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Bonjour/bonsoir à tous et à toutes.

I've two questions related to something on which I'm working. I've already tried to discuss about them elsewhere, but it hasn't been fruitful so far.

Edit (4 Dic 2011). Let me simplify the original text according to the comments of fedja and Michael Greinecker.

Let $\mathcal{X} \equiv (X,\mathcal{O}_X)$ and $\mathcal{Y} \equiv (Y,\mathcal{O}_Y)$ be topological spaces, $I \ne \emptyset$ an index set, $\{X_i\}_{i \in I}$ a chain of $(2^X, \subseteq)$ such that $\textstyle \bigcup_{ i \in I} X_i = X$ and $X_i$ is dense in $\mathcal{X}$ for all $i \in I$, $\{f_i\}_{i \in I}$ a family of continuous functions $(X_i, X_i \cap \mathcal{O}_X) \to \mathcal{Y}$ such that $f_j$ extends $f_i$, for $i,j \in I$, if $X_i \subseteq X_j$. Then, set $f := \textstyle\bigcup_{i \in I} f_i$ (identifying each $f_i$ with its graph). Now come the questions.

Question 1. Is $f$ a continuous function $(X, X \cap \mathcal{O}_X) > \to \mathcal{Y}$?

I've some clues that the answer to Question 1 may be negative, but so far I wasn't able to find out any counterexample by myself. In any case, if the answer is really "No", it will make still sense to ask for the following:

Question 2. Is $f$ a Borel function $(X, \mathfrak{B}(X \cap > \mathcal{O}_X)) \to (Y, > \mathfrak{B}(\mathcal{O}_Y))$? Here, provided $(W,\mathcal{O}_W)$ is a topological space, I'm denoting by $\mathfrak{B}(\mathcal{O}_W)$ the Borel algebra on $W$ generated by the open sets of $\mathcal{O}_W$.

Partial results (updated to 6 Dec 2011). fedja proved below in the comments that the answer to Question 1 is affirmative if the topology of $Y$ is such that points can be separated from closed sets. On another hand, Yulia Kuznetsova showed that Q1 is false assuming that $\mathcal{X}$ is the interval $[0,1]$ (with its usual subspace topology) and $\mathcal{Y}$ is the Sierpiński space (here).

Thank you in advance for any hint.

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    $\begingroup$ An obvious remark is that if you can separate a point from a closed set in $Y$, then $f$ is continuous. Take $x$, take a neighborhood $U$ of $y=f(x)=f_i(x)$; choose a neighborhood $W$ of $y$ such that its closure is contained in $U$. Choose a neighborhood $V$ of $x$ such that $f(D_i\cap V)\subset W$. Then for every $j$ and $x'\in V\cap D_j$, the image $f(x')$ is contained in the closure of $W$ by density of $D_i$ in $V$ and continuity of $f_j$. So, you need bad topological spaces for a counterexample. Also, the one-to-one condition is useless: you can always change $f$ to $(id,f)$. $\endgroup$
    – fedja
    Commented Dec 4, 2011 at 15:01
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    $\begingroup$ I think your notation is needlessly complicated. One can simplify the problem a lot if one identifies a function with its graph. Then, $(f_i)$ is a chain of sets itself, $f=\bigcup_i f_i$ and one can assum w.l.o.g. that $X=D=\bigcup_i\text{dom}f_i$. $\endgroup$
    – Michael Greinecker
    Commented Dec 4, 2011 at 15:24
  • $\begingroup$ @fedja. Obvious?! I'd say it's a great remark. @Michael. You're clearly right. I'm going to edit the OP according to the comments of you both and withdraw what is actually unnecessary. $\endgroup$
    – Salvo Tringali
    Commented Dec 4, 2011 at 22:51
  • $\begingroup$ Look at Engelking 2.1.13 and at the exercise 2.1.J. It claims that a sufficient condition is that $X_i$ are locally finite, and that there is a counterexample if they are not. $\endgroup$
    – Yulia Kuznetsova
    Commented Dec 5, 2011 at 14:41
  • $\begingroup$ Sorry, that was for closed $X_i$, and since yours are dense, this will not work. $\endgroup$
    – Yulia Kuznetsova
    Commented Dec 5, 2011 at 15:19

2 Answers 2

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Look if this works as a counterexample to Q1: Let $Y=\{0,1\}$ be a two-point set such that 1 is open and 0 is not, and let $X=[0,1]$ in the usual topology. Let $X_i = [0,1] \setminus \{2^{-k}: k>i\}$. Put $f_i(0)=1$, $f_i(2^{-k})=0$ for all $k$, and $f_i(x)=1$ for other $x$. Then every $f_i$ is continuous but $f$ is not. Right?

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  • $\begingroup$ (continue because of problems with TeX) .. and let $X_i=\{0\}\cup [2^{-k},1]$. Put $f_i(0)=1$, $f_i(2^{-k})=0$ for all $k$, and $f_i(x)=1$ for other $x$. Then every $f_i$ is continuous but $f$ is not. Right? $\endgroup$
    – Yulia Kuznetsova
    Commented Dec 5, 2011 at 16:27
  • $\begingroup$ Sorry, need to correct this: obviously $X_i$ are not dense like this, one should put $X_i = [0,1] \setminus \{2^k: k>i\}$. $\endgroup$
    – Yulia Kuznetsova
    Commented Dec 5, 2011 at 16:33
  • $\begingroup$ Yes, I know how to type TeX... Done. This time it worked, but before the same text didn't want to show after multiple edits. $\endgroup$
    – Yulia Kuznetsova
    Commented Dec 5, 2011 at 16:58
  • $\begingroup$ If $\{1\}$ is open, then $\{0\}$ is closed, hence it can’t be dense. But you don’t really need the latter in the argument, it works with $Y$ being the Sierpiński space. $\endgroup$
    – Emil Jeřábek
    Commented Dec 5, 2011 at 17:36
  • $\begingroup$ Also, the $U_k$ are not used anywhere, and there is a missing $-$ in the definition of $X_i$. $\endgroup$
    – Emil Jeřábek
    Commented Dec 5, 2011 at 17:39
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Sorry if this doesn't work - but: I think one can modify the example to yieald a non-measurable function. The task is to find such a chain $F_i$ that: every $F_i$ is closed; $E=\cup F_i$ is not Borel; $X\setminus E$ is dense in $X$. Then the counterexample will be given by $f_i|_{F_i}=0$, $f_i|_{X\setminus E}=1$ where $\{0,1\}$ is the Sierpinski space with 1 open.

This is most easily constructed in a non-metrizable dyadic space: $X=\{0,1\}^A$ (where $\{0,1\}$ is a Hausdorff two-point set), $A$ is uncountable. Split $A=\cup_{i\in S} A_i$ into a noncountable disjoint union of subsets $A_i$, every $A_i$ being infinite and countable. For $B\subset A$, let $p_B:X\to \{0,1\}^B$ be the natural projection. On $S$, choose a linear order without a maximal element. For every $i\in S$, put $F_i = \{ x\in X: p_{A_k}(x)=0 \forall k\ge i\}$.

If $E=\cup F_i$, then clearly $X\setminus E$ is dense. I realize also that $E$ is not Baire, but I cannot say whether it is Borel.

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