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I am currently going through Apostol's Calculus Book. Apostol is explaining the upper bound axiom. In example 3, it is stated that the set does not contain a Maximum bur rather only bounded above. My Question is why is 0.999999999999999 not the upper bound making it also the Maximum element ? Upper Bound example

samuel
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    $0.999999999999999 <1$ – geetha290krm Jul 10 '22 at 08:31
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    do you mean 0.999...? It equals to 1 exactly but it is not in the sequence set. – auntyellow Jul 10 '22 at 08:41
  • Also, for a proof, say $x$ is any number the set. Then $x<\frac{x+1}2<1$, showing that you can always find a bigger number in the set, so there can be no maximal number. – Milten Jul 10 '22 at 09:16
  • @auntyellow Yes that's exactly what I meant. I did not know what it would be equal to exactly one until Adrian and yourself pointed out now. But now that I know this I guess the example makes sense. – samuel Jul 10 '22 at 09:39

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Well, for starters: $0.999999…<1$ as pointed out by @geetha290krm.

Now, the interval $I=[0,1)$ is the subset of $\mathbb{R}$: all $x \in \mathbb{R}$ such that $0\leq x <1$. Note that $1$ is an upper bound of the interval (a number that is larger that the biggest number of the set). For example: $2$, $\pi$, $2022$ and $10^{10}$ are all upper bounds of $I$ since each of those real numbers are less than $1$.

The supremum of an interval is the smallest upper bound of a (upper) bounded set. If the supremum of the set is in the set itself, then the supremum is called the maximum element of the set. For example, in the interval $J=[0,1]$ has $1$ as its maximum element; in $I$, the supremum is also $1$, but since $1 \notin I$ then $1$ is not the maximum element.

Also, $0.999999999999999$ is not an upper bound since $0.999999999999999<1$. You can put as many $9$’s as you like and it still wouldn’t an upper bound of $I$, since $0.999999999999999…9<1$.

The only way is to put an infinite number of $9$‘s, since $0.\overline{9}=1$.

James Garrett
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  • Hi, Adrian, First thank you for the answer. However, I am still a bit confused because why can't 0.999 with infinite number of 9's be the upper bound. Furthermore I don't understand how 0.99 with infinite 9's would equal 1. – samuel Jul 10 '22 at 09:36
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    The proof is not easy. Some answers from here may not be strict. Please refer to here and here, and need some knowledge about real analysis. – auntyellow Jul 10 '22 at 11:22