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 $\tan x$ (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 $a\ne b$, then $\forall x\in (a,b),$
$$ f(x)=\tan\left({\pi\cdot (x-a)\over b-a}-{\pi\over 2}\right)$$
supplies this mapping.
Since it is possible to pick any real number $r$ and find $x_r$ such that $f(x_r)=r$, 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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