Ytukyg

I'm reading the book General Topology by S. Willard and I came across the following problem.
We have to find an example of a noncontinuous function $F:mathbb R^mathbb Rto mathbb R$ with the property that
whenever $f_nto f$ in $mathbb R^mathbb R$, then $F(f_n)to F(f).$ Here $mathbb R^mathbb R$ denote a space
with the product topology and $mathbb R$ is equipped with the usual topology. Please give
me some hint.

Edit (T.B.) The problem comes from section 10, specifically as a problem illustrating the points in (10.6).

The more I look into this, the more I suspect that Willard slipped up here: it appears that the existence of such function depends on the existence of certain large cardinals and hence on one’s set theory. It is possible to find such a function from a subspace of $mathbb{R}^mathbb{R}$ to $mathbb{R}$; one subspace that works is $X = Sigma_0 cup Sigma_1$, where $$Sigma_0 = {f in {0,1}^mathbb{R}:{x in mathbb{R}:f(x) = 0} mbox{ is countable}}$$ and $$Sigma_1 = {f in {0,1}^mathbb{R}:{x in mathbb{R}:f(x) = 1} mbox{ is countable}}.$$ It’s not hard to check that $Sigma_0$ and $Sigma_1$ are dense, sequentially closed subsets of $X$. Now just define $F:X to mathbb{R}$ by $F(f) = 0$ if $f in Sigma_0$ and $F(f) = 1$ if $f in Sigma_1$. If $langle f_n rangle_n to f in Sigma_i$ (where of course each $f_n in X$), there is some $n_0$ such that $f_n in Sigma_i$ for all $n ge n_0$, whence $F(f_n) = i = F(f)$ for all $n ge n_0$, and $langle F(f_n) rangle_n to F(f)$, so $F$ is sequentially continuous. However, $F^{-1}[(-1/2,1/2)] = Sigma_0$ is not open in $X$, so $F$ isn’t continuous.

If you want to try a substitute exercise based on the material in Section 10, find a sequentially continuous, non-continuous function from $mathbf{Omega}$ to $mathbb{R}$, i.e., a function $f:mathbf{Omega} to mathbb{R}$ such that $f$ is not continuous, but $langle f(x_n) rangle_n to f(x)$ in $mathbb{R}$ whenever $langle x_n rangle_n to x$ in $mathbf{Omega}$. Hint: This can be done with a function that is constant on the set ${x in mathbf{Omega}:omega_0 le x < omega_1}$.

In Willard’s notation $mathbfOmega$ denotes the set of ordinals ${alpha:1le alphaleomega_1}$ with the order topology (i.e. the sets $(gamma,omega_1]={alphainmathbfOmega: alpha>gamma}$, $[1,gamma)={alphainmathbfOmega: alpha<gamma}$ for $gammainmathbfOmega$ form a subbase).

It is not an answer, but this example may be funny.

Let $X$ be the subspace of $]0,1[^{]0,1[}$ consisting of measurable functions (with the product topology), and let $F : X rightarrow ]0,1[, f mapsto int f$.

Then $F$ is sequentially continuous by the Dominated Convergence Theorem, but nowhere continuous since it is surjective on every open subset.