TrueFalse The function that assigns to each bit string the #of ones in that string has asdomain thesetofallbit

Question

The function that assigns to each bit string the #of ones in that string has as domain the set of all bit strings

Answer 1

If you are thinking of f : ? -> { 0, 1, 2, 3, ... } (i.e., the domain is mapped into the set of non-negative integers) then clearly the domain "?" contains [at least] all finite bit strings. If you keep the same restriction on the range, you can even extend the domain to include those infinite bit strings (however you wish to define that) which contain only finitely many ones. If you extend the range to also include infinite cardinals and/or ordinals then you can include even more infinite bit strings into the domain.

Answer 2

Hmmm...
lets say the function is f(B).
if bit string B1 has "1" ocurring in it n1 times then the function would spit out f(B1) = n1
for B2 the answer might be n2, for B3 the answer might be n3 and so on. So for the function f, the domain is (B1, B2, B3, B4........) and the range is (n1, n2, n3, n4........)

The requirement for a function is there cannot be two values from the range to match with a single value from the domain. In other words if f(x) = y, and f(x1) = y1 and at the same time f(x1) = y2 then f(x) cannot be a function. However more than one x values can have the same y value, like f(x1) = y1 and f(x2) = y1 is allowed.
Going back to the question, does your function violate this rule? That is for any bitstring, can there be two answers? No. Clearly it is a function.

So in other words your question is: Is the set {B1, B2...} the complete set of all bit strings possible?
Now n1, n2, n3 are all integers, because you can't have a fractional number of "1" in a string. So the set {n1, n2...} is a set of an infinite number of integers. Is the set {B1, B2....} a set of the same order?
No. Because you can have two bitstrings B1 and B2 that have the same number of "1" but the strings are entirely different. for example, 0110 and 1010. Therefore the number of bitstrings that are possible for the same number of "1" in the string is more than one. How big is the set {B1, B2...}?

I will give you a hint: The number of strings with the same # of 1's are countably infinite, that is you can assign an integer for each of the strings. Let this be group m; now there are countably infinite number of groups like m. I can't use the Aleph symbol but as you can see the total set of bitstrings is a different order of infinity than integers. But it doesn't seem to me that any of these orders leads to a set that is uncountably infinite like the set of real numbers.

So using that argument you may say the answer is true.