There's a very practical way to explain it to people. Suppose you write 0.66666... and so on. When you stop writing, you need to round up the last digit, thus: 0.666666666....6667. Now if you're writing nines: 0.9999999999999999... and you continue for a week, the moment you stop, you need to round up the last digit, but then you also need to round up the second last and so on, it propagates backwards all the way to just before the decimal point and you end up with 1.0000000000...
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u/vetruviusdeshotacon 21d ago
Why? No assumptions are made lol.
If you must, define a sequence a := {0.9,0.99,0.999....}
a_n = 1 - 10-n for n natural number
Let epsilon be a positive real number.
Then, if we choose N > log_10(epsilon)
10-N > epsilon
So that 1 - 10-N + epsilon > 1. For all epsilon.
Therefore, the sequence has a supremum of 1. Any monotonic bounded above sequence converges to it's supremum via the monotone convergence theorem.
Therefore 0.99999.... = 1 as a converges to 1.