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Cardinality of infinite sequences of real numbers 
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#1
Feb2613, 03:42 PM

P: 19

I have to prove that the cardinality of the set of infinite sequences of real numbers is equal to the cardinality of the set of real numbers. So:
[tex]A := \mathbb{R}^\mathbb{N}=\mathbb{R} =: B[/tex] My plan was to define 2 injective maps, 1 from A to B, and 1 from B to A. B <= A is trivial, just map a real number x on the sequence (xxxxxxxxx...). But I can't find a injective map from A to B. Can someone help? 


#2
Feb2613, 04:29 PM

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You could reduce it to an easier problem first, for example. Something like [tex][0,1]^\mathbb{N}=[0,1][/tex]



#3
Feb2613, 04:48 PM

P: 19

I would start by thinking about why the cardinality of ##[0,1)^2## is equal to the cardinality ##[0,1)##. To do this you think realize that any element of ##[0,1)^2## can be written as ##(x,y)## where ##x## and ##y## have infinite decimal expansions ##x = a_1 a_2 a_3 ...## and ##y = b_1 b_2 b_3 ... ##, then you can combine these into a unique real number ##z = a_1 b_1 a_2 b_2 a_3 b_3 ...## .
From here, you can generalize this proof to show that ##[0,1)^\mathbb{N} = [0,1)## by recalling the proof that the rational and natural numbers have the same cardinality. At this point you should be almost home. Good Luck! 


#4
Feb2613, 05:03 PM

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Cardinality of infinite sequences of real numbers



#5
Feb2613, 05:10 PM

P: 19

(By the way, I'm sorry that I have edited and updated my post multiple times. I'm still trying to figure out how to use the Tex features properly.) 


#6
Feb2613, 08:03 PM

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#8
Feb2713, 11:12 AM

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"the cardinality of the set of ([countable] infinite) sequences of real numbers" I added "()" to clarify the structure. 


#9
Feb2713, 11:35 AM

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#10
Feb2713, 08:54 PM

P: 595

How about this:
If you accept that the Reals are uncountable and the rationals are countable, and that a number is rational iff it has an eventuallyperiodic exoansion: First show that the set , say S , of sequences in ℝ^{N} that are eventuallyperiodic are countable, and then set up a bijection between ℝ^{N}\S and the irrationals, sending a sequence to its "natural" decimal expansion. 


#11
Feb2813, 12:51 AM

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#12
Feb2813, 07:03 PM

P: 19




#13
Feb2813, 08:43 PM

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Edit: just because "transfinite sequence" has the word sequence in it, does not mean it is in fact a "sequence" in the technical sense. 


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