6.3: Uncountable Sets
In this section we prove one of the most remarkable results of modern mathematics. There are sets which are not countable. When this result was first communicated in 1878 by Georg Cantor, it astonished the mathematical world. It follows from this result that, in a most meaningful sense, there are different sizes of infinity. Suppose \(X\) is not a countable set. By Theorem 6.6, \(\mathbb{N} \preceq X\) . By the Schröder-Bernstein Theorem, if \(X \preceq \mathbb{N}\) , then \(|\mathbb{N}|=|X|\) , and \(X\) would be countable. So if \(X\) is not countable, \(X \npreceq \mathbb{N}\) , and \(\mathbb{N} \prec X\) . That is, \[\aleph_{0}<|X| \text {. }\] DEFINITION. Uncountable A set that is not countable is called uncountable.
Of course, we have yet to show that there are any uncountable sets.
NOTATION. \(\prec\) Let \(X\) and \(Y\) be sets. Then \(X \prec Y\) provided that \[X \preceq Y\] and \[|X| \neq|Y| .\] We write \(|X| \leq|Y|\) if \(X \preceq Y\) , and \(|X|<|Y|\) if \(X \prec Y\) .
Definition. Power set, \(P(X)\) Let \(X\) be a set. Then \[P(X)=\{Y \mid Y \subseteq X\} .\] \(P(X)\) is called the power set of \(X\) . It is the set of all subsets of \(X\) .
The next theorem, due to G. Cantor, is one of the most remarkable results of mathematics. It not only proves the existence of an uncountable set, it implies that the power set necessarily generates sets of larger cardinality and thereby provides a means of constructing infinitely many different infinite cardinalities. THEOREM 6.7. Let \(X\) be a set. Then \[|X|<|P(X)| .\] Discussion. To prove this result we need to show that a bijection between a set and its power set is impossible. How does one show the impossibility of such a function? We can assume that such a bijection exists and derive a contradiction. Alternatively, we can show that any function from a set to its power set necessarily fails to be a surjection - which is nearly the same thing, and more elegant. We need to show that any function from a set to its power set "misses" some elements of the power set. We shall use a technique known as a diagonal argument to construct an element of the power set that is not in the range of the function. The domain, \(X\) , acts like an index to keep track of the elements of the range of the function (this is another use for functions). We construct an element \(Y \in P(X)\) not in the range of the function by adding \(x \in X\) to \(Y\) iff \(x\) is not in the element of \(P(X)\) indexed by \(x\) (that is, \(x\) is not in the image of \(x\) under the function). It is easy to show that this subset \(Y\) of \(X\) is not in the range of the function, and the function therefore fails to be a surjection onto \(P(X)\) .
Proof. We observe that the function \(g: X \rightarrow P(X)\) defined by \[g(x)=\{x\}\] is an injection. In the case the \(X=\emptyset, g\) is the empty function that is, the function whose graph is the \(\emptyset\) . (You should check that the empty function is an injection.) Therefore \[|X| \leq|P(X)| \text {. }\] Let \[f: X \rightarrow P(X)\] We define \[Y:=\{x \in X \mid x \notin f(x)\} .\] Discussion. Recall that for every \(x \in X, f(x)\) is a subset of \(X\) . Thus it makes sense to consider whether \(x\) is an element of \(f(x)\) . The self-referential flavor of this argument makes it challenging on the first reading!
Clearly, \[Y \subseteq X\] Is \(Y \in f[X]\) ? Suppose it were, so \(Y=f\left(x_{0}\right)\) for some \(x_{0}\) in \(X\) . But then, \(x_{0}\) would be in \(Y\) iff \(x_{0}\) were not in \(f\left(x_{0}\right)=Y\) . This is impossible, contradicting the assumption that \(Y\) is in \(f[X]\) .
You might try to repair \(f\) by modifying it to include in its range the diagonal set we constructed. Applying the diagonal argument again will identify a new element missing from the range of the modified function. In fact most elements of the codomain are missing from the range of the function, although this is not immediately obvious from the proof.
You still might be confused by why this is called a diagonal argument. This will be obvious when we apply the technique to infinite binary sequences.
If \(X\) is finite the theorem is obvious. Indeed, if there is \(n \in \mathbb{N}\) such that \(|X|=n\) , then \(|P(X)|=2^{n}\) . Theorem \(6.7\) implies that any set, including an infinite set, is strictly smaller than its power set. In fact, by iterating the applications of the power set function to \(\mathbb{N}\) , it is easily seen that there are infinitely many infinite sets of distinct cardinality in the sequence of sets \[\langle\mathbb{N}, P(\mathbb{N}), P(P(\mathbb{N})), \ldots\rangle\] (What is the cardinality of the union over this sequence? What is the cardinality of the power set of that union?)
We shall prove that two more sets are uncountable. Both of these are sets of mathematical interest. We shall show first that infinite binary sequences are uncountable. Infinite binary sequences are functions from \(\mathbb{N}\) to \(\ulcorner 2\urcorner\) . As we shall see, there is a very close relationship between infinite binary sequences and the power set of \(\mathbb{N}\) . More generally, the collection of all functions from one set to another can be of mathematical interest. We introduce a notation for such collections.
NOTATION. \(Y^{X}\) Let \(X\) and \(Y\) be sets. The set of all functions with domain \(X\) and codomain \(Y\) is written \(Y^{X}\) .
Do not confuse this with exponentiation. However if \(X\) and \(Y\) are finite, \[\left|Y^{X}\right|=|Y|^{|X|} .\] The set of all functions from some set \(X\) into \(\ulcorner 2\urcorner\) is in bijective correspondence with \(P(X)\) :
Proposition 6.8. Let \(X\) be a set, and define \(F:\ulcorner 2\urcorner X \rightarrow P(X)\) by: for \(\chi \in\ulcorner 2\urcorner X\) , \[F(\chi)=\chi^{-1}(1) .\] That is \(F(\chi)=\{x \in X \mid \chi(x)=1\}\) . Then \(F:\ulcorner 2\urcorner^{X} \rightarrow P(X)\) is a bijection.
PROOF. The proof is left as an exercise.
The existence of this bijection allows us to easily prove the following theorem.
THEOREM 6.9. The set of infinite binary sequences is bijective with \(P(\mathbb{N})\) and is therefore uncountable.
PROOF. By Proposition \(6.8\) \[\left|\ulcorner 2\urcorner^{\mathbb{N}}\right|=|P(\mathbb{N})| .\] By Theorem 6.7, \[|\mathbb{N}| \prec|P(\mathbb{N})| .\] Therefore \(\ulcorner 2\urcorner^{\mathbb{N}}\) is uncountable.
Notation. \(2^{\mathbb{N}_{0}}\) We use \(2^{\mathbb{N}_{0}}\) for the cardinality of \(\left\ulcorner 22^{\mathbb{N}}\right.\) . It is worth illustrating by an application to infinite binary sequences why the technique used to prove Theorem \(6.7\) is called a diagonal argument (sometimes called the second diagonal argument, to distinguish from the "first diagonal argument" in Section 6.4). Let \(f: \mathbb{N} \rightarrow\ulcorner 2\urcorner \mathbb{N}\) . We prove that \(f\) is not a surjection by direct application of the diagonal argument in the proof of Theorem 6.7. We enumerate all the elements in the range of \(f\) ; each one is a sequence of 0 ’s and 1 ’s. \[\begin{aligned} & f(0)=a_{00} \quad a_{01} \quad a_{02} \quad a_{03} \quad \ldots \quad a_{0 j} \ldots \\ & f(1)=a_{10} \quad a_{11} \quad a_{12} \quad a_{13} \quad \ldots \quad a_{1 j} \quad \ldots \\ & f(2)=a_{20} \quad a_{21} \quad a_{22} \quad a_{23} \quad \ldots \quad a_{2 j} \quad \ldots \\ & f(3)=a_{30} \quad a_{31} \quad a_{32} \quad a_{33} \quad \ldots \quad a_{3 j} \quad \ldots \end{aligned}\] We now create a sequence by altering the diagonal elements of this infinite array. Let \(s\) be the sequence of diagonal elements \[\left\langle 1-a_{00}, 1-a_{11}, \ldots, 1-a_{i i}, \ldots\right\rangle .\]
The sequence \(s\) is an element of \(\ulcorner 2\urcorner^{\mathbb{N}}\) , and differs from every element in the range of \(f\) : indeed, \(s\) differs from \(f(i)\) in at least the \(i^{\text {th }}\) slot. Hence, \[s \notin f[\mathbb{N}],\] and so \(f\) is not a surjection. We leave it as an exercise to show that \(s\) is the diagonal set \(Y\) constructed in the proof of Theorem 6.7, where \(X=\mathbb{N}\) . More precisely, \(s\) is the image of \(Y\) under the natural bijection from \(P(\mathbb{N})\) to \(\ulcorner 2\urcorner^{\mathbb{N}}\) of Proposition 6.8.
We consider another set of mathematical interest, the set of all infinite decimal sequences. This set has a close relationship with the closed interval \([0,1]\) . Understanding this relationship requires a deeper, more formal understanding of the real numbers than most students have been exposed to in calculus, and we postpone the detailed discussion of this relationship until Section 8.9. With some modifications, the following theorem will prove that \([0,1]\) is uncountable, and therefore \(\mathbb{R}\) is uncountable (see Section 8.9).
THEOREM 6.11. The set of infinite decimal expansions is uncountable. In fact, \[\left|\ulcorner 10\urcorner^{\mathbb{N}}\right|=2^{\aleph_{0}} .\] DISCUSSION. The identity function on the infinite binary sequences into the infinite decimal sequences is clearly an injection. We shall construct an injection from the infinite decimal sequences to infinite binary sequences. The theorem will follow from the Schröder-Bernstein Theorem.
PROOF. It is obvious that \[\left|\ulcorner 2\urcorner^{\mathbb{N}}\right| \leq\left|\ulcorner 10\urcorner^{\mathbb{N}}\right| .\] (Why?) We shall define an injection \[f:\ulcorner 10\urcorner^{\mathbb{N}} \rightarrow\ulcorner 2\urcorner^{\mathbb{N}} .\] Let \(x \in\ulcorner 10\urcorner^{\mathbb{N}}\) . So \[x=\left\langle x_{j} \mid j \in \mathbb{N}\right\rangle\] where \(x_{j}\) is the \(j^{t h}\) member of the sequence \(x\) and \[0 \leq x_{j} \leq 9 .\] We want to define a binary sequence \(s(x)\) that "encodes" \(x\) . There are many ways to do it. One is to look at blocks of 10 bits (short for "binary digits"), and, in the \(j^{\text {th }}\) such block, have nine of the bits 0 , and put a 1 in the \(x_{j}^{\text {th }}\) slot. Formally, given an infinite decimal sequence \(x\) , we define a binary sequence \[f(x)=\left\langle y_{i} \mid i \in \mathbb{N}\right\rangle\] so that \(y_{i}=1\) if there is \(j \in \mathbb{N}\) such that \[i=10 j+x_{j} .\] Otherwise \(y_{i}=0\) . We thereby define a function \[f:\ulcorner 10\urcorner \rightarrow\ulcorner 2\urcorner .\] We show that \(f\) is an injection. Let \(x\) and \(y\) be distinct elements of \(\ulcorner 10\urcorner^{\mathbb{N}}\) . Then there is some \(j \in \mathbb{N}\) such that \[x_{j} \neq y_{j} .\] Then \[10 j+x_{j} \neq 10 j+y_{j} .\] Let \(i=10 j+x_{j}\) . Then \(f(x)\) and \(f(y)\) differ in the \(i^{\text {th }}\) component. That is, \[(f(x))_{i}=1 \neq 0=(f(y))_{i} .\] Therefore \(f\) is an injection and \[\left|\ulcorner 10\urcorner^{\mathbb{N}}\right| \preceq\left|\ulcorner 2\urcorner^{\mathbb{N}}\right| .\] By the Schröder-Bernstein Theorem, (6.12) and (6.13) yield \[\left|\ulcorner 10\urcorner^{\mathbb{N}}\right|=2^{\aleph_{0}} .\] We prove in Section \(8.9\) that \[|[0,1]|=\left|\ulcorner 10\urcorner^{\mathbb{N}}\right|,\] essentially by identifying a real number with its decimal expansion. If we assume this result, we can easily prove that the real numbers are uncountable. COROLLARY 6.14. \(\mathbb{R}\) is uncountable.