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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 $F:\mathbb R^\mathbb R\to \mathbb R$ with the property that whenever $f_n\to f$ 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).

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Is $R=\mathbb{R}$? If yes, please edit your question for better readabilty (use \mathbb{R} to get $\mathbb{R}$) – Florian Jul 26 '11 at 9:35
@Florian: What else could it possibly be? – t.b. Jul 26 '11 at 11:10
On a side note for those reading German: You may want to have a look at A. Tychonoff, Über einen Funktionenraum, Math. Ann. 111 (1) (1935), 772-776, where this space was first introduced. @Asaf: ... – t.b. Jul 26 '11 at 11:22
If your instution does not have subscription, the paper mentioned by Theo is freely available here: – Martin Sleziak Jul 26 '11 at 11:57
up vote 4 down vote accepted

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 $\mathbf\Omega$ denotes the set of ordinals $\{\alpha:1\le \alpha\le\omega_1\}$ with the order topology (i.e. the sets $(\gamma,\omega_1]=\{\alpha\in\mathbf\Omega: \alpha>\gamma\}$, $[1,\gamma)=\{\alpha\in\mathbf\Omega: \alpha<\gamma\}$ for $\gamma\in\mathbf\Omega$ form a subbase).

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What sort of large cardinals you think need to be involved in the general solution, and how? – Asaf Karagila Jul 26 '11 at 22:04
@Asaf: I stumbled over similar things when I tried to find an answer to this question (submeasurable and measurable cardinals). See e.g. here (Balcar, Hušek, 2000) and here (Noble 1970). Google for sequential continuity product spaces – t.b. Jul 26 '11 at 22:39
@Asaf: A few high spots from what Theo and I have found: A cardinal $\kappa$ is sequential if there is a sequentially continuous, non-continuous, real-valued function on the Cantor space $2^\kappa$. Let $m_s,m_R$, and $m_2$ be the first sequential, real-valued measurable, and measurable cardinals. Then $m_s\le m_R$, and if $m_s > 2^\omega$ (e.g., under MA), then $m_s=m_2$. Moreover, every sequentially continuous function from a product of $<m_s$ separable metric spaces to a metric space is continuous. – Brian M. Scott Jul 27 '11 at 5:24
@Martin: Thanks very much. I added just a little extra clarification; that Willard starts at $1$ was implicit in your subbase, but it’s odd enough that I wanted to make it explicit. – Brian M. Scott Jul 27 '11 at 8:03
@Brian: Thank you very much for all the information. This example is enough to me. – Dusan Jul 27 '11 at 15:16

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.

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Nice and quick. +1! – Giuseppe Negro Jul 28 '11 at 17:26

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