4 Properties of \(\mathbb{R}\)

Now that we established the axiomatic definition of the Real Numbers \(\mathbb{R}\) as a complete ordered field, let us investigate some of the properties of \(\mathbb{R}\). These will be consequence of the axioms of the real numbers, particulaly of the Axiom of Completeness.

4.1 Archimedean Property

The Archimedean property is one of the most useful properties of \(\mathbb{R}\), and it essentially states that the set of natural numbers \(\mathbb{N}\) is not bounded above in \(\mathbb{R}\).

More precisley, the Archimedean Property says two things:

For any \(x \in \mathbb{R}\) we can always find a natural number \(n \in \mathbb{N}\) such that \[ n > x \,. \]
For any \(x \in \mathbb{R}\) with \(x>0\), we can always find a natural number \(m \in \mathbb{N}\) such that \[ 0< \frac1m < x \,. \]

The situation is depicted in Figure 4.2.

Figure 4.1: For any \(x>0\) we can find \(n,m \in \mathbb{N}\) such that \(1/m<x<n\).

Remark 1

The Archimedean property might sound trivial. However there are examples of ordered fields \(K\) that satisfy:

\(\mathbb{N}\subseteq K\).
\(K\) does not have the Archimedean property.
In particular, \(\mathbb{N}\) is bounded above in \(K\).

Of course such fields \(K\) cannot be complete.

If \(K\) is complete, then \(K\) is essentially \(\mathbb{R}\), and we are going to prove the Archimedean Property holds in \(\mathbb{R}\).

Let us proceed with the precise statement of the Archimedean property in \(\mathbb{R}\).

Theorem 2: Archimedean Property

Let \(x \in \mathbb{R}\) be given. Then:

There exists \(n \in \mathbb{N}\) such that \[ n > x \,. \]
Suppose in addition that \(x>0\). There exists \(n \in \mathbb{N}\) such that \[ \frac1n < x \,. \]

Proof

Part 1. Let \(x \in \mathbb{R}\). Suppose by contradiction that there is no \(n \in \mathbb{N}\) such that \[ n > x \,. \] This means that \[ n \leq x \, \quad \forall n \in \mathbb{N}\,. \tag{4.1}\] The above is saying that the set \(\mathbb{N}\) is bounded above. Since \(\mathbb{N}\) is not empty, by the Axiom of Completeness there exists \[ \alpha = \sup \mathbb{N}\,. \]

Claim: \((\alpha-1)\) is not an upper bound for \(\mathbb{N}\).

Proof of Claim. Indeed, we have \[ (\alpha - 1) < \alpha \,. \tag{4.2}\] Therefore \(\alpha-1\) cannot be an upper bound for \(\mathbb{N}\). Indeed, if by contradiction \(\alpha - 1\) was an upper bound for \(\mathbb{N}\), then we would have \[ \alpha \leq (\alpha - 1)\,, \] since \(\alpha\) is the smallest upper bound for \(\mathbb{N}\). This contradicts (4.2). Therefore \(\alpha - 1\) is not an upper bound for \(\mathbb{N}\).

Conclusion. Since \(\alpha - 1\) is not an upper bound for \(\mathbb{N}\), there exists \(n_0 \in \mathbb{N}\) such that \[ \alpha - 1 < n_0 \,. \] The above implies \[ \alpha < n_0 + 1 \,. \] Since \[ (n_0+1) \in \mathbb{N}\,, \] we have obtained a contradiction, given that \(\alpha\) was the supremum of \(\mathbb{N}\). Thus (4.1) is false, meaning that there exists \(n \in \mathbb{N}\) such that \[ n > x \,. \]

Part 2. Suppose \(x \in \mathbb{R}\) with \(x>0\). We can define \[ y:=\frac1x \,. \] By Part 1 there exists \(n \in \mathbb{N}\) such that \[ n > y = \frac{1}{x} \,. \] Using that \(x>0\), we can rearrange the above inequlaity to obtain \[ \frac1n < x \,, \] which is the desired thesis.

There is another formulation of the Archimedean Property which, depending on the situation, might be more useful. This formulation says the following: If \(x,y \in \mathbb{R}\) are such that \[ 0 < x < y \,, \] then there exists \(n \in \mathbb{N}\) such that \[ nx > y \,. \] In other words, if one does \(n\) steps of size \(x\) in the positive numbers direction, then the resulting number \(nx\) will be larger than \(y\). The situation is depicted in Figure 4.2.

Figure 4.2: For \(0<x<y\) there exists \(n \in \mathbb{N}\) such that that \(nx>y\). In the picture \(n=3\).

Theorem 3: Archimedean Property (Alternative formulation)

Let \(x,y \in \mathbb{R}\), with \(0<x<y\). There exists \(n \in \mathbb{N}\) such that \[ nx > y \,. \]

Proof

Suppose by contradiction that there does not exist some \(n \in \mathbb{N}\) such that \[ nx > y \,. \] This means that \[ nx \leq y \,, \quad \forall \, n \in \mathbb{N}\,. \tag{4.3}\] Define the set \[ A := \{ nx \, \colon \,n \in \mathbb{N}\} \,. \] Condition (4.3) is saying that \(A\) is bounded above by \(y\). Morever \(A\) is trivially non-empty. By the Axiom of Completeness there exists \[ \alpha = \sup A \,. \] Since \(\alpha\) is the supremum of \(A\), by definition of supremum and of the set \(A\), we have \[ nx \leq \alpha \,, \quad \forall \, n \in \mathbb{N}\,. \tag{4.4}\] As (4.4) holds for every \(n \in \mathbb{N}\), then it also holds for \((n+1)\), meaning that \[ (n+1) x \leq \alpha \,. \] The above implies \[ nx \leq \alpha - x \,. \] As \(n\) was arbitrary, we conclude that \[ nx \leq \alpha - x \,, \quad \forall \, n \in \mathbb{N}\,. \] The above is saying that \((\alpha - x)\) is an upper bound for \(A\). Since \(\alpha\) is the supremum of \(A\), in particular \(\alpha\) is the smallest upper bound. Thus it must hold \[ \alpha \leq \alpha - x \,. \] The above is equivalent to \[ x \leq 0 \,, \] which contradicts our assumption of \(x>0\). Therefore (4.3) is false, and there exists \(n \in \mathbb{N}\) such that \[ nx > y \,, \] concluding the proof.

4.2 Nested Interval Property

Another consequence of the axiom of completeness is the Nested Interval Property. This is yet another way of saying the same thing: \(\mathbb{R}\) does not have gaps.

Let us look at a construction. Suppose given some closed intervals \[ I_n := [a_n,b_n] = \{ x \in \mathbb{R}\, \colon \,a_n \leq x \leq b_n \} \,, \] where the end points are ordered in the following way: \[ a_1 \leq a_2 \leq \ldots \leq a_n \leq \dots \leq b_n \leq \ldots b_n \leq b_1 \,, \] as shown in Figure 4.3.

Figure 4.3: Nested intervals \(I_n = [a_n,b_n]\).

The intervals \(I_n\) are nested, meaning that \[ I_1 \supset I_2 \supset I_3 \supset \ldots I_n \supset \ldots \] For finite intersections we clearly have \[ \bigcap_{n=1}^k I_n = I_k \,, \] that is, intersecting the first \(k\) intervals yields \(I_k\), the smallest interval in the sequence.

Question 4

Consider the infinite intersection

\[ \bigcap_{n=1}^\infty I_n := \{ x \in \mathbb{R}\, \colon \,x \in I_n \,, \forall \, n \in \mathbb{N}\} \,. \]

What can we say about it? Is it empty? Is it not empty?

The answer is that the infinite intersection is not empty, because \(\mathbb{R}\) was constructed in a way that it does not have gaps.

Theorem 5: Nested Interval Property

For each \(n \in \mathbb{N}\) assume given a closed interval \[ I_n := [a_n,b_n] = \{ x \in \mathbb{R}\, \colon \,a_n \leq x \leq b_n \} \,. \] Suppose that the intervals are nested, that is, \[ I_n \supset I_{n+1}\,, \quad \forall \, n \in \mathbb{N}\,. \] Then \[ \bigcap_{n=1}^\infty I_n \neq \emptyset \,. \tag{4.5}\]

Proof

By definition we have \[ \bigcap_{n=1}^\infty I_n := \{ x \in \mathbb{R}\, \colon \,x \in I_n \,, \forall \, n \in \mathbb{N}\} \,. \] We want to prove (4.5). This means we need to find a real number \(x\) such that \[ x \in I_n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{4.6}\]

Idea of the Proof: Condition (4.6) is saying that it should hold \[ x \geq a_n \,, \quad \forall \, n \in \mathbb{N}\,. \] We might be tempted to choose \(x\) to be any of the \(b_n\). This choice would indeed satisy the above. However (4.6) also implies that \[ x \leq b_n \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \(x\) has to be larger than all the \(a_n\), but not too large. This suggests that \(x\) should be defined as a supremum.

Define the set \[ A:= \{ a_n \, \colon \,n \in \mathbb{N}\} \,. \] The set \(A\) is non-empty and is bounded above by any of the \(b_n\). Therefore there exists \[ x = \sup \ A \,. \] By definition of supremum and definition of the set \(A\), we have \[ a_n \leq x \,, \quad \forall \, n \in \mathbb{N}\,. \] On the other hand, consider an arbitrary number \(b_n\). By construction we have \[ a_i \leq b_n \,, \quad \forall \, i \in \mathbb{N}\,. \] Therefore \(b_n\) is an upper bound for \(A\). Since the supremum is the smallest upper bound, we conclude that \[ x \leq b_n \,. \] The index \(n\) was chosen arbitrarily, and therefore \[ x \leq b_n \,, \quad \forall \, n \in \mathbb{N}\,. \] In total we have \[ a_n \leq x \leq b_n \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \(x\) satisfies (4.6). Therefore (4.5) holds and the proof is concluded.

Important

The assumption that \(I_n\) is closed is crucial in Theorem 5. Without such assumption the thesis of Theorem 5 does not hold in general, as seen in Example 6 below.

Example 6

Consider the open intervals \[ I_n := \left( 0, \frac1n \right) \,. \] These are clearly nested \[ I_n \supset I_{n+1}\,, \quad \forall \, n \in \mathbb{N}\,. \] For this choice of \(I_n\) we have \[ \bigcap_{n=1}^\infty I_n = \emptyset \,. \tag{4.7}\] Indeed, suppose by contradiction that the intersection is non-empty. Then there exists \(x \in \mathbb{N}\) such that \[ x \in I_n \,, \quad \forall \, n \in \mathbb{N}\,. \] By definition of \(I_n\) the above reads \[ 0 < x < \frac1n \,, \quad \forall \, n \in \mathbb{N}\,. \tag{4.8}\] Since \(x>0\), by the Archimedean Property in Theorem 2 Point 2, there exists \(n_0 \in \mathbb{N}\) such that \[ 0 < \frac{1}{n_0} < x \,. \] The above contradicts (4.8). Therefore (4.7) holds.

4.3 Revisiting Sup and inf

We now investigate some of the properties of supremum and infimum in \(\mathbb{R}\). The first property is an alternative characterization of the supremum, which we will often use. A sketch of such characterization is in Figure 4.4 below.

Proposition 7: Characterization of Supremum

Let \(A \subseteq \mathbb{R}\) be a non-empty set. Suppose that \(s \in \mathbb{R}\) is an upper bound for \(A\). They are equivalent:

\(s = \sup A\)
For every \(\varepsilon> 0\) there exists \(x \in A\) such that \[ s - \varepsilon< x \,. \]

Figure 4.4: Let \(s = \sup A\). Then for every \(\varepsilon>0\) there exist \(x \in A\) such that \(s-\varepsilon<x\).

Proof: Proof of Proposition 7

Step 1. Assume that \[ s = \sup A \,. \] Let \(\varepsilon>0\) be arbitrary. We clearly have that \[ s - \varepsilon< s \,. \tag{4.9}\] Therefore \((s-\varepsilon)\) cannot be an upper bound of \(A\). Indeed, if by contradiction \((s-\varepsilon)\) was an upper bound, then we would have \[ s \leq (s - \varepsilon)\,, \] since \(s\) is the smallest upper bound. The above contradicts (4.9), and therefore \((s-\varepsilon)\) is not an upper bound for \(A\). Hence there exists some \(x \in A\) such that \[ s - \varepsilon< x \, , \] concluding.

Step 2. Assume that Point 2 in the statement of Proposition 7 holds. By assumption we have that \(s\) is an upper bound for \(A\). Suppose by contradiction that \[ s \neq \sup A \,. \] This is equivalent to the statement \[ s \, \mbox{ is not the smallest upper bound of } \, A. \tag{4.10}\] Hence there exists an upper bound \(b\) of \(A\) such that \[ b < s \,. \] Let \[ \varepsilon:= s - b \,. \] By assumption there exists \(x \in A\) such that \[ s - \varepsilon< x \,. \] Substituting the definition of \(\varepsilon\) we get \[ s - s + b < x \quad \implies \quad b < x \,. \] Since \(b\) is an upper bound for \(A\) and \(x \in A\), the above is a contradiction. Therefore (4.10) is false, and \(s\) is the smallest upper bound of \(A\). Thus \(s = \sup A\).

The analogue of Proposition 7 is as follows. The proof is left as an exercise.

Proposition 8: Characterization of Infimum

Let \(A \subseteq \mathbb{R}\) be a non-empty set. Suppose that \(i \in \mathbb{R}\) is a lower bound for \(A\). They are equivalent:

\(i = \inf A\)
For every \(\varepsilon\in \mathbb{R}\), with \(\varepsilon> 0\), there exists \(x \in A\) such that \[ x < i + \varepsilon\,. \]

A sketch of the characterization in Proposition 8 can be found in Figure 4.5 below.

Figure 4.5: Let \(i = \inf A\). Then for every \(\varepsilon>0\) there exist \(x \in A\) such that \(x<i+\varepsilon\).

With the above characterizations of supremum and infimum, it is now easier to prove that some candidate number is the supremum or infimum of some set. As an example, let us characterize supremum and infimum of an open interval of \(\mathbb{R}\).

Proposition 9

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= (a,b) = \{ x \in \mathbb{R}\, \colon \,a < x < b \}\,. \] Then \[ \inf A = a \,, \quad \sup A = b \,. \]

Proof

We will only prove that \[ \inf A = a \,, \] since the proof of \[ \sup A = b \] is similar.

By definition of \(A\), we have that \[ a < x \,, \quad \forall \, x \in A \,. \] The above says that \(a\) is a lower bound for \(A\).

Claim. \(a\) is the largest lower bound of \(A\).

Proof of Claim. Let \(L\) be a lower bound for \(A\), that is, \[ L \leq x \,, \quad \forall \, x \in A \,. \] We have to prove that \[ L \leq a \,. \tag{4.11}\] Indeed suppose by contradiction that (4.11) does not hold, namely that \[ a < L \,. \] Since \(a<b\) we have \[ a = \frac{2a}{2} < \frac{a+b}{2} < \frac{2b}{b} = b \,, \] showing that the midpoint between \(a\) and \(b\) satisfies \[ \frac{a+b}{2} \in A \,. \] Since \(L\) is a lower bound for \(A\) we have \[ L \leq \frac{a+b}{2} < b \,. \tag{4.12}\] Consider the midpoint \[ M := \frac{a + L}{2} \,. \] We have that \[ M \in A \,. \]

Indeed, recalling that \(a<L\), we have \[ a = \frac{2a}{2} < \frac{a+L}{2} = M \,. \] Moreover by (4.12) we have \(L<b\). Thus \[ M = \frac{a+L}{2} \leq \frac{a+b}{2} < b \,. \] This shows \(M \in A\).

Moreover \[ M < L \,. \]

This is because \(a<L\), and therefore \[ M = \frac{a + L}{2} < \frac{2L}{2} = L \,. \]

This is a contradiction, since by assumption \(L\) is a lower bound for \(A\), and thus we should have \[ L \leq M \,. \] Therefore (4.11) holds, showing that \(a\) is the largest lower bound of \(A\). Thus \(a = \inf A\).

As a corollary of the above we have that the maximum and minimum of an open interval do not exist.

Corollary 10

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= (a,b) = \{ x \in \mathbb{R}\, \colon \,a < x < b \}\,. \] Then \(\min A\) and \(\max A\) do not exist.

Proof

Suppose by contradiction that \(\min A\) exists. We have shown that if the minimum of a set exists, then it must be \[ \min A = \inf A \,. \] Since \[ \inf A = a\,, \] by Proposition 9, we would obtain that \[ \min A = a \,. \] By definition \(\min A \in A\), so that \(a \in A\). This is contradiction. Then \(\min A\) does not exist.

The proof that \(\max A\) does not exist is similar, and is left as an exercise.

We can also consider intervals for which one or both of the sides are closed.

Corollary 11

Let \(a, b \in \mathbb{R}\) with \(a<b\). Let \[ A:= [a,b) = \{ x \in \mathbb{R}\, \colon \,a \leq x < b \}\,. \] Then \[ \min A = \inf A = a \,, \quad \sup A = b \,, \] \(\max A\) does not exist.

The proof is very similar to the ones above, and is left to the reader for exercise. Let us now compute supremum and infimum of a set which is not an interval.

Proposition 12

Define the set \[ A := \left\{ \frac1n \, \colon \,n \in \mathbb{N}\right\} \,. \] Then \[ \inf A = 0 \,, \quad \sup A = \max A = 1 \,. \]

Proof

Part 1. We have \[ \frac1n \leq 1 \,, \quad \forall \, n \in \mathbb{N}\,. \] Therefore \(1\) is an upper bound for \(A\). Let us prove it is the least upper bound: let \(b\) be an upper bound for \(A\). Since \(1 \in A\) and \(b\) is an upper bound, we have \(1 \leq b\). Hence \(1\) is the least upper bound, and \[ \sup A = \max A = 1 \,. \]

Part 2. We have \[ \frac1n > 0 \,, \quad \forall \, n \in \mathbb{N}\,, \] showing that \(0\) is a lower bound for \(A\). Suppose by contradiction that \(0\) is not the infimum. Therefore \(0\) is not the largest lower bound. Then there exists \(\varepsilon\in \mathbb{R}\) such that:

\(\varepsilon\) is a lower bound for \(A\), that is, \[ \varepsilon\leq \frac1n \,, \quad \forall \, n \in \mathbb{N}\,, \tag{4.13}\]
\(\varepsilon\) is larger than \(0\): \[ 0 < \varepsilon\,. \]

As \(\varepsilon>0\), by the Archimedean Property there exists \(n_0 \in \mathbb{N}\) such that \[ 0 < \frac{1}{n_0} < \varepsilon\,. \] This contradicts (4.13). Thus \(0\) is the largest lower bound of \(A\), that is, \(0 = \inf A\).

Part 3. We have that \(\min A\) does not exist. Indeed suppose by contradiction that \(\min A\) exists. Then \[ \min A = \inf A \,. \] As \(\inf A = 0\) by Part 2, we conclude \(\min A = 0\). As \(\min A \in A\), we obtain \(0 \in A\), which is a contradiction.

4.4 Density of \(\mathbb{Q}\) in \(\mathbb{R}\)

A set \(S\) is dense in \(\mathbb{R}\) if the elements of \(S\) are arbitrarily close to the elements of \(\mathbb{R}\).

Definition 13: Dense set

Let \(S \subseteq \mathbb{R}\). We say that \(S\) is dense in \(\mathbb{R}\) if for every \(x \in \mathbb{R}\) and \(\varepsilon\in \mathbb{R}\) with \(\varepsilon>0\), there exist \(q \in \mathbb{Q}\) such that \[ |x - q| < \varepsilon\,. \]

In other words, the above definition is saying that \(S\) and \(\mathbb{R}\) are tightly knitted together. An equivalent definition of dense set is given below.

Remark 14

Let \(S \subseteq \mathbb{R}\). They are equivalent:

\(S\) is dense in \(\mathbb{R}\).
For every pair of numbers \(x,y \in \mathbb{R}\) with \(x<y\), there exists \(s \in S\) such that \[ x < s < y \,. \]

We now prove that \(\mathbb{Q}\) is dense in \(\mathbb{R}\).

Theorem 15: Density of \(\mathbb{Q}\) in \(\mathbb{R}\)

Let \(x,y \in \mathbb{R}\), with \(x<y\). There exists \(q \in \mathbb{Q}\) such that \[ x < q < y \,. \]

Figure 4.6: Let \(n \in \mathbb{N}\) be such that \(1/n< y-x\). Then take \(m\) so that \(m/n \in (x,y)\).

Proof

We need to find \(q \in \mathbb{Q}\) such that \[ x < q < y \,. \tag{4.14}\] By definition of \(\mathbb{Q}\), we have that \(q\) has to be \(q=m/n\) for \(m \in \mathbb{Z}\) and \(n \in \mathbb{N}\). Therefore (4.14) is equivalent to finding \(m \in \mathbb{Z}\) and \(n \in \mathbb{N}\) such that \[ x < \frac{m}{n} < y \,. \tag{4.15}\] The idea is to proceed as in Figure 4.6: We take \(n\) such that \(1/n\) is small enough so that we can make \(m\) jumps of size \(1/n\) and end up between \(x\) and \(y\).
To this end, let \(n \in \mathbb{N}\) be such that \[ \frac{1}{n} < y - x \,. \tag{4.16}\] Such \(n\) exists thanks to the Archimedean Property in Theorem 2 Point 2. Inequality (4.15) is equivalent to \[ nx < m < ny \,. \] Take \(m \in \mathbb{Z}\) such that \[ m-1 \leq nx < m \,. \tag{4.17}\]

Why does such \(m\) exist? Because by Archimedean Property in Theorem 2 Point 1, there exists \(m' \in \mathbb{Z}\) such that \[ m' > nx \] We can then choose \(m\) to be the smallest element in \(\mathbb{Z}\) such that \(m > nx\). Such \(m\) satisfies (4.17).

The second inequality in (4.17) implies \[ x < \frac{m}{n} \,, \] which is the first inequality in (4.15). Now note that (4.16) is equivalent to \(x < y - 1/n\). We can use the latter and the first inequality in (4.17) to estimate \[\begin{align*} m & \leq 1 + nx \\ & < 1 + n \left( y - \frac{1}{n} \right ) \\ & = ny \,, \end{align*}\] which yields \[ \frac{m}{n} < y \,. \] Therefore the second inequality in (4.15) is proven, concluding the proof.

We have constructed the real numbers \(\mathbb{R}\) so that they would fill the gaps of \(\mathbb{Q}\). Formally, these gaps are the numbers in \(\mathbb{R}\smallsetminus \mathbb{Q}\). Let us give a name to this set.

Definition 16: Irrational numbers

The set of irrational numbers in \(\mathbb{R}\) is \[ \mathcal{I} := \mathbb{R}\smallsetminus \mathbb{Q}\,. \]

Question 17

How many gaps does \(\mathbb{Q}\) have? In other words, how many irrational numbers are out there?

The answer is quite surprising, and is a corollary of the density result of Theorem 15: The irrational numbers are dense in \(\mathbb{R}\).

Corollary 18

Let \(x,y \in \mathbb{R}\), with \(x<y\). There exists \(t \in \mathcal{I}\) such that \[ x < t < y \,. \]

Proof

Consider \[ \widetilde{x}:=x - \sqrt{2}\,, \quad \widetilde{y} :=y - \sqrt{2} \,. \] Since \(x<y\), we have \[ \widetilde{x} < \widetilde{y} \,. \] By Theorem 15 there exists \(q \in \mathbb{Q}\) such that \[ \widetilde{x} < q < \widetilde{y} \,. \] Adding \(\sqrt{2}\) to the above inequalities we get \[ x < t < y \,, \quad t := q + \sqrt{2} \,. \tag{4.18}\] We claim that \(t \in \mathcal{I}\). Indeed, suppose by contradiction \(t \in \mathbb{Q}\). Then \[ \sqrt{2} = t - q \in \mathbb{Q}\,, \] since \(t,q \in \mathbb{Q}\), and \(\mathbb{Q}\) is closed under summation. Since \(\sqrt{2} \in \mathcal{I}\), we obtain a contradiction. Thus \(t \in \mathcal{I}\) and (4.18) is our thesis.

4.5 Existence of \(k\)-th Roots

We have started our discussion by proving that \[ \sqrt{2} \notin \mathbb{Q}\,. \tag{4.19}\] We have shown that (4.19) implies that the set \[ A := \{ q \in \mathbb{Q}\, \colon \,q^2 < 2 \} \] does not have a supremum in \(\mathbb{Q}\). We then introduced the Real Numbers \(\mathbb{R}\) so that each non-empty and bounded above set would have a supremum. As the set \(A\) is non-empty and bounded above, there exists \(\alpha \in \mathbb{R}\) such that \[ \alpha = \sup A \,. \] We are going to prove that \[ \alpha^2 = 2 \,, \] which means that in \(\mathbb{R}\) we can take the square root of \(2\). More in general, with the same fudamental idea, we can prove that for each \(x \in \mathbb{R}\) with \(x \geq 0\) and \(k \in \mathbb{N}\), there exists \(\alpha \in \mathbb{R}\) such that \[ \alpha^k = x \,. \]

Theorem 19: Existence of \(k\)-th roots

Let \(x \in \mathbb{R}\) with \(x \geq 0\) and \(k \in \mathbb{N}\). There exists a unique \(\alpha \in \mathbb{R}\) such that \[ \alpha^k = x \,. \]

The proof of Theorem 19 rests on similar ideas to the ones used to prove that \(\mathbb{Q}\) does not have the cut property.

Proof: Proof of Theorem 19

Part 1: Uniqueness.

Suppose \(\alpha_1,\alpha_2 \in \mathbb{R}\) are such that \[ \alpha_1^k = \alpha_2^k = x \,. \] If \(\alpha_1 \neq \alpha_2\), then \[ \alpha_1^k \neq \alpha_2^k \,, \] obtaining a contradiction. Therefore \(\alpha_1 = \alpha_2\).

Part 2: Existence.

Let \(x \in \mathbb{R}\) with \(x \geq 0\). If \(x = 0\) there is nothing to prove, as \[ 0^k = 0 \,, \] so that \(\alpha = 0\). Therefore we can assume \(x>0\). Define the subset of \(\mathbb{R}\) \[ A := \{ t \in \mathbb{R}\, \colon \,t^k < x \} \,. \] Clearly \(A\) is non-empty and bounded above.

An upper bound is given, for example, by \(b := x + 1\). Indeed, since we are assuming \(x \geq 0\), then \(x+1 \geq 1\). In particular we have \[ (x+1)^k \geq x + 1 \,. \] Let \(t \in A\). Then \[ t^k < x < x + 1 < (x+1)^k , \] showing that \(t< x+1\).

By the Axiom of Completeness of \(\mathbb{R}\), there exists \(\alpha \in \mathbb{R}\) such that \[ \alpha = \sup A \,. \] We claim that \[ \alpha^k = x \,. \tag{4.20}\] Suppose by contradiction that (4.20) does not hold. We will need the formula: For all \(a,b \in \mathbb{R}\) it holds \[ a^k-b^k = (a - b)( a^{k-1} + a^{k-2} b + a^{k-3}b^k + \ldots + a b^{k-2} + b^{k-1} ) \,. \tag{4.21}\] Formula (4.21) can be easily proven by induction on \(k\). Since we are assuming that (4.20) does not hold, we have two cases:

\(\alpha^k < x\): We know that \(\alpha\) is the supremum of \(A\). We would like to violate this, by finding a number \(L\) which is larger than \(\alpha\), but still belongs to \(A\). This means \(L\) has to satisfy \[ \alpha < L \,, \quad L^k < x \,. \] We look for \(L\) of the form \[ L_n := \alpha + \frac{1}{n} \] for \(n \in \mathbb{N}\) to be chosen later. Clearly \[ \alpha < L_n \,, \tag{4.22}\] for all \(n \in \mathbb{N}\). We now search for \(n_0 \in \mathbb{N}\) such that \[ L_{n_0}^k < x \,. \] Using formula (4.21) with \(a = \alpha\) and \(b = L_{n_0}\) we obtain \[ L_{n_0}^k - \alpha^k = \frac{1}{n_0} \left( L_{n_0}^{k-1} + L_{n_0}^{k-2} \alpha + \ldots + L_{n_0}\alpha^{k-2} + \alpha^{k-1} \right) \,. \tag{4.23}\] Now notice that (4.22) implies \[ \alpha^j < L_{n_0}^j \] for all \(j \in \mathbb{N}\). Using this estimate on all the terms \(\alpha^j\) appearing in the RHS of (4.23) we obtain \[\begin{align*} L_{n_0}^k - \alpha^k & = \frac{1}{n_0} \left( L_{n_0}^{k-1} + L_{n_0}^{k-2} \alpha + \ldots + L_{n_0}\alpha^{k-2} + \alpha^{k-1} \right) \\ & < \frac{1}{n_0} \left( L_{n_0}^{k-1} + L_{n_0}^{k-2} L_{n_0} + \ldots + L_{n_0} L_{n_0}^{k-2} + L_{n_0}^{k-1} \right) \\ & = \frac{k}{n_0} \, L_{n_0}^{k-1} \end{align*}\] Rearranging the above we get \[ L_{n_0}^k < \frac{k}{n_0} \, L_{n_0}^{k-1} + \alpha^k \,. \tag{4.24}\] Now note that \[ L_{n_0} = \alpha + \frac{1}{n_0} < \alpha + 1 \,. \] Therefore \[ L_{n_0}^{k-1} < (\alpha + 1)^{k-1} \,, \] and from (4.24) we obtain \[ L_{n_0}^k < \frac{k}{n_0} \, (\alpha + 1)^{k-1} + \alpha^k \,. \] We wanted to find \(n_0 \in \mathbb{N}\) so that \(L_{n_0}^k < x\). Therefore we impose \[ \frac{k}{n_0} \, (\alpha + 1)^{k-1} + \alpha^k < x \,, \] and find that the above is satisfied for \[ n_0 > \frac{ k (\alpha + 1)^{k-1} }{ x - \alpha^k } \,. \tag{4.25}\] Notice that the RHS in (4.25) is a positive real number, since \(\alpha^k < x\) by assumption. Therefore, by the Archimedean Property of Theorem 2 Point 1, there exists \(n_0 \in \mathbb{N}\) satisfying (4.25).
We have therefore shown the existence of \(n_0 \in \mathbb{N}\) such that \[ \alpha < L_{n_0} \,, \quad L_{n_0}^k < x \,. \] The above says that \(L_{n_0} \in A\) and that \[ L_{n_0} > \alpha = \sup A \,, \] which is a contradiction, as \(\sup A\) is an upper bound for \(A\).
\(\alpha^k > x\): We know that \(\alpha\) is the supremum of \(A\). We would like to find a contradiction, by finding an upper bound \(L\) for \(A\) which is smaller than \(\alpha\). This means \(L\) has to satisfy \[ L < \alpha \,, \quad L^k > x \,. \]

Such \(L\) is an upper bound for \(A\): If \(t \in A\) then \[ t^k < x < L^k \quad \implies \quad t < L \,. \]

We therefore look for \(L\) of the form \[ L_n := \alpha - \frac{1}{n} \] for \(n \in \mathbb{N}\) to be chosen later. In this way \[ L_n < \alpha \,, \tag{4.26}\] for all \(n \in \mathbb{N}\). We now search for \(n_0 \in \mathbb{N}\) such that \[ L_{n_0}^k > x \,. \] Using formula (4.21) with \(a = L_{n_0}\) and \(b = \alpha\) we obtain \[ \alpha^k - L_{n_0}^k = \frac{1}{n_0} \left( \alpha^{k-1} + \alpha^{k-2} L_{n_0} + \ldots + \alpha L_{n_0}^{k-2} + L_{n_0}^{k-1} \right) \,. \tag{4.27}\] Now notice that (4.26) implies \[ L_{n_0}^j < \alpha^j \] for all \(j \in \mathbb{N}\). Using this estimate on all the terms \(L_{n_0}^j\) appearing in the RHS of (4.27) we obtain \[\begin{align*} \alpha^k - L_{n_0}^k & = \frac{1}{n_0} \left( \alpha^{k-1} + \alpha^{k-2} L_{n_0} + \ldots + \alpha L_{n_0}^{k-2} + L_{n_0}^{k-1} \right) \\ & < \frac{1}{n_0} \left( \alpha^{k-1} + \alpha^{k-2} \alpha + \ldots + \alpha \alpha^{k-2} + \alpha^{k-1} \right) \\ & = \frac{k}{n_0} \, \alpha^{k-1} \end{align*}\] Rearranging the above we get \[ L_{n_0}^k > \alpha^k - \frac{k}{n_0} \, \alpha^{k-1} \,. \] We wanted to find \(n_0 \in \mathbb{N}\) so that \(L_{n_0}^k > x\). Therefore we impose \[ \alpha^k - \frac{k}{n_0} \, \alpha^{k-1} > x \,, \] and find that the above is satisfied for \[ n_0 > \frac{ k \alpha^{k-1} }{ \alpha^k - x} \,. \tag{4.28}\] Notice that the RHS in (4.28) is a positive real number, since \(\alpha^k > x\) by assumption. Therefore, by the Archimedean Property of Theorem 2 Point 1, there exists \(n_0 \in \mathbb{N}\) satisfying (4.28).
We have therefore shown the existence of \(n_0 \in \mathbb{N}\) such that \[ L_{n_0} < \alpha \,, \quad L_{n_0}^k > x \,. \] Condition \(L_{n_0}^k > x\) says that \(L_{n_0}\) is an upper bound for \(A\). At the same time it holds \[ L_{n_0} < \alpha = \sup A \,, \] which is a contradiction, as \(\sup A\) is the smallest upper bound for \(A\).

Therefore, both cases \(\alpha^k > x\) and \(\alpha^k < x\) lead to a contradiction. Hence \(\alpha^k =x\), concluding.

Definition 20: \(k\)-th root of a number

Let \(x \in \mathbb{R}\) with \(x \geq 0\) and \(k \in \mathbb{N}\). The real number \(\alpha\) such that \[ \alpha^k = x \] is called the \(k\)-th root of \(x\), and is denoted by \[ \sqrt[k]{x} := \alpha \,. \]

4.6 Cardinality

We have proven that the sets or rational numbers \(\mathbb{Q}\) and irrational numbers \(\mathcal{I}\) are both dense in \(\mathbb{R}\), with \[ \mathbb{R}= \mathbb{Q}\cup \mathcal{I} \,. \] From this result we might think that \(\mathbb{R}\) is obtained by mixing \(\mathbb{Q}\) and \(\mathcal{I}\) in equal proportions. This is however false. We will see that \(\mathbb{R}\) has much more elements than \(\mathbb{Q}\). Therefore also the set of irrational numbers \(\mathcal{I}\) is much larger than \(\mathbb{Q}\).

To make the above discussion precise, we need to define what we mean by size of a set. For this, we need the concept of bijective function.

Definition 21: Bijective function

Let \(X,Y\) be sets and \(f \colon X \to Y\) be a function. We say that:

\(f\) is injective if it holds: \[ f(x)=f(y) \quad \implies \quad x = y \,. \]
\(f\) is surjective if it holds: \[ \forall y \in Y \,, \,\, \exists \, x \in X \, \text{ s.t. } \, f(x) = y \,. \]
\(f\) is bijective if it is both injective and surjective.

In other words: A function \(f \colon X \to Y\) is

injective if any two different elements in \(X\) are mapped into two different elements in \(Y\).
surjective if every element in \(Y\) has at least one element in \(X\) associated via \(f\).
bijective if to each element in \(X\) we associate one and only one element in \(Y\) via \(f\).

Example 22: Injectivity

Consider the sets \[ X = \{ 1, 2, 3 \} \,, \quad Y = \{ a,b,c,d,e\} \,. \]

The function \(f \colon X \to Y\) defined by \[ f(1)=c \,, \quad f(2) = a \,, \quad f(3) = e \,, \] is injective.
The function \(g \colon X \to Y\) defined by \[ g(1)=c \,, \quad g(2) = a \,, \quad g(3) = c \,, \] is not injective, since \[ g(1) = g(3) = c \,, \quad 1 \neq 3 \,. \]
The function \(h \colon \mathbb{R}\to \mathbb{R}\) defined by \[ h(x) = x^2 \] is not injective, since \[ h(1) = h(-1) = 1 \,, \quad 1 \neq - 1 \,. \]
The function \(l \colon \mathbb{R}\to \mathbb{R}\) defined by \[ l(x) = 2x \] is injective, since \[ l(x) = l(y) \quad \implies \quad 2x = 2y \quad \implies \quad x = y \,. \]

Example 23: Surjectivity

Consider the sets \[ X = \{ 1, 2, 3 ,4\} \,, \quad Y = \{ a,b,c\} \,. \]

The function \(f \colon X \to Y\) defined by \[ f(1)=c \,, \quad f(2) = a \,, \quad f(3) = a \,, \quad f(4) = b \,, \] is surjective.
The function \(g \colon X \to Y\) defined by \[ g(1)=a \,, \quad g(2) = a \,, \quad g(3) = c \,, \quad g(4) = a \,, \] is not surjective, since there is no element \(x \in X\) such that \[ g(x) = b \,. \]
The function \(h \colon \mathbb{R}\to \mathbb{R}\) defined by \[ h(x) = x^2 \] is not surjective, since there is no \(x \in \mathbb{R}\) such that \[ h(x) = x^2 = -1 \,. \]
The function \(l \colon \mathbb{R}\to [0 , \infty)\) defined by \[ l(x) = x^2 \] is surjective, since for every \(y \geq 0\) there exists \(x \in \mathbb{R}\) such that \[ l(x)= x^2 = y\,. \] This is true by Theorem 19.

Example 24: Bijectivity

Let \(X = \{ 1, 2, 3 \}, Y = \{ a,b,c\}\). The function \(f \colon X \to Y\) defined by \[ f(1)=c \,, \quad f(2) = a \,, \quad f(3) = b \,, \] is bijective, since it is both injective and surjective.
Let \(X = \{ 1, 2, 3 \}, Y = \{ a,b\}\). The function \(g \colon X \to Y\) defined by \[ g(1)=a \,, \quad g(2) = b \,, \quad g(3) = b \,, \] is not bijective, since it is not injective: we have \[ g(2) = g(3) = b \, \quad 2 \neq 3 \,. \]
Let \(X = \{ 1, 2 \}, Y = \{ a,b,c\}\). The function \(h \colon \mathbb{R}\to \mathbb{R}\) defined by \[ h(1) = a \,, \quad h(2) = c \,, \] is not bijective, since it is not surjective: there is no \(x \in X\) such that \[ h(x) = b \,. \]
Let \(X = \{ 1, 2 , 3\}, Y = \{ a,b,c\}\). The function \(l \colon \mathbb{R}\to [0 , \infty)\) defined by \[ l(1) = a \,, \quad l(2) = a \,, \quad l(3) = b \,, \] is not bijective, as it is neither injective nor surjective.

We are ready to define the size of a set.

Definition 25: Cardinality, finite, countable, uncountable

Let \(X\) be a set. The cardinality of \(X\) is the number of elements in \(X\). We denote the cardinality of \(X\) by \[ |X| := \# \, \mbox{ of elements in } \, X \,. \] Further, we say that:

\(X\) is finite if there exists a natural number \(n \in \mathbb{N}\) and a bijection \[ f \colon X \to \{1,2,\ldots, n \} \,. \] In particular \[ |X| = n \,. \]
\(X\) is countable if there exists a bijection \[ f \colon X \to \mathbb{N}\,. \] In this case we denote the cardinality of \(X\) by \[ |X| = |\mathbb{N}| \,. \]
\(X\) is uncountable if \(X\) is neither finite, nor countable.

In other words: A set \(X\) is

finite, if \(X\) can be listed as \[ X = \{ x_1 , \ldots, x_n \} \] for some \(n \in \mathbb{N}\).
countable, if \(X\) can be listed as \[ X = \{ x_n \, \colon \,n \in \mathbb{N}\} \,. \]
uncountable, if \(X\) cannot be listed.

Question 26

Is there an intermediate cardinality between finite and countable?

The answer is no, as shown in the next proposition.

Proposition 27

Let \(X\) be a countable set and \(A \subseteq X\). Then either \(A\) is finite or countable.

Proof

If \(A\) is finite we are done. Therefore suppose \(A\) is not finite. Since \(X\) is countable there exists a bijection \(f \colon \mathbb{N}\to X\). Let \(n_1 \in \mathbb{N}\) be such that \[ n_1 = \min \{ n \in \mathbb{N}\, \text{ s.t. } \, f(n) \in A \} \,. \] Note that \(n_1\) exists since \(f\) is surjective. Define \[ g(1) := f(n_1) \,. \] Now let \[ n_2 = \min \{ n \in \mathbb{N}\, \text{ s.t. } \, n > n_1 \,, \,\, f(n) \in A \} \,. \] Notice that \(n_2\) exists, since \(f\) is surjective and \(A\) is not finite. Set \[ g(2):=f(n_2) \,. \] Iterating, we define \[ n_k = \min \{ n \in \mathbb{N}\, \text{ s.t. } \, n>n_{k-1} \,, \,\, f(n) \in A \} \] and \[ g(k) := f(n_k) \,. \] In this way we have defined a function \(g \colon \mathbb{N}\to A\). We have:

\(g\) is injective: This is because \(g\) was defined thourgh \(f\), and \(f\) is injective.
\(g\) is surjective: If \(x \in A\), by surjectivity of \(f\) there exists \(\widetilde{n} \in \mathbb{N}\) such that \(f(\widetilde{n})=x\). Therefore \[ \widetilde{n} \in \{ n \in \mathbb{N}\, \text{ s.t. } \, f(n) \in A \} \,, \] so that \(n_k= \widetilde{n}\), for some \(k \geq \widetilde{n}-1\).

Hence \(g\) is bijective, showing that \(A\) is countable.

Example 28

Let \(X = \{a,b,c\}\) and \(Y=\{ 1,2,3\}\). The function \(f \colon X \to Y\) defined by \[ f(1)=a \,, \quad f(2) = b \,, \quad f(3) = c \,, \] is bijective. Therefore \(X\) is finite, with \(|X| = 3\).
Let \(X = \mathbb{N}\). The function \(f \colon X \to \mathbb{N}\) defined by \[ f(n):=n \,, \] is bijective. Therefore \(X = \mathbb{N}\) is countable.
Let \(X\) be the set of even numbers \[ X = \{ 2n \, \colon \,n \in \mathbb{N}\} \,. \] Define the map \(f \colon X \to \mathbb{N}\) by \[ f(m):= \frac{m}{2} \,. \] We have that:
- \(f\) is injective: \(f(m)=f(k)\) implies that \(m/2=k/2\) which implies \(m=k\).
- \(f\) is surjective: If \(n \in \mathbb{N}\), then \(f(2n)=n\).
Therefore \(f\) is bijective, showing that \(X\) is countable and \(|X|=|\mathbb{N}|\).
Let \(X = \mathbb{Z}\) the set of integers. Define \(f \colon \mathbb{N}\to \mathbb{Z}\) by \[ f(n):= \begin{cases} \dfrac{n}{2} & \,\, \mbox{ if } \, n \, \mbox{ even } \\ -\dfrac{n+1}{2} & \,\, \mbox{ if } \, n \, \mbox{ odd } \end{cases} \] For example \[\begin{align*} f(0) & = 0 \,, \quad f(1) = -1 \,, \quad f(2) = 1 \,, \quad f(3) = -2 \,, \\ f(4) & = 2 \,, \quad f(5) = -3 \,, \quad f(6) = 3 \,, \quad f(7) = -4 \,. \end{align*}\] We have:
- \(f\) is injective: Indeed, suppose that \(m \neq n\). If \(n\) and \(m\) are both even or both odd we have, respectively \[\begin{align*} f(m) = \frac{m}{2} & \neq \frac{n}{2} = f(n) \\ f(m) = -\frac{m+1}{2} & \neq - \frac{n+1}{2} = f(n) \,. \end{align*}\] If instead \(m\) is even and \(n\) is odd, we get \[ f(m) = \frac{m}{2} \neq - \frac{n+1}{2} = f(n) \,, \] since the LHS is positive and the RHS is negative. The case when \(m\) is odd and \(n\) even is similar.
- \(f\) is surjective: Let \(z \in \mathbb{Z}\). If \(z \geq 0\), then \(m:=2z\) belongs to \(\mathbb{N}\), is even, and \[ f(m) = f(2z) = z \,. \] If instead \(z < 0\), then \(m := -2z -1\) belongs to \(\mathbb{N}\), is odd, and \[ f(m) = f(-2z-1) = z \,. \]
Therefore \(f\) is bijective, showing that \(\mathbb{Z}\) is countable and \[ |\mathbb{Z}| =| \mathbb{N}| \,. \]

We have seen that the sets \(\mathbb{N}\) and \(\mathbb{Z}\) are countable. What about \(\mathbb{Q}\)? To study this case, we need the following result.

Proposition 29

Let the set \(A_n\) be countable for all \(n \in \mathbb{N}\). Define \[ A = \bigcup_{n \in \mathbb{N}} \, A_n \, . \] Then \(A\) is countable.

Proof

Since each \(A_i\) is countable, we can list their elements as \[ A_i = \{ a_k^i \, \colon \,k \in \mathbb{N}\} = \{ a_1^i , \, a_2^i , \, a_3^i , \, a_4^i , \, \ldots \} \,. \] The proof that \(A\) is countable is based on a diagonal argument by Georg Cantor, see Wikipedia page. The idea of th is that we can list the elements of the sets \(A_i\) in an infinite square: In the first row we put the elements of \(A_1\), in the second row the elements of \(A_2\), and so on. Therefore the \(i\)-th row contains the elements of \(A_i\). This procedure is illustrated in Figure 4.7. Therefore this infinite square contains all the elements of \(A\). We then list all the elements of the square by looking at the diagonals, as shown in Figure 4.7. This procedure defines a function \(f \colon \mathbb{N}\to A\). For example the first few terms of \(f\) are \[ f(1) = a_1^1 \,, \quad f(2) = a_2^1 \,, \quad f(3) = a_1^2 \,, \quad f(4) = a_1^3 \,. \] Since \(f\) is injective and surjective, we have that \(A\) is countable.

Figure 4.7: The i-th row contains all the elements \(a_1^i,a_2^i,a_3^i, \ldots\) of the countable set \(A_i\). We define the function \(f \colon \mathbb{N}\to A\) by going throgh the square diagonally.

Theorem 30: \(\mathbb{Q}\) is countable

The set of rational numbers \(\mathbb{Q}\) is countable.

Proof

For \(i \in \mathbb{N}\) define the sets \[ L_i := \left\{ \frac{m}{i} \, \colon \,m \in \mathbb{Z}\right\} \,. \] We have that \(f \colon L_i \to \mathbb{Z}\) defined by \[ f \left(\frac{m}{i}\right) := m \] is a bijection. As \(\mathbb{Z}\) is countable, we deduce that \(L_i\) is countable. Therefore the set \(L\) defined by \[ L:= \bigcup_{i \in \mathbb{N}} \, L_i \] is countable by Proposition 29. Clearly we have \[ \mathbb{Q}\subseteq L \,. \] Since \(\mathbb{Q}\) is not finite, by Proposition 27 we conclude that \(\mathbb{Q}\) is countable.

We have proven that the sets \[ \mathbb{N}\,, \quad \mathbb{Z}\,, \quad \mathbb{Q}\,, \] are all countable. What about \(\mathbb{R}\)?

Theorem 31: \(\mathbb{R}\) is uncountable

The set of Real Numbers \(\mathbb{R}\) is uncountable.

Proof

Suppose by contradiction \(\mathbb{R}\) is countable. Then there exists a bijection \(f \colon \mathbb{N}\to \mathbb{R}\), meaning that we can list the elements of \(\mathbb{R}\) as \[ \mathbb{R}= \{ x_1 , x_2 , x_3 , x_4 , x_5 , \ldots \} \,. \] Let \(I_1\) be a closed interval such that \[ x_1 \notin I_1 \,. \] Let \(I_2\) be another closed interval, contained in \(I_1\), and such that \(x_2 \notin I_2\). Such interval exists, because \(I_1\) contains two disjoint closed intervals: hence \(x_2\) can be at most in one of these two intervals. To summarize, we have \[ x_1 \notin I_1 \,, \quad x_2 \notin I_2 \,, \quad I_2 \subseteq I_1 \,. \] We can iterate this procedure, and construct a sequence of nested intervals \(I_n\) such that \[ I_{n+1} \subseteq I_n \,, \quad x_n \notin I_n \,, \] for all \(n \in \mathbb{N}\), see Figure 4.8. Since \(x_k \notin I_k\), we conclude that \[ x_k \notin \bigcap_{n =1}^\infty \, I_n \,, \quad \forall \, k \in \mathbb{N}\,. \] As the points \(x_k\) are all the elements of \(\mathbb{R}\), we conclude that \[ \bigcap_{n =1}^\infty \, I_n = \emptyset \,. \] However the Nested Interval Property of Theorem 5 implies that \[ \bigcap_{n =1}^\infty \, I_n \neq \emptyset \,, \] yielding a contradiction. Therefore \(\mathbb{R}\) is uncountable.

Figure 4.8: The intervals \(I_n\) are nested, and can be chosen so that \(x_n \notin I_n\).

As a corollary we obtain that also the irrational numbers are uncountable.

Theorem 32

The set of irrational numbers \[ \mathcal{I}:=\mathbb{R}\smallsetminus \mathbb{Q} \] is uncountable.

Proof

In Theorems 30, 31 we have shown that \(\mathbb{R}\) in uncountable and \(\mathbb{Q}\) is countable. Suppose by contradiction that \(\mathcal{I}\) is countable. Then \[ \mathbb{Q}\cup \mathcal{I} \] is countable by Proposition 29, being union of countable sets. Since by definition \[ \mathbb{R}= \mathbb{Q}\cup \mathcal{I} \,, \] we conclude that \(\mathbb{R}\) is countable. Contradiction.