It appears that there is a path for each binary string, and there is a binary string for any path. Then there exists a bijection from $\mathbb{N}$ to $[0, 1]$ . 4. In other words, we can create an infinite list which contains every real number. It appears that there is a path for each binary string, and there is a binary string for any path. Clearly $[0, 1]$ is not a finite set, so we are assuming that $[0, 1]$ is countably infinite. The set of all such sequences is uncountable because the set of real numbers, between 0 and 1, is uncountable. To do it, just consider any sequence as binary fraction with integer part 0. Naive Set Theory breaks down. Let R Be The Set Of All Binary Sequences. Your example: take your infinite set to be $\mathbf{Q}$ and look at approximations to reals. Such a sequence is NOT "uncountable" (a "sequence", pretty much by definition, is countable). • Axiomatic Set Theory introduced to control these paradoxes. An infinite binary sequence can represent a formal language (a set of strings) by setting the n th bit of the sequence to 1 if and only if the n th string (in shortlex order) is in the language. Hence the isomorphism. Doesn't Cantor's diagonal argument mean that you cannot "list" all real numbers between 0 and 1? Just for the fun, we can use continued fractions to map the sequences of positive integers injectively to [0,1] the sequence may end with $\infty$ meaning that we get a finite fraction (a rational number). In the above derivation, the fact that tree nodes (cons cells) have no values is … And here's where the fact that caused years of controversy comes in. Use Cantors Argument To Show That R Is Not Countable. Countable and Uncountable Sets. The set S of all infinite binary sequences is uncountable. Infinite binary strings correspond to other paths. While perhaps the most classic example is the set of reals, R, the easiest proof lies in the set of infinite binary sequences. (By the way, your initial reference to "the sequence of infinitely many coin tosses" confused me for a moment. Hence the isomorphism. Take your infinite set to be the nodes of an infinite binary tree, and take your sets to be the paths from the root to infinity. Although most of the answers have already explained the concept, let me try and add another perspective. Proof.Theproofisbycontradiction.Supposethat S iscountable.Since S isclearlynota finiteset,thismeansthatthereisa1-1correspondence f :Z + → S .Hence,everyinfinite You can conduct a direct proof by constructing a map from the considered set of 0-1 sequences into [0,1] segment. The set of all binary strings is known to be uncountable. Every two such paths will go separate ways after finitely many steps. Since R is uncountable, R is not the union of two countable sets. Ask Math: Uncountable set. Let S = All Sets of Type 2 = Set of all sets not containing themselves as Elements • S∈S → S is Type 1 → S∉S CONTRADICTION • S∉S → S is Type 2 → S∈S CONTRADICTION But every element must either be in S or not in S! Statement: the set of infinite binary (0,1) sequences is uncountable. The set C = {0, 1} ∞ of all infinite binary sequences is sometimes called the Cantor space. A "sequence of infinitely many coin tosses" can be thought of as a infinite sequence of "1"s and "0"s and, putting a decimal point in front of it, as a real number, between 0 and 1, expressed in binary. We simply show that a subset of such an infinite sequence is uncountable. The Set of all Subsets of Natural Numbers is Uncountable Theorem 1: The set $\mathcal P (\mathbb{N})$ of all subsets of $\mathbb{N}$ is uncountable. That is, there exist a one-to-one and onto function from the set of such sequences to the real numbers between 0 and 1. The set of all binary strings is known to be uncountable. The set of all such sequences is uncountable because the set of real … In the proof below, we use the famous diagonalization argument to show that the set of all subsets of $\mathbb{N}$ is uncountable.