Cardnality of Infinite Sets (4)

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SUMMARY

The cardinality of the set of all functions from the natural numbers (N) to the set {1, 2, 7} is equal to c, the cardinality of the continuum. This conclusion is derived by establishing a bijection between the sequences of functions from N to {1, 2, 7} and the set of ternary sequences, which correspond to the interval [0, 1]. The discussion also covers proving properties of cardinality, specifically that if |A1|=|B1| and |A2|=|B2|, then |A1 x A2|=|B1 x B2| and |A1 U A2|=|B1 U B2| when A1 and A2 are disjoint.

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  • #31
After all of these questions, you still can't think of a way to make a correspondence between ternary number is [0,1] and maps from N to {1,2,7} without posting a question and then waiting around for someone to tell you?
 
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  • #32
V={{s1,s2,s3,...}|si=0 or 1 or 2}
To each sequence in V, let it correspond to the subset of N consisting of {j|sj=2}
e.g.{1,2,0,1,2,...} <-> {2,5,...}

|V|=|P(N)|=c

But this only shows 2^|N|=c
 
  • #33
kingwinner said:
V={{s1,s2,s3,...}|si=0 or 1 or 2}
To each sequence in V, let it correspond to the subset of N consisting of {j|sj=2}
e.g.{1,2,0,1,2,...} <-> {2,5,...}

|V|=|P(N)|=c

But this only shows 2^|N|=c

Sigh. If you meant e.g. {1,2,0,1,2,..}<->{2,7,1,2,7,..}. Yes. No, it DOESN'T show 2^|N|=c. Why would it show that? What does it show?
 
  • #34
Dick said:
Sigh. If you meant e.g. {1,2,0,1,2,..}<->{2,7,1,2,7,..}. Yes. No, it DOESN'T show 2^|N|=c. Why would it show that? What does it show?

OK, I think I get the overall argument:

|{f| f: N -> {1,2,7} }|
=|{sqeuences of 1,2,7}|
=|{sequences of 0,1,2}|
=|{ternary decimals in [0,1]}|
=|[0,1]|
=c

I know this is easy for you, but not so for a student with only a 2-lecture intro to this topic without many practical examples. Thanks for your patience!
 
Last edited:
  • #35
That's it exactly. You're welcome!
 

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