10-327/Classnotes for Monday September 27 - Revision history

D at 16:28, 2 October 2010

2010-10-02T16:28:30Z

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	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
	****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)		****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)

			[[User:D\|D]] 12:28, 2 October 2010 (EDT) Note on clopen (closed and open) set: In ANY topological space, an empty set is clopen. Similarly, the whole set is clopen. However it is not true that any clopen set is trivial (although it is true in a connected space, since it cannot be represented by two disjoint open sets). For example, consider Q = {all rational numbers} and its subset A = {all rational numbers smaller than pi}. Then the boundary of A is empty in Q, because pi is not in Q. Hence A is clopen in Q. Of course, A is not open nor closed in R.

Jdw at 20:19, 30 September 2010

2010-09-30T20:19:21Z

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	{{10-327/Navigation}}		{{10-327/Navigation}}

	See some blackboard shots at {{BBS Link\|10_327-100927-142655.jpg}}.		See some blackboard shots at {{BBS Link\|10_327-100927-142655.jpg}}.

Drorbn at 13:41, 28 September 2010

2010-09-28T13:41:21Z

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	**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
	****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)		****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)

Drorbn at 13:40, 28 September 2010

2010-09-28T13:40:04Z

← Older revision		Revision as of 09:40, 28 September 2010
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	**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
	****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours.		****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)

Drorbn at 13:39, 28 September 2010

2010-09-28T13:39:48Z

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	** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 September 2010 (EDT)
			****Yes, <math>k</math> and <math>n</math> here are integers. To represent integers, you use ''finite'' sequences of 0s and 1s, and there are countably many of those. To represent real numbers in <math>[0,1]</math> you use ''infinite'' sequences of 0s and 1s, and there are uncountably many of those. [[User:Drorbn\|Drorbn]] 09:39, 28 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		*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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
			****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours.

Xwbdsb at 13:12, 28 September 2010

2010-09-28T13:12:20Z

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	*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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 27 September 2010 (EDT)
	** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|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? -Kai[[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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		*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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
			***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)

Xwbdsb at 13:06, 28 September 2010

2010-09-28T13:06:58Z

← Older revision		Revision as of 09:06, 28 September 2010
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	*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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 27 September 2010 (EDT)
	** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|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? -Kai[[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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		*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

Drorbn at 00:51, 28 September 2010

2010-09-28T00:51:38Z

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	[[User:Xwbdsb\|Xwbdsb]] 20:26, 27 September 2010 (EDT)		[[User:Xwbdsb\|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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 27 September 2010 (EDT)
	*Question1:
			** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto
	the unit interval [0,1] 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
	N union {0}? (the first number stands for 2^0, second stands for 2^1, etc.)
	-Kai

			*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
	[[User:~~Xwbdsb~~\|~~Xwbdsb~~]] 20:37, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	*Question2:
	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

Xwbdsb at 00:37, 28 September 2010

2010-09-28T00:37:31Z

← Older revision		Revision as of 20:37, 27 September 2010
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	[[User:Xwbdsb\|Xwbdsb]] 20:26, 27 September 2010 (EDT)		[[User:Xwbdsb\|Xwbdsb]] 20:26, 27 September 2010 (EDT)
			*Question1:
	*Question:
	Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto		Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto
	the unit interval [0,1] and hence is not countable.Is it true that every real number in the unit		the unit interval [0,1] 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		interval has more than one binary expansion? Is it possible to map the set of all permutations onto
	N union {0}? (the first number stands for 2^0, second stands for 2^1, etc.)		N union {0}? (the first number stands for 2^0, second stands for 2^1, etc.)
			-Kai

			[[User:Xwbdsb\|Xwbdsb]] 20:37, 27 September 2010 (EDT)
			*Question2:
			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		-Kai

Xwbdsb at 00:33, 28 September 2010

2010-09-28T00:33:06Z

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	[http://katlas.math.toronto.edu/drorbn/images/0/01/10-327-lec05p03.jpg Lecture 5 page 3]		[http://katlas.math.toronto.edu/drorbn/images/0/01/10-327-lec05p03.jpg Lecture 5 page 3]

	[[User:Xwbdsb\|Xwbdsb]] 20:26, 27 September 2010 (EDT)~~Question:~~		[[User:Xwbdsb\|Xwbdsb]] 20:26, 27 September 2010 (EDT)
			*Question:
	Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto		Dror you said in class the set of permutations of 0's and 1's could be mapped "bijectively" onto
	the unit interval [0,1] and hence is not countable.Is it true that every real number in the unit		the unit interval [0,1] and hence is not countable.Is it true that every real number in the unit

← Older revision		Revision as of 12:28, 2 October 2010
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	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
	****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)		****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)

			[[User:D\|D]] 12:28, 2 October 2010 (EDT) Note on clopen (closed and open) set: In ANY topological space, an empty set is clopen. Similarly, the whole set is clopen. However it is not true that any clopen set is trivial (although it is true in a connected space, since it cannot be represented by two disjoint open sets). For example, consider Q = {all rational numbers} and its subset A = {all rational numbers smaller than pi}. Then the boundary of A is empty in Q, because pi is not in Q. Hence A is clopen in Q. Of course, A is not open nor closed in R.

← Older revision		Revision as of 09:41, 28 September 2010
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	**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
	****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)		****No, it does not imply that <math>n=\infty</math> is true. Yet I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours. [[User:Drorbn\|Drorbn]] 09:40, 28 September 2010 (EDT)

← Older revision		Revision as of 09:39, 28 September 2010
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	** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 September 2010 (EDT)
			****Yes, <math>k</math> and <math>n</math> here are integers. To represent integers, you use ''finite'' sequences of 0s and 1s, and there are countably many of those. To represent real numbers in <math>[0,1]</math> you use ''infinite'' sequences of 0s and 1s, and there are uncountably many of those. [[User:Drorbn\|Drorbn]] 09:39, 28 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		*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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
	***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)		***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)
			****No. I cannot give a full review of induction here; you may want to read [http://en.wikipedia.org/wiki/Mathematical_induction Wikipedia: Mathematical Induction] and/or come to my office hours.

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	*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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 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 <math>[0,1]</math> 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 <math>N</math> union <math>{0}</math>? (the first number stands for <math>2^0</math>, second stands for <math>2^1</math>, etc.) -Kai [[User:Xwbdsb\|Xwbdsb]] 20:37, 27 September 2010 (EDT)
	** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		** Except for the "dyadic rationals", the numbers of the form <math>\frac{k}{2^n}</math>, 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. [[User:Drorbn\|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? -Kai[[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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? -Kai [[User:Xwbdsb\|Xwbdsb]] 09:06, 28 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		*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. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)		**Induction will only go to finite numbers. [[User:Drorbn\|Drorbn]] 20:51, 27 September 2010 (EDT)
			***If n = 1 is true, n=k is true implies n=k+1 is true. Why wouldn't this logic imply that n=infinity is true?-Kai [[User:Xwbdsb\|Xwbdsb]] 09:12, 28 September 2010 (EDT)