# Honours Analysis - Math Problem Example

Summary
Second remove the central interval I1 =). Next, subdivide each of the remaining intervals into three equal open subintervals and from each remove the central third denoted by I2. This can…

## Extract of sample "Honours Analysis"

Download file to see previous pages Characterizing of the Cantor set in terms of ternary is done when a real number from the closed real interval [0, 1] belongs to Cantor set with a ternary expansion containing digits 0 and 2. To construct this expansion, one has to consider the points in closed real interval [0, 1] in terms of base 3 notation.
When constructing the Cantor Middle Third set, we start with the interval [0, 1] removing the middle thirds , this leaves [0,1/3] [2/3,1].The next step is to also remove the middle thirds (1/9, 2/9) and (7/9, 8/9) from the remaining two intervals. This process is repeated continuously. From the results, we can note that all the endpoints remain, which are the Cantor set.
The total length of the intervals removed in the construction of the Cantor set can be determined as follows. From interval [0, 1] we first remove a middle third interval 1/3 second step we remove two middle intervals of 1/9. We continue with the process so that at the nth stage we remove 2n-1 intervals with the length 3-n. The total sum of the removed intervals is
It can be proven that the Cantor set is perfect and totally disconnected. In this case, x and y are two distinct points in the cantor set. Since x ≠ y therefore │x - y│>. As we can see there is a natural number N that exists in the interval. Next we identify that Cantor set ⫋ Ck for all k, such that x, y Ck. For each 2N disjoint closed interval from CN there is. Therefore, x and y are inside distinct closed intervals in CN. The two intervals should have an open interval between them, which is not part of the Cantor set otherwise this would be a single closed interval. The chosen point can be represented by z, therefore z Cantor set and it is between x and y. (Gordon, 1994, p. 301)
If we put f in its inverse: If x Q, then also - x Q. therefore f o f (x) = f ( f (x)) = f ( - x) = - (-x) = x. If x ∉ Q then f o f (x) = f (f (x)) = f (x) = x. Thus for all x R, we have that f o f ...Download file to see next pagesRead More
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