The Cantor Set Is Uncountable

Professor Zap Sketches the proof that the Cantor set is uncountable.

Category:

Education

Tags:

• Math
• Advanced Calculus
• Cantor Set
• Uncountable
• Real numbers
• infinite binary sequences

[cavoj]

nomber 1 = absolute number ........proof

[ProfessorElvisZap]

@cavoj Huh?

[markov2b1]

is there a way we can transform this cantor uncountability proof to an example with binary decimals. like what would say, LLRL correspond to in binary decimals. But LLRL is not a number. The real number would be the infinite sequence of L's,R's

[ProfessorElvisZap]

@markov2b1 Just change the notation: Set L=0, and R=1. It makes sense since 0 is to the left of 1.

[markov2b1]

you wrote a1 = a11a12a13...

What is the purpose of the double scripts. They correspond to decimals like

a1 = .032485

a2 = .4539358

... except that we are in base 2 basically? what would the decimal number look like?

[ProfessorElvisZap]

@markov2b1 In binary notation these would be 0s and 1s. So if a_{11}=0, a_{12}=0, a_{13}=1, then

a_1=001...

More generally one of the elements in the list could be _k=0.0000110101010100100001111­10011001010...

[gpetty91]

If you used open intervals instead of closed intervals, would the cantor set be empty?

[8PlanA]

To make this a rigorous proof you could have just proved injectivity from the set of 0-1 sequences into the cantor set using the Nested Interval Property - that actually takes less work than having to mess around with decimal's and the diagonal argument, which are actually quite subtle to make rigorous. {0,1}^N is homeomorphic to C also - a great addition to YouTube, keep providing more material!

[cavoj]

111111111111111111111111

[cavoj]

number 1 = absolute number .........number 2,3,96,348,etc. = abstract number ......

[joeglimmix]

7:45 elvis has left the lecture hall

[tylerscottsmith]

Just a quick question. For ternary, instead of saying choose 0 or 1 for the ann place, you would choose 0 or 2, correct? And the point would differ from the list in the 3^(-n)th place?

[tommyrjensen]

@ProfessorElvisZap The usual difficulty here is that two different binary sequences, say, 0111111... and 1000000... represent the same number .0111111... = .100000... = 1/2. Thus there ought to be "fewer" distinct binary decimals than there are elements of the Cantor set. The Cantor-Bernstein-Schröder Theorem can be used to show that there is in fact a one-to-one correspondence, but it is not completely trivial.

[barnettb901]

isnt this some sort of fractal monster??