like an answer in ZFC.

Answer 1

Let $X=\beta\omega\setminus\omega$; $X$ is compact Hausdorff. Moreover, $X$ has no non-trivial convergent sequences, so every function on $X$ is sequentially continuous. Finally, $w(X)=2^\omega$, so let $\mathscr{B}=\{B_\xi:\xi<2^\omega\}$ be a base for $X$.

Let $\{\langle\alpha_\xi,i_\xi\rangle:\xi<2^\omega\}$ enumerate $2^\omega\times 2$. Given $\eta<2^\omega$ and distinct points $x_\xi\in X$ for $\xi<\eta$, let $x_\eta$ be any point of $B_{\alpha_\eta}\setminus\{x_\xi:\xi<\eta\}$; this is possible, since $|B_{\alpha_\eta}|=2^{\mathfrak{c}}$. Thus, we can recursively construct $X_0=\{x_\xi:\xi<2^\omega\}$ such that the points $x_\xi$ are distinct, and $x_\xi\in B_{\alpha_\xi}$ for each $\xi<2^\omega$.

Now define

$$f:X\to 2:x\mapsto\begin{cases} i_\xi,&\text{if }x=x_\xi\\ 0,&\text{if }x\in X\setminus X_0\;. \end{cases}$$

Then $f^{-1}[\{0\}]$ and $f^{-1}[\{1\}]$ are both dense in $X$, so $f$ is not continuous.

Answer 2

Let $X$ be any topological space and let $Y$ have the same underlying set as $X$ but the "sequential topology" (i.e., a subset of $Y$ is closed iff it is sequentially closed in $X$). The identity map $f:X\to Y$ is then sequentially continuous, but is only continuous at $x\in X$ if $X$ is "locally sequential" at $x$, meaning that $x\in \overline{A}$ implies $x$ is in the sequential closure of $A$. Note that this example is universal in the sense that any sequentially continuous map $g:X\to Z$ factors as a composition $g=hf$ for some continuous map $h:Y\to Z$, and so if there is any such $g$ that is nowhere continuous then $f$ must also be nowhere continuous.

It remains to give an example of a (locally) compact Hausdorff space that is nowhere locally sequential. This is not so hard. For instance, if $(K_i)$ is an uncountable family of compact Hausdorff spaces with more than one point, it is easy to show the product $\prod K_i$ is nowhere locally sequential (if $(x_i)\in \prod K_i$ and $y_i\neq x_i$ for each $i$, consider the set $A$ of points which are $y_i$ for all but countably many coordinates and $x_i$ on the remaining coordinates).

For another example, let $Q$ be a countably saturated dense linear order and let $X$ be its Dedekind completion. Since $Q$ is countably saturated, no point of $X$ can have countable cofinality on both sides, and so $X$ is nowhere locally sequential (if $x$ has uncountable cofinality from below, say, you can take $A=\{y:y<x\}$).