1. Well, so far I have'nt been able to do this one so any solutions will be new to me!

How do you prove that the Cantor set is nowhere dense?

I suspect I know who's going to come with a quick reply and make this look fiendishly simple!

Just to be clear, anyone who feels the need to chip in with their two cents worth when they don't really know what they're talking about, with things like "it's how we use 'numbers' that makes it 'dense'" or any of that of that meaningless dribble, please save it for another thread! I would like to see genuine solutions to the problem please.

2.

3. Hey Naz, there are two steps to your question and the easiest way to solve each one is using a different representation of the Cantor Set.

1. Show that the Cantor set is closed : Well to do that consider the middle third deletion method of constructing the set.

A<sub>0</sub> = [0, 1]. A<sub>1</sub>= A<sub>0</sub> \ (1/3, 2/3). A<sub>2</sub> = A<sub>1</sub> \ ( (1/9, 2/9) U (7/9, 8/9) ) ...

A = intersection of all the A<sub>i</sub>. Now each A<sub>i</sub> is a closed set intersected with the compliment of the union of a whole bunch of open sets so is closed itself. So A is closed i.e. the Cantor set is closed.

2. Show the Cantor Set has empty interior : To do that its easier to argue from the trinary expansion construction of the cantor set. Here the cantor set is defined as the set of all real numbers in [0, 1] which when expanded in the form Sum a<sub>k</sub> 3<sup>-k</sup> k > 0 do not have a<sub>k</sub> = 1 for all k > 0.

Now suppose t is in int(A), so we can find an open interval (t-e, t+e) contained in A for some e > 0. Now in any interval (a, b) you can find a number k such that the trinary expansion of k does not consist of 0 and 2 only. This number in (t-e, t+e) contradicts the construction of A so A contains no intervals i.e The Cantor set has empty interior.

So the int(A) = int(cl(A)) is empty i.e. The cantor set is nowhere dense.

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