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Does there exist an explicit function $f:\mathbb{R}\to\mathbb{R}$, such if $X$ is the graph of $f$ (i.e., $X=\left\{(x,f(x)):x\in\mathbb{R}\right\}$) then for each point $p$ on the graph of $f$, I want to find some $\varepsilon$-ball around $p$ whose intersection with the graph of $f$ is precisely $\left\{p\right\}$

Attempt:

Does the conway base-13 function answer the question?

Edit: I assume the statement is false but I don't know how to prove this. It seems there has to be at least one sequence $(x_r)$ of real numbers, where as $x_r\to x$, we get $f(x)\to f(x_r)$, but I don't know how to proceed. Any tips would be nice.

Arbuja
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  • Please tell me this makes sense. – Arbuja Jul 27 '23 at 17:07
  • Please show the attempt to prove the statement – Sgg8 Jul 27 '23 at 17:12
  • @Sgg8 I'm not sure how to prove the statement, but let me try an explanation. – Arbuja Jul 27 '23 at 17:14
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    I am having a hard time understanding this. What is $V$, and why do you not mention it again? It seems to me that for each point $p$ on the graph of $f$, you want to find some $\epsilon$-ball around $p$ whose intersection with the graph of $f$ is precisely ${ p }$? Is that right? – terran Jul 27 '23 at 17:14
  • @Lewis Yes, that is correct. – Arbuja Jul 27 '23 at 17:24
  • What is the relevance of the metric space $(V, d)$? – terran Jul 27 '23 at 17:32
  • @Lewis I wasn't sure whether $X$ and $V$ should be different. If so, then $X$ should be a subset of $V$. – Arbuja Jul 27 '23 at 17:34
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    You have a function from the reals to the reals. Why do you need any other metric space, and why would you use $X$ for both some mentioned metric space which is never seen nor alluded to again after the first sentence, and for the graph of $f$? – Arturo Magidin Jul 27 '23 at 17:38
  • $f(x)$ with $f(x)=1$ when $x$ is rational and $f(x)=0$ otherwise, I thing is an example. – Piquito Jul 27 '23 at 17:40
  • @ArturoMagidin I assumed if I mentioned $d$, I have to refer to a metric space. – Arbuja Jul 27 '23 at 17:41
  • @ArturoMagidin I got rid of it, hopefully everything is clear now. – Arbuja Jul 27 '23 at 17:43
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    But you never mentioned $d$! And the reals already have a metric. It's like saying "Let $X$ represent a country on Earth. Is there a way to get from 2nd avenue in Manhattan to 3rd avenue in Manhattan?" – Arturo Magidin Jul 27 '23 at 17:44
  • @Piquito certainly not. Ever ball around a point in the graph includes lots of other points in the graph. – Arturo Magidin Jul 27 '23 at 17:45
  • @ArturoMagidin Sorry, you have to look at my previous edits to understand how I mentioned $d$. I apologize for my stupidity. – Arbuja Jul 27 '23 at 17:46
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    Interesting question (despite the issues with its formulation). I think this should answer it. – milore Jul 27 '23 at 17:56
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    $\Bbb{R}^n$ with usual topology has no uncountable discrete subset – Soumik Mukherjee Jul 27 '23 at 17:59
  • @ArturoMagidin Precisaly for what you say, the example i got is good. Well, maybe my English goes against my understanding of problem. – Piquito Jul 27 '23 at 18:18
  • @Piquito Yes, it seems that you're not understanding the problem. The OP is asking for a graph where every point is isolated. – jjagmath Jul 27 '23 at 18:24
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    @ArturoMagidin If proving that a no hyper-continuous $f:[0,1]\to\mathbb{R}$ is the same as proving no hyper-continuous $f:\mathbb{R}\to\mathbb{R}$ exists, then the linked post answers my question. – Arbuja Jul 27 '23 at 18:48
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    Any such function from $\mathbb{R}\to\mathbb{R}$ restricts to one on $[0,1]$. And essentially the problem is that a discrete uncountable subset of $\mathbb{R}^2$ would allow you to find an injection from that set to $\mathbb{Q}^2$, which is patently impossible. – Arturo Magidin Jul 27 '23 at 19:01

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