A bit of background on why I'm asking this:
Take the sequence of ℤ+, which has the same cardinality as ℤ:
1, 2, 3, 4, 5, 6, ...
Suppose we create a number in base 11 using x
as the extra symbol, by concatenating the numbers, and delimiting with x
.
1x2x3x4x5x6x...
This is a perfectly legal number in our base 11 scheme.
Now scale it:
0.1x2x3x4x5x6x...
This would be a real number between 0 and 1 [0, 1).
In other words, this real number represents the sequence ℤ+ and consists of a single point p on the real number line, 0 ≤ p < 1.
Take any other sequence in ℤ+, transform it the same way, and it can be represented as a point p on the real number line, 0 ≤ p < 1:
0.10x20x30x40x... times ten
0.1x4x9x16x25x... squares
0.41x8x20x9x5x... some random sequence
All finite sequences of similar length can also be represented similarly. For example, all finite sequences of length 2: {1, 1} {1, 2} {1, 3} ...
can be ordered and represented by a single number in ℝ:
0.1x1x1x2x1x3x...
So we have the set of all sequences, finite and infinite, representable as a point p, 0 ≤ p < 1.
These sequences can be scaled to any arbitrary [q, r) such that r → q, but always r > q.
Given that [q, r) is an arbitrarily small but bounded interval in ℝ, and that ℤ+ is represented as a single point p', q ≤ p' < r, can it be argued that the size of this set is larger than ℤ but smaller that ℝ?
I guess it all hinges on whether the cardinality of any bounded interval of ℝ = |ℝ|, hence the title of the question.
Answer
The function tanx (among others, but tangent is easy to use) can be mapped from any specific range in the real numbers to the entire set of reals. In particular, let the range [a,b] be given where a≠b, then ∀x∈(a,b),
f(x)=tan(π⋅(x−a)b−a−π2)
supplies this mapping.
Since it is possible to pick any real number r and find xr such that f(xr)=r, and vice versa, this function is one to one and onto, and is thus a complete mapping AKA bijection from a bounded interval of the reals to the entire set, which means that the cardinalities are the same.
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