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.


This is a perfectly legal number in our base 11 scheme.

Now scale it:


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:


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 rq, 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', qp' < 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.

  • 1
    $\begingroup$ Consider the function $\tan x$, which maps from a finite range to the entire set of real numbers. Note that $\tan x$ over its finite domain is one to one and onto and thus proves the cardinalities are equal. $\endgroup$
    – abiessu
    Sep 11, 2013 at 0:51
  • $\begingroup$ The interval $[q,r)$ is neither infinitely small nor finite, for any $r > q$. The intersection or "limit set" of all these sets is just the single point $q$, but you can't map all the integer sequences to it. This is basically another example of behavior of a sequence not matching the behavior of the limit. $\endgroup$ Sep 11, 2013 at 0:54
  • $\begingroup$ @abiessu, thank you for such a clear and concise answer. $\endgroup$
    – user94213
    Sep 11, 2013 at 1:04
  • $\begingroup$ Generally, the question has been asked ad nauseum on this site. math.stackexchange.com/questions/200180/… is a particular example, and there are more - much more - examples for closed, open, left-closed and right-open, and right-closed and left-open, and so on. $\endgroup$
    – Asaf Karagila
    Sep 11, 2013 at 6:00
  • $\begingroup$ @Asaf Karagila - If the question is asked ad nauseum then that might be an indication that the matching done by SO isn't working so well. It was only when I revised the title to use "bounded interval" did proper matches show up. $\endgroup$
    – user94213
    Sep 11, 2013 at 13:30

1 Answer 1


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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    $\begingroup$ Yes, but you and the OP should not be saying "finite range" when you mean "bounded interval". Also when you say "complete mapping" you mean "1-1 and onto mapping" or "bijection". Apologies for being pedantic. $\endgroup$
    – Rob Arthan
    Sep 11, 2013 at 1:52
  • $\begingroup$ No apologies necessary, I have forgotten much of what it means to be exact and I am relearning by attempting to answer questions like this, and by getting constructive criticism such as you have provided. $\endgroup$
    – abiessu
    Sep 11, 2013 at 2:14

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