1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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
2
$\begingroup$

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 $\frak c$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.