# More numbers between $[0,1]$ or $[1,\infty)$?

There are infinitely many real numbers between any two real numbers, therefore there are infinitely many real numbers in the range $[0,1]$ as there are in $[1, \infty)$. In a mathematical sense, are there more numbers in one range as opposed to the other? What about $[0,1]$ and $[0, \infty)$ ?

I'd assume this is akin to how limits work in that the limit as $x\to\infty$ of $\frac{x^x}{\log(x)}$ is $\infty$, even though both the numerator and denominator approach infinity.

• @Casteels Thanks for the formatting help. – dberm22 Jun 16 '15 at 12:30
• Two sets are said to have the same size (technical jargon: same cardinality) if we can construct a one-to-one correspondence between them. – Akiva Weinberger Jun 16 '15 at 12:37
• Hint: $[1,\infty)=\{1\}\cup(1,2]\cup(2,3]\cup(4,5]\cup\dotsb$, and $[0,1]=\{1\}\cup[0,\frac12)\cup[\frac12,\frac34)\cup[\frac34,\frac78)\cup\dotsb$. – Akiva Weinberger Jun 16 '15 at 12:45
• ^corrected version – Akiva Weinberger Jun 16 '15 at 12:48
• Another hint: Show that for any $a,b,c,d$, the sets $[a,b)$, $[c,d)$, and $(c,d]$ have the same cardinality. – Akiva Weinberger Jun 16 '15 at 14:45

The sets $[0,1]$ and $[1,\infty)$ have the same cardinality (i.e., there are exactly as many numbers in one as in the other). You can construct an explicit bijection between them with some tinkering on the tangent function (which provides a bijection between $[0,\pi/2)$ and $[0,+\infty)$).

In fact, you can easily construct a bijection between $[0,1)$ and $[1,\infty)$, which I suppose is sufficient for you at this point. It is more complicated to do it from $[0,1]$, see for example this question.

• So [0,1] and [0,∞) also have the same cardinality, even though one is a subset of another? – dberm22 Jun 16 '15 at 12:38
• Yes, a characteristic of infinite sets is that an infinite set can have the same cardinality as one of its proper subsets. Ever heard of Hilbert's hotel? – fkraiem Jun 16 '15 at 12:39
• The hard part is finding a bijection between $[0,\frac\pi2)$ and $[0,1]$. Notice that one is half-closed and the other is closed. – Akiva Weinberger Jun 16 '15 at 12:43
• I have not until just now. Thanks for the resource. Kind of hurts my brain to think about, though. – dberm22 Jun 16 '15 at 12:43
• As @columbus8myhw mentioned in his comment: which number in the second set is mapped from the 1 in the first set? – Gottfried Helms Jun 16 '15 at 12:44

On top we have a circle centered at, say, $(0,2)$ with radius one. Extend a radius through the circle until it hits the $x$-axis and we'll always get two points: $x$ and $x'$, in the graph.

Tracing out from left to right, we trace out both the semicircle and the whole $x$-axis, i.e. the real line.

But the former only has length $\pi$; with some rigor we can show that this means there are exactly as many numbers in and subset of the reals as there are reals.

• This compares $[1,\infty)$ with $[0,1)$, but the hard part is dealing with the extra point $\{1\}$! – Akiva Weinberger Jun 16 '15 at 12:42
• @columbus8myhw I think it compares $(0,1)$ to $(-\infty,\infty)$, but your point is valid – MichaelChirico Jun 16 '15 at 12:46
• As you said, this doesn't answer the exact question, but it does help me understand the issue much better. I'm definitely a visual learner. +1 – dberm22 Jun 16 '15 at 12:46
• Very cool way of looking at it. I have never seen this before. – anakhro Jun 16 '15 at 12:49
• Note that this is almost exactly the tangent function (modulo the fact that the line is not tangent to the circle). ;) Indeed, keeping in mind the visual aspect of the trigonometric functions is very useful. – fkraiem Jun 16 '15 at 12:54

You might read later in your education the sense in which $A:=[0,1]$ is almost finite while $B:=[1,\infty)$ is not. This almost finiteness is known as compactness and sets that are compact are in a different way smaller than non-compact sets.

For example, given any $\varepsilon>0$, you can quite easily find a finite number of elements of $A$, $\mathcal{C}:=\{x_1,x_2,\dots,x_N\}$ such that for all $x\in A$ there exists an $x_i\in \mathcal{C}$ such that

$$|x-x_i|<\varepsilon.$$

This is not the case for $B$.

You are a little confuse with the two concepts about infinity involve. The first one is about the "number of elements" in the sets. Each non degenerate interval has the same "number of elements" that other.

In the real line there are two types of infinity that counts "the numbers of elements". The first is named countable, this is the "number of elements" that has the natural numbers, and as the named said the intuitive idea is that you can count it. For example, the integers, an finite set, or the rational are countable.

The other concept is name no countable. The sets that are no countable has more elements than those who are countable. As and example each non degenerate interval is countable.

If you are interested in this topic you can read for example the third chapter of Analysis of Apostol.

The other concept involve is the infinity as a behavior, this means that as the variable get closer to some number the function gets greater, but the function itself is not infinity.

Another alternative:

There is an injection $[0,1] \to [1,\infty)$ given by $x \mapsto x + 1$. Hence the cardinalities $\|\,[0,1]\, \| \leq \|\,[1,\infty)\,\|$.

And there is an injection $[1,\infty) \to [0,1]$ given by $x \mapsto 1/x$. Hence $\|\, [1,\infty)\, \| \leq \|\,[0,1]\,\|$.

Therefore

$$\| [0,1] \| = \|[1,\infty)\|$$

Hint: Let $f(x)=\dfrac1x$ and $g(x)=\dfrac1{x+1}$ . Apply the former on $[1,\infty)$, and the latter on $[0,\infty)$.