Talk:Smith–Volterra–Cantor set

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Nowhere dense

The same content is present at nowhere dense but I felt this deserves a separate page. I'll be creating a page on Volterra's function in the near future if someone doesn't beat me to it :-) - Gauge 06:07, 23 Aug 2004 (UTC)

There is a slight inconsistency here. Either the intervals removed at each step are the "middle quarter of the remaining intervals" or they are centred on a/2^n. But they cannot be both. The first leads to measure of 0.5, the second to measure of 0.53557368... --Henrygb 17:17, 25 Apr 2005 (UTC)

Thanks for noticing. I noticed that the intervals you gave seemed to be off, so I corrected them (hopefully :-)). - Gauge 01:42, 30 Apr 2005 (UTC)

You are right - somehow I multipled 3 by 2 and got 12. Thanks --Henrygb 22:53, 30 Apr 2005 (UTC)

Can somebody please help me understand?

Let the set be called S. By construction, S contains no intervals (i.e. S contains points that are seperate from each other.). And the measure of a single point is 0. So how can the total measure be 1/2? On the other hand, the total length of removal is 1/2, hence remaining length must be 1/2. Hence, S must contains intervals of length greater than 0. Can somebody please help me resolve this? 108.162.157.141 (talk) 01:53, 28 November 2013 (UTC)[reply]

Intuition must adapt to facts! Yes, it is hard. For now, your intuition tells you that the measure of a set is the sum of lengths of intervals. And your logic already tells you the opposite. Your intuition must adapt. It is a hard internal work. For even harder case, see Weierstrass function. Such is the life. Boris Tsirelson (talk) 06:18, 28 November 2013 (UTC)[reply]