From Polynomial Roots to Category
Transcendental numbers are defined by what they are not: they are real numbers that are not roots of any nonzero polynomial with integer coefficients. This definition involves infinitely many polynomials, so it does not immediately tell us how frequently transcendental numbers occur in the real line. Countability and the Baire theorem give a strong answer: every nonempty open interval contains transcendental numbers, and their set is a dense G-delta.
Throughout, \(\mathbb{R}\) has its standard topology. A real number is algebraic if it is a root of a nonzero polynomial with integer coefficients; it is transcendental if it is not algebraic. Write \(\mathcal{A}\) for the set of algebraic real numbers and \(\mathcal{T}\) for the set of transcendental real numbers. Thus \(\mathcal{T}=\mathbb{R}\setminus\mathcal{A}\).
Countability is the first step. There are countably many integer polynomials, and each nonzero polynomial has only finitely many real roots. The next result verifies both facts in the form needed here.
Proof. For \(n,m\in\mathbb{N}\), consider the nonzero polynomials of degree at most \(n\) whose integer coefficients all have absolute value at most \(m\). There are only finitely many such polynomials: each is specified by at most \(n+1\) coefficients, and each coefficient has only \(2m+1\) possible values. Excluding the all-zero coefficient list removes the zero polynomial.
Each nonzero polynomial of degree \(n\) has at most \(n\) distinct real roots. To justify this root bound, use induction on \(n\). A nonzero constant has no roots. If \(p\) has positive degree and \(r\) is a root, the factor theorem gives \(p(x)=(x-r)q(x)\), where \(q\) has degree \(n-1\). Every root \(s\ne r\) of \(p\) satisfies \(0=p(s)=(s-r)q(s)\); since \(s-r\ne0\), \(q(s)=0\). By induction there are at most \(n-1\) such roots, so \(p\) has at most \(n\) distinct roots in total. The factor theorem here follows by writing each difference \(x^k-r^k\), for \(k\ge1\), as \((x-r)(x^{k-1}+x^{k-2}r+\cdots+r^{k-1})\).
Consequently, for each pair \(n,m\), the union of the root sets of the finitely many polynomials just described is finite. Every nonzero integer polynomial belongs to one of these finite collections: take \(n\) at least its degree and \(m\) at least the absolute value of each of its finitely many coefficients. Therefore \(\mathcal{A}\) is a countable union of finite sets, indexed by the countable set \(\mathbb{N}\times\mathbb{N}\), and is countable. \(\square\)
Worked Example: Checking a Polynomial Root
The polynomial \(p(x)=x^2-2\) is nonzero and has integer coefficients. Solving \(p(x)=0\) gives \(x^2=2\), hence the two real roots \(x=\sqrt{2}\) and \(x=-\sqrt{2}\). Substitution checks each root: \(p(\sqrt{2})=(\sqrt{2})^2-2=2-2=0\), and \(p(-\sqrt{2})=(-\sqrt{2})^2-2=2-2=0\). Since a degree-two polynomial has at most two distinct roots, these are all its real roots. In particular, both numbers are algebraic by definition.
Why the Algebraic Numbers Are Meager
In the standard topology on \(\mathbb{R}\), a singleton is closed: if \(x\ne a\), the open interval \((x-|x-a|/2,x+|x-a|/2)\) contains \(x\) and does not contain \(a\). A singleton also has empty interior, because every open interval around its point contains other real numbers. Thus each singleton is nowhere dense. The countability proposition now implies that \(\mathcal{A}\) is meager: enumerate its elements, allowing repetitions if the set is finite, and express it as a countable union of nowhere-dense singletons.
Worked Example: A Small Block of Polynomials
Consider all nonzero polynomials \(ax+b\) with \(a,b\in\{-1,0,1\}\). If \(a=1\), the three choices \(b=-1,0,1\) give \(x-1,x,x+1\), with roots \(1,0,-1\), respectively. Substitution gives \(1-1=0\), \(0=0\), and \(-1+1=0\). If \(a=-1\), the three polynomials are \(-x-1,-x,-x+1\), with roots \(-1,0,1\); substitution gives \(1-1=0\), \(0=0\), and \(-1+1=0\). If \(a=0\), the nonzero choices \(b=-1\) and \(b=1\) give nonzero constant polynomials, which have no roots. The choice \(a=b=0\) is excluded because it gives the zero polynomial.
Thus this finite block has root set exactly \(\{-1,0,1\}\). The calculation illustrates how a finite collection of polynomials contributes only finitely many algebraic numbers. Allowing all degrees and all integer coefficients adds countably many such finite root sets, not an uncountable collection.
The key category theorem applies to the complement of a countable union of nowhere-dense sets. We use its complete-metric form, with its hypotheses stated explicitly.
The real line with its usual metric is complete. Enumerate the algebraic numbers as \((a_n)_{n\in\mathbb{N}}\), repeating elements if necessary, and set \(U_n=\mathbb{R}\setminus\{a_n\}\). Each \(U_n\) is open because \(\{a_n\}\) is closed. It is dense because its complement, a singleton, has empty interior: every nonempty open interval contains a point other than \(a_n\).
Proof. With \(U_n=\mathbb{R}\setminus\{a_n\}\) as above, a real number belongs to every \(U_n\) exactly when it is different from every algebraic number in the enumeration. Therefore $$ \mathcal{T}=\mathbb{R}\setminus\mathcal{A} =\bigcap_{n\in\mathbb{N}}\bigl(\mathbb{R}\setminus\{a_n\}\bigr) =\bigcap_{n\in\mathbb{N}}U_n. $$ This expresses \(\mathcal{T}\) as a countable intersection of open sets, so it is a G-delta. Each \(U_n\) is open and dense, and \(\mathbb{R}\) is complete, so the complete-metric Baire theorem says their intersection is dense. Finally, \(\mathbb{R}\setminus\mathcal{T}=\mathcal{A}\) is meager, as proved above. By definition, \(\mathcal{T}\) is residual. \(\square\)
Worked Example: Transcendentals in a Specified Interval
Consider the interval \((1,2)\) in the standard real line. It is nonempty and open, since it contains \(3/2\) and is an open interval. The theorem says that the dense set \(\mathcal{T}\) intersects every nonempty open set, so \((1,2)\cap\mathcal{T}\ne\varnothing\). Hence there is at least one transcendental number strictly between \(1\) and \(2\).
This conclusion does not identify a particular number in the interval. It is an existence statement obtained from density. Its hypotheses have been checked: the ambient metric space is the complete real line, and the sets \(U_n=\mathbb{R}\setminus\{a_n\}\) are all open and dense. As a check on the distinction between avoiding finitely many roots and avoiding all algebraic numbers, \(3/2\) itself is not such a transcendental example: it is a root of \(2x-3\), because \(2(3/2)-3=3-3=0\).
What Residual Does—and Does Not—Mean
The conclusion is stronger than saying that the algebraic numbers are countable: it gives a topological sense in which transcendental numbers occur throughout \(\mathbb{R}\). Density means that every nonempty open interval contains a transcendental number. The G-delta description records the set as a countable intersection of open sets, each obtained by excluding one algebraic number.
A common mistake is to infer that a countable union of nowhere-dense sets must itself be nowhere dense. For a concrete counterexample, enumerate the rational numbers as \((q_n)_{n\in\mathbb{N}}\). Each singleton \(\{q_n\}\) is nowhere dense by the singleton argument above, so \(\mathbb{Q}=\bigcup_{n\in\mathbb{N}}\{q_n\}\) is meager. But \(\overline{\mathbb{Q}}=\mathbb{R}\): every nonempty open interval contains a rational number, by density of \(\mathbb{Q}\) in \(\mathbb{R}\). Therefore \(\operatorname{int}\overline{\mathbb{Q}}=\mathbb{R}\), which is nonempty, and \(\mathbb{Q}\) is not nowhere dense. The correction is that a countable union of nowhere-dense sets is meager; it need not be nowhere dense. The Baire theorem says that this meager set cannot fill any nonempty open interval of the complete real line.
Check Your Understanding
Use the definitions and the complete-metric Baire theorem to answer the following.
- Why must the zero polynomial be excluded from the definition of an algebraic number?
- Explain why the roots of all integer polynomials of degree at most \(3\), with coefficients of absolute value at most \(2\), form a finite set.
- Show that \(\mathbb{R}\setminus\{a\}\) is open and dense for any \(a\in\mathbb{R}\).
- Write the transcendental numbers as a countable intersection of open dense subsets of \(\mathbb{R}\).
- Does meagerness imply that a set is nowhere dense? Use \(\mathbb{Q}\) to justify your answer.