10-327/Classnotes for Monday September 27
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See some blackboard shots at BBS/10_327-100927-142655.jpg.
Video: Topology-100927
Dror's notes above / Student's notes below |
Here are some lecture notes..
Xwbdsb 20:26, 27 September 2010 (EDT)
- Question 1: Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto the unit interval and hence is not countable. Is it true that every real number in the unit interval has more than one binary expansion? Is it possible to map the set of all permutations onto union ? (the first number stands for , second stands for , etc.) -Kai Xwbdsb 20:37, 27 September 2010 (EDT)
- Except for the "dyadic rationals", the numbers of the form , real numbers have a unique binary expansion, and there is only a countable set of dyadic rationals. Also, we are talking about sequences of 0s and 1s, not "permutations". And just like decimals, binary number may have a "decimal point", where the (finite) part of the sequence ahead of the point represents the integer part of the number and the (possibly infinite) sequence after the point represents the fractional part. Drorbn 20:51, 27 September 2010 (EDT)
- So here both k and n are integers? Also what if I don't use decimal point to represent a sequence of 0s and 1s just mapping the sequence into the natural numbers? Then wouldn' the set be countable? -KaiXwbdsb 09:06, 28 September 2010 (EDT)
- Except for the "dyadic rationals", the numbers of the form , real numbers have a unique binary expansion, and there is only a countable set of dyadic rationals. Also, we are talking about sequences of 0s and 1s, not "permutations". And just like decimals, binary number may have a "decimal point", where the (finite) part of the sequence ahead of the point represents the integer part of the number and the (possibly infinite) sequence after the point represents the fractional part. Drorbn 20:51, 27 September 2010 (EDT)
- Question 2: Suppose we have a collection of sets which is closed under intersection. Does it mean that it is closed under arbitrary intersecions(i.e. uncountable intersections?) Also, suppose this set satisfies the condition: for any A,B in the set, A intersect B is in the set. What can mathematical induction give us? A.the set is closed under finite intersections B.the set is closed under countable intersections C.the set is closed under arbitrary intersections. And Why? -Kai
- Induction will only go to finite numbers. Drorbn 20:51, 27 September 2010 (EDT)