Tuesday, 6 October 2015

real analysis - Proving that R is uncountable

Is the following proof for the uncountability of R sufficient? We first assume that the interval (0,1) is countable. So we can define a bijection f:N(0,1)



x1=x11x12x13x2=x21x22x23x3=x31x32x33...



Where xij is the digit in the jth decimal place of the ith number in the list. Now construct some number y whose jth decimal place yj=xii+1 when xii9 and 0 otherwise. But y, while clearly in (0,1), is not in the list, for it differs from x1 in the first decimal, x2 in the second, and so on. So f:N(0,1) is not surjective, and so not a bijection. (0,1) is therefore not countable, and so neither is R.

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