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.


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.

  • $\begingroup$ If you just want a counterexample to the main question, you can just let $Y$ be $X$ with the discrete topology and let $f$ be the identity. $\endgroup$ Sep 20 '15 at 22:40
  • $\begingroup$ Awesome example! $\endgroup$
    – hot_queen
    Sep 21 '15 at 21:02

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 $A=\{y:y<x\}$).


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