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The somewhat intuitive idea of 'holes' all depends on the definition of the word of course. In a much more naive notion of 'holes', $\mathbb{Q}$ doesn't have any, as for any two $p, q \in \mathbb{Q}$ ($p < q$) one can always find another $r \in \mathbb{Q}$ such that $p < r < q$. Of course, it is well-known that $\mathbb{Q}$ is not complete, while $\mathbb{R}$ is. In that sense, $\mathbb{Q}$ does have holes, whereas $\mathbb{R}$ does not.

My question is if there is any sense in which $\mathbb{R}$ does have holes. I know little about 'extensions' of $\mathbb{R}$ (not sure if that's the right term) like the surreal or hyperreal numbers, but as far as I know, those don't fill up any 'gaps' in $\mathbb{R}$, they simply 'add more'. In other words, I am wondering if there is a property that $\mathbb{R}$ (with its standard topology and all that sort of thing) does not have, which could be described as $\mathbb{R}$ having holes (in the sense of that property).

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  • $\begingroup$ Many infinitesimal systems fill the "hole" between $0.999\ldots$ and $1$. Although there is no hole there on the real number line, so I sort of understand what you say by "add more". $\endgroup$ – Arthur Jun 19 '17 at 8:19
  • $\begingroup$ @Arthur, exactly right. Whichever way we interpret the "absence of gaps" we have to be careful not to misinterpret this as the statement of impossibility of infinitesimals. $\endgroup$ – Mikhail Katz Jun 19 '17 at 8:20
  • $\begingroup$ One may wonder if it makes sense to say that $i$ or $\pm\infty$ fill some kind of "hole" (that of algebraic or toppological completion). $\endgroup$ – Hagen von Eitzen Jun 19 '17 at 8:39
  • $\begingroup$ @S. van NIgtevecht Just want to point out that under your definition, $(-\infty,0)\cup (1,\infty)$ doesn't have holes. $\endgroup$ – themaker Jun 19 '17 at 9:12
  • $\begingroup$ As far as I understand, in the geometric sense of a 'gap', R is complete. There are however numbers that can be conceived of from R but are not themselves in R - sqrt(-1) for example. $\endgroup$ – ZyTelevan Jun 19 '17 at 9:17
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Once the rational numbers $\mathbb Q$ are extended to a Dedekind-complete field, there is no room to add any further numbers, so long as one is working with an Archimedean number system. The intuitive picture that "there are no more holes" is perhaps helpful in grasping this mathematical fact, but it can be detrimental if extended further to some kind of notion of impossibility of even larger (and "more closely packed") number systems. Indeed such systems do exist; one of them that's particularly useful in analysis is the hyperreal number system. In this sense $\mathbb R$ does have holes.

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