-1

The least upper bound of $(0,1)$ is 1 and from what I learned it does not have a maximum.
What is wrong with the statement that a maximum of this interval is $0.\overline{9}$?

fituldo
  • 1
  • 1
  • 2
  • 6
    $1$ is not in the interval and $0.\overline{9}=1$. – Randall Jun 23 '22 at 20:49
  • $0.\overline{9}$ actually represents the limit of an infinite sequence, each of whose terms are less than $(1)$. Despite this, the limit of the sequence is in fact $(1)$, and so is not in the interval. – user2661923 Jun 23 '22 at 20:52
  • 2
    Related $;-;$ Is it true that $0.999999999\ldots=1$? – dxiv Jun 23 '22 at 20:54
  • Thank you all for the clarification. – fituldo Jun 23 '22 at 20:58
  • 1
    The standard argument is as follows: Suppose $x$ is the maximum of $(0, 1)$ and let $y = (1 + x)/2$. Note $x \in (0, 1)$ implies $x < 1$. This implies $x + 1 < 2$, and so $\frac{x + 1}{2} < 1$, i.e. $y < 1$. Also, because $x \in (0, 1)$, we have $x > 0$, so $y = \frac{x + 1}{2} > \frac{1}{2} > 0$. So, $y \in (0, 1)$. But, $x$ is the maximum of $(0, 1)$, so $y \le x$. This means $\frac{x + 1}{2} \le x \implies x + 1 \le 2x \implies x \ge 1$, a contradiction, QED. You can try following this argument with $x = 0.\overline{9}$. Consider this: what would $y$ be in this case? – Theo Bendit Jun 23 '22 at 20:59

1 Answers1

1

The answer is simply that the maximum value of a set must lie in the set (the same is not true for a supremum, however).

$0.999\cdots$ is equal to $1$, and hence not in the set. While $0.999\cdots9$ is, for finitely many $9$'s, infinitely many $9$'s represents the limit of an infinite sequence:

$$0.999\cdots = \lim_{N \to \infty} \sum_{n=1}^N \frac{9}{10^n} = 9 \cdot \lim_{N \to \infty} \sum_{n=1}^N \left( \frac{1}{10} \right)^n=9 \cdot \frac 1 9 = 1$$

thanks to the geometric series. (Or you can use whichever argument of your preference you like that $0.999\cdots = 1$.)

Hence, $1$ cannot be the maximum. Of course, that doesn't necessarily mean a maximum does not exist.

By way of contradiction, suppose $M$ is the maximum. Then $M \in (0,1)$ by definition.

However, the midpoint between $M$ and $1$ (the point $(M+1)/2$) is also in the set, and is strictly larger than $M$, a contradiction.

Hence, $(0,1)$ has no maximum.

Largely just posting this since everything was already put in the comments; hence I'm doing this to prevent this question from being in the "unanswered" queue. Making it Community Wiki since I don't really have anything to add myself.

PrincessEev
  • 43,815
  • 1
    Instead of answering a question just so that it is answered, maybe take a few minutes to look for duplicates. The question "Why don't open intervals have maxima?" is a pretty elementary question (I found a nearly identical question with a quick google search), and the fact that $0.\overline{9}=1$ is a question which has been asked-and-answered here many, many times before. – Xander Henderson Jun 23 '22 at 21:13