Let C be the standard Cantor set, a subset of the interval [0,1]. One can easily verify that this set is
- uncountable
- closed [i.e., all Cauchy sequences in C converge to points in C]
- totally disconnected [i.e., contains no contiguous interval (a,b) ]
- has Lebesgue measure 0 [i.e., its "total length" adds up to 0]
If these aren't obvious, it's definitely worthwhile to verify them; it's not hard. Note that for some time mathematicians were wondering if any set with properties (1) and (4) even exists.
Put E = [0,1] \ C; that is, E is the complement of C in the interval [0,1]. Then E is an open set in [0,1]. Now any open subset of the real line is a countable union of disjoint open segments; again, if this is news to you, do take the time to convince yourself. [A glossary of topological terms may be found here, but if you're seeing these terms for the first time, it might be too early to attempt this problem.]Thus one might reason as follows: "Every open segment comprising E corresponds to two points in C. But the segments of E are a countable collection, while the points of C are uncountable - we have a contradiction." Resolve the apparent contradiction. You will be enlightened, or your money back.
Solutions/discussion welcome in comments.
3 comments:
I'm somewhat confused as to why this is supposed to be a contradiction. Each segment in E corresponds to two points in C, but the converse is not necessarily true.
Another way to think about it: the construction of C is not (apologies if my quasi-formal notation is incomprehensible) {xeC|En s.t. x is a boundary of a region excluded from the nth step of the construction of C}, but rather {xeC|~En s.t. x is excluded from C in the nth step of the construction of C}.
Well, it's not a real contradiction -- I just tried hard to word it as one :)
And yes, you're absolutely right. Can you exhibit a specific point x in [0,1] that belongs to C but is not the endpoint of any segment comprising E?
1/4, 3/4, 1/13... I'll leave the formula as a puzzle for people who haven't taken college math yet. :)
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