 It is not hard to check that the real number constructed in this way is the limit of the sequence. To see how this works in a specific case, we can consider the sequence \(1/2, 3/4, 7/8, \dots, 1-1/2^n, \dots\) This sequence is increasing, since each term is greater than the one that came before, and it is bounded, since 2 is a number that is strictly greater than every term in the sequence. The Dedekind cut constructed as above will correspond to the number 1. To see this, note that for any rational number q less than 1, we can let \(\epsilon=1-q\). Then there is some \(N\) such that \(1/2^N&lt;\epsilon\). The \(N\)th term in the sequence will then be greater than \(q\), so \(q\) will be in its left set, and thus in the left set we constructed above. But 1 itself, and every rational number greater than it, are all in the right set constructed above. This reasoning also shows that the sequence converges to 1 according to the definition of a limit. For any \(\epsilon\), and an \(N\) such that \(1/2^N&lt;\epsilon\), the \(N\)th term in the sequence is within \(\epsilon\) of 1, and all later terms of the sequence are greater than the \(N\)th term, but still less than 1, so they must also all be within \(\epsilon\) of 1.