We say $A\subset\Bbb R$ is $\mathfrak c$-dense if $|A\cap (a,b)|=\mathfrak c,$ for all $a<b.$ where $||$ denotes to the cardinality of the set and $\mathfrak c$ denotes to the cardinality of $\Bbb R.$ Let $\{A_y\colon y\in\Bbb Q\}$ be a family of pairwise disjoint $\mathfrak c$-dense subsets of $\Bbb R$. Define $f\colon\Bbb R\to\Bbb R$ as follows $$f:=\sum_{y\in\Bbb Q} y\, \chi_{A_y},$$ where $\chi_{A_y}$ is the characteristic function.

Question: Show that $f^{-1}(U)$ is $\mathfrak c$-dense for every nonempty open set $U$?

My proof: For this, let $U$ be a nonempty open set and $I$ be an open interval, then we will show that there is a set with cardinality $\mathfrak c$ contained in $f^{-1}(U)\cap I.$

Indeed, there exists a $y\in U\cap\Bbb Q$ and a set $P$ with cardinality $\mathfrak c$ contained in $I\cap A_y,$ since $A_y$ is $\mathfrak c$-dense, such that $f(x)\in U$ for all $x\in P.$ Then, $P\subset f^{-1}(U)\cap I.$ So, $f^{-1}(U)$ is $\mathfrak c$-dense.

Is that right? Any hint, comments, or different proof will be appreciated greatly.

  • $\begingroup$ In your definition of $\mathfrak c$-dense, you should mention what $a$ and $b$ are. Do you mean that $|A \cap (a, b)| = \mathfrak c$ for all $a < b$? Or that there exist some $a, b \in \Bbb R$ such that $|A \cap (a, b)| = \mathfrak c$? $\endgroup$ Jul 8 '21 at 12:15
  • $\begingroup$ @AryamanMaithani, I fixed it. I meant for all $a<b$ $\endgroup$
    – 00GB
    Jul 8 '21 at 12:18
  • $\begingroup$ Is $f$ constant? It looks like $y$ is a dummy variable. $\endgroup$
    – Joe
    Jul 8 '21 at 12:30
  • 1
    $\begingroup$ @Joe, $y\in\Bbb Q$, and $f$ is not constant on $\Bbb R$ but it would be constant on each $A_y.$ $\endgroup$
    – 00GB
    Jul 8 '21 at 12:35
  • $\begingroup$ In your question, I assume you want $U$ to be open. (That's what you assume in your solution, at least.) Otherwise the question is not true. $\endgroup$ Jul 8 '21 at 12:43

Looks fine but here's a simpler way of doing it: First note that $f(A_y) = \{y\} \subset U$. Thus, $A_y \subset f^{-1}(U)$. In turn, $A_y \cap I \subset f^{-1}(U) \cap I$.
Since $A_y$ is $\frak c$-dense, it follows that $|A_y \cap I| = \frak c$ and we are done.

(Essentially, this is your proof but I've shown that I can take $P = A_y \cap I$ itself, which I know to be of cardinality $\frak c$.)


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