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So I was looking at the question above. There was a pre-question which was the same for circles, but I believe this is easy to prove uncountable by considering circles $C_{r} $ of radius r centred at the origin, and $S= \{ C_{r} : r \in \mathbb{R} \} $.

I tried a similar technique for this question- Consider a figure of 8 on the x axis lying horizontally with most -ve x point going through origin. Now we can construct 'concentric' figures of 8 containing one another, each of which has most -ve x point going through a different real, from 0 to -ve infinity. (i.e. let $F_{r} $ be figure of 8 with most -ve x point going through r, and if $s>r$ , $F_{r}$ contains $F_{s}$.

So if this is a fair argument, its clear none of them intersect. But I came across this question here Countable or uncountable set 8 signs where the answer is it must be countable.

So what is incorrect with my solution?

Consider the set of $F_{r}$ for all $r<0$

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  • What does "with most -ve x point going through origin" mean? – lulu Sep 17 '16 at 10:07
  • You will have problems managing the right ends of your eights – Hagen von Eitzen Sep 17 '16 at 10:09
  • @lulu Imagine drawing an infinity symbol, with the far most left point going through point (0,0). All other figures of 8 contain this one, and have left most point going through (k,0) for k<0 – Isomorphism Sep 17 '16 at 10:21
  • @HagenvonEitzen Ok, please can you explain what you mean? I just thought we could stretch the next figure of 8 in the sequence enough so that it encompasses all previous figures, and we can do this as the x axis is infinitely long. So is the problem that the length of $F_{r}$ tends to infinity as r tends to infinity? – Isomorphism Sep 17 '16 at 10:24
  • Ok...but your post refers to "-ve x point" which isn't defined. "-ve" isn't even a word. Can you edit your post to include actual definitions? I also think a picture would be good. – lulu Sep 17 '16 at 10:24
  • @lulu Ok, but isnt it well known that "-ve" means negative?? – Isomorphism Sep 17 '16 at 10:25
  • I've never seen it before, for what that's worth. But include a picture. The only way I know to make "concentric" figure $8's$ is to put each successor inside one of the loops of the predecessor. Otherwise, they have to intersect. – lulu Sep 17 '16 at 10:28

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If $F_r$ is one of your figure eights, its righthand loop defines an open interval $I_r$ on the $x$-axis between the crossover point of the $8$ and its most positive point on the $x$-axis. These open intervals are clearly pairwise disjoint and non-empty. But any family of non-empty open intervals of real numbers is necessarily countable, as each must contain a different rational number, and there are only countably many rational numbers. Thus, you can’t build an uncountable nested family of the kind that you’re describing.

Brian M. Scott
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