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How should I rewrite the beginning of this post?

It seems people aren’t commenting since it’s difficult to understand.

Any suggestions are welcomed.

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Maybe break up the first paragraph, something like this:

Does there exist a function $f:[0,1]\to[0,1]$ such that

  1. the graph of $f$ is dense in $[0,1]\times[0,1]$, and

  2. the collection of all the pre-images of each of the subsets of the range of $f$ under $f$, such that the pre-images are non-measurable in the sense of Caratheodory, is a [non-perfect dense set][1] in the collection of all subsets of $[0,1]$, and

  3. $f$ is non-uniform (i.e. without [complete spacial randomness][2]) in $[0,1]\times[0,1]$?

I had trouble parsing 2), I don't know whether what I have written conveys what you intended.

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  • $\begingroup$ What you’ve written is correct. $\endgroup$
    – Davis Johnson
    May 16 at 12:35
  • $\begingroup$ You stated you have trouble parsing 2. What is your interpretation of this? $\endgroup$
    – Davis Johnson
    May 16 at 12:36
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    $\begingroup$ The way I wrote it reflects my interpretation. I'd say, you look at the range of $f$ – let's call it $R$; you look at a subset of $R$, let's call it $S$; you look at $T$ defined by $T=f^{-1}(S)$; you insist that $T$ be "non-measurable in the sense of Caratheodory" (I don't know what that means, but that's my problem, not yours); and then you ask that the collection of all such $T$ be a non-perfect dense set in the collection of all subsets of $[0,1]$. I guess it's not clear to me what topology you are putting on the collection of all subsets of $[0,1]$. $\endgroup$
    – Gerry Myerson
    May 16 at 13:04
  • $\begingroup$ Did you click the link on “non-perfect dense set”. It should lead you to Dave L. Renfro’s answer. Perhaps the topology can be specified there. $\endgroup$
    – Davis Johnson
    May 16 at 13:06
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    $\begingroup$ No, I didn't click on the link. I'm not actually interested in the question, I'm just trying to help you get it to where it might be reopened. $\endgroup$
    – Gerry Myerson
    May 16 at 13:11
  • $\begingroup$ As a last note, I added that $f$ should be “measurable in the sense of caratheodory”? Is the question clear or should I make changes. $\endgroup$
    – Davis Johnson
    May 16 at 13:24
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    $\begingroup$ I think it reads much better now than it did before, but I'll leave final judgement to those who have expertise in the subject matter. $\endgroup$
    – Gerry Myerson
    May 16 at 13:43

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