In this lecture, it is proved that a set S of all infinite binary sequences is uncountable. I have two questions regarding that. i) If one considers B, a set of binary numbers? Each binary number is a binary sequence and can be mapped to a natural number. Does it imply that B is countable? ii) Since B includes all binary numbers, can we say that S is a subset of B?
There is a difference between a binary sequence and a binary number. A binary number can be regarded as a binary sequence by making all entries after a certain stage zero. On the other hand, not every infinite binary sequence is a binary number. Thus B is a proper subset of S.
Even if B is empty, that does not change the main argument. We are not discussing bijection from A to B. g is assumed to be a bijection from A to P(A).
@21:50 how does f(A)=f(B) imply A=B?. If we have same binary sequence for A and B, that need not imply that the elements of A are same as elements of B. It might be the case that if f(A)={0,1}=f(B), then A={a1,a2} while B={a3,a4} s.t. a2 is in A and a4 is in B.
g(x) is not assumed to be a bijection, g is assumed to be a bijection. There is a difference between g and g(x). g is a map from A to its power set. Hence for each x in A, g(x) is some subset of A. Thus we can pick those x that do not belong to g(x).
g(x) isn't mapping all elements of A. It's mapping certain elements of A. Hence, the map won't contain all elements of 2^A, infact A. And the ones it doesn't contain are the elements we will pick.
Zooming in is necessary to help viewers clearly see writing on the board, thats why they are moving i guess. Improvement will be good though.. but this is ok too.
Incomplete explanation at 30:00 - 30:40. Decimal expansion of a number may not be unique. One needs to show that the cardinality of the subset of real numbers with multiple (2) decimal expansion is countable, and then needs to use the fact that since the union of two countable sets is countable, set formed by removing a countable number of elements from an uncountable set will be uncountable.
what about infinite product of countable sets ? i guess that results in an uncountable set .,.,., couldn't that be used as tool to show a set uncountable ?
33:15 Cardinal Numbers
finally made it to lecture number 5, after 1.5 months of learning- Set, relation and functions - Algebraic structures- group thoery- lattice theory.
In this lecture, it is proved that a set S of all infinite binary sequences is uncountable. I have two questions regarding that.
i) If one considers B, a set of binary numbers? Each binary number is a binary sequence and can be mapped to a natural number. Does it imply that B is countable?
ii) Since B includes all binary numbers, can we say that S is a subset of B?
There is a difference between a binary sequence and a binary number. A binary number can be regarded as a binary sequence by making all entries after a certain stage zero. On the other hand, not every infinite binary sequence is a binary number. Thus B is a proper subset of S.
Thanks for the explanation, Professor.
11:24 Cantor's Theorem. If B is empty, then no element of A will map to the empty set in B, and thus there will be no bijection from A to B.
Even if B is empty, that does not change the main argument. We are not discussing bijection from A to B. g is assumed to be a bijection from A to P(A).
@21:50 how does f(A)=f(B) imply A=B?. If we have same binary sequence for A and B, that need not imply that the elements of A are same as elements of B. It might be the case that if f(A)={0,1}=f(B), then A={a1,a2} while B={a3,a4} s.t. a2 is in A and a4 is in B.
Got the proof!
At 11:25, how can we pick all those x which do not belong to g(x)? g(x) is assumed to be a bijection, meaning that every element of A maps to P(A)?
g(x) is not assumed to be a bijection, g is assumed to be a bijection. There is a difference between g and g(x). g is a map from A to its power set. Hence for each x in A, g(x) is some subset of A. Thus we can pick those x that do not belong to g(x).
g(x) isn't mapping all elements of A. It's mapping certain elements of A. Hence, the map won't contain all elements of 2^A, infact A.
And the ones it doesn't contain are the elements we will pick.
The camera panning is quite distracting. The frame is always moving and it is difficult to see the board.
Zooming in is necessary to help viewers clearly see writing on the board, thats why they are moving i guess. Improvement will be good though.. but this is ok too.
11:04 how can g(x) be subset of A?? It should be subset of P(A).
Am I wright?
Same question.
g(X) is not subset of power set of A ...it is the element of P(A) .....thus, g(X) is subset of A
Incomplete explanation at 30:00 - 30:40.
Decimal expansion of a number may not be unique.
One needs to show that the cardinality of the subset of real numbers with multiple (2) decimal expansion is countable, and then needs to use the fact that since the union of two countable sets is countable, set formed by removing a countable number of elements from an uncountable set will be uncountable.
ok, but its unique since 999999 its excluded
I made it to lecture 5!
Very good teaching
is this course for BSc or MSc??
Best❤
what about infinite product of countable sets ? i guess that results in an uncountable set .,.,., couldn't that be used as tool to show a set uncountable ?
An infinite product of countable sets can be uncountable.
Sir please refer a book on Real analysis for competatative exams in Post Graduation level
royden
Too good
Thank you Sir
That cough tho !!
11:24 Cantor's Theorem. If B is empty, then no element of A will map to the empty set in B, and thus there will be no bijection from A to B.
cardinal set cannot be empty