Adapt the proof of Theorem 2.8 to show that \(\#\{1,2\}^{\mathbb{N}} \leqslant \#(0,1) \leqslant \#\{0,1\}^{\mathbb{N}}\) and conclude that \(\#(0,1)=\#\{0,1\}^{\mathbb{N}}\).

Remark. This is the reason for writing \(\mathfrak{c}=2^{\aleph_{0}}\).

[ interpret \(\{0,1\}^{\mathbb{N}}\) as base- 2 expansions of all numbers in \((0,1)\) while \(\{1,2\}^{\mathbb{N}}\) are all infinite base-3 expansions lacking the digit 0 .]

Data from theorem 2.8

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The interval (0, 1) is uncountable; its cardinality c:=#(0, 1) is called the continuum. Proof We can write each x (0, 1) as a decimal fraction, i.e. x=0.yy2y3... with yi {0, 1,...,9}. If x has a finite decimal representation, say x= 0.y1y2y3...yns yn #0, we replace the last digit yn by yn - 1 and fill it up with trailing 9s. For example, 0.24 0.2399.... This yields a unique representation of x by an infinite decimal expansion. Assume that (0, 1) were countable and let {x,x2,...} be an enumeration (containing no element more than once!). Then we can write x=0.a1,191,2a1,3a1,4... x2=0.92,192,292,392,4... x3 = 0.a3,193,293,393,4... X4=0.a4,194,294,394,4... and construct a new number x :=0.y1y2y3... (0, 1) with digits 1 if ai,i=5, 5 if aii #5. Yi := (2.8) (2.9) By construction, xx; for any x; from the list (2.8): x and x; differ at the ith decimal. But then we have found a number x = (0, 1) which is not contained in our supposedly complete enumeration of (0, 1) and we get a contradiction. By (0, 1) we denote the set of all sequences (xi)iN where x; (0, 1).