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    I know it is uncountable. But what is wrong with this proof?
    I use the lemma that a countable union of countable sets is countable.

    Let f(0)=0. Then for each function f, construct [a0, a1, a2....] as following: ai is the number of times the function f takes the value i. For example, for x^2 the set would be [1,1,0,0,1,0,0,0,0,1,......].

    There's a 1 to 1 map from each function to such a set. Thus, the set is a countable union of natural numbers, so is countable.

    Then we vary the value of f(0) to = 0,1,2,3.... There is a bijection from the set of increasing functions for which f(0)=0 and for which f(0)=i (the latter just shifts to the right by (i+1)) so the number of functions for which f(0) = 0,1,2... is just a countable union of the countable sets we made earlier, which is countable.
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    (Original post by Omghacklol)
    Let f(0)=0. Then for each function f, construct [a0, a1, a2....] as following: ai is the number of times the function f takes the value i. For example, for x^2 the set would be [1,1,0,0,1,0,0,0,0,1,......].
    I'm confused here, but surely you're treating this as some kind of ordered list or array instead of an actual set where listing the same element twice or thrice doesn't make sense.
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    (Original post by Omghacklol)
    I know it is uncountable. But what is wrong with this proof?
    I use the lemma that a countable union of countable sets is countable.

    Let f(0)=0. Then for each function f, construct [a0, a1, a2....] as following: ai is the number of times the function f takes the value i. For example, for x^2 the set would be [1,1,0,0,1,0,0,0,0,1,......].

    There's a 1 to 1 map from each function to such a set. Thus, the set is a countable union of natural numbers, so is countable.

    Then we vary the value of f(0) to = 0,1,2,3.... There is a bijection from the set of increasing functions for which f(0)=0 and for which f(0)=i (the latter just shifts to the right by (i+1)) so the number of functions for which f(0) = 0,1,2... is just a countable union of the countable sets we made earlier, which is countable.
    You seem rather... confused about a few things.

    There's a 1 to 1 map from each function to such a set. Thus, the set is a countable union of natural numbers, so is countable.
    What does this mean? What is a map from a function to a set? (Or to an ordered list, since that isn't a set). What does it mean for such a map to be 1 to 1? That thing is obviously countable: it's a sequence indexed by the natural numbers. It's also irrelevant, since that's a sequence of numbers, not a set of functions.



    Now, on to the actual problem:
    Thus, the set is a countable union of natural numbers, so is countable.
    This set (assuming you mean the set of all sequences of that form) is not a countable union of countable sets.
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    (Original post by Omghacklol)
    ,,,
    Your argument seems to be:

    1) There is 1-to-1 mapping from, the set of increasing functions from N to N, to a set of binary sequences. I say "a" set, as this is not the set of all binary sequences, since it does not include sequences with only a finite number of "1"s.

    2) You claim this set is countable. But is it?
 
 
 
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