0
$\begingroup$

What is the first number in an open interval? For example, if I have the open interval (0, 1), what is the lowest number in that interval? Does this question even make sense with real numbers?

$\endgroup$
  • $\begingroup$ No, it doesn't make sense. There is no lowest number. $\endgroup$ – rae306 Oct 19 '14 at 17:54
  • 1
    $\begingroup$ If you are talking about real numbers then there in no a lowest number in the interval. In your case, $(0,1),$ if $a$ were the lowest number then you would have $0<a/2<a$ which makes no sense. $\endgroup$ – mfl Oct 19 '14 at 17:55
  • 1
    $\begingroup$ If you use the classic ordering on reals, then no. If not, and you assume Zorn valid, then any set has a well-ordering. $\endgroup$ – Milly Oct 19 '14 at 17:58
  • 1
    $\begingroup$ The question does make sense. The answer is that such number does not exist. There is no smallest number in $(0,1)$, even if you restrict yourself to rational numbers (the argument provided by mfl above works in this case as well). $\endgroup$ – posilon Oct 19 '14 at 18:07
  • 1
    $\begingroup$ You will probably want to look at the concepts of infimums and supremums. As opposed to minimum or maximum, these will exist for open intervals. en.wikipedia.org/wiki/Infimum_and_supremum $\endgroup$ – Kelvin Soh Oct 19 '14 at 18:39
2
$\begingroup$

There is no lowest number. This is because, by definition, given a point $x\in(0,1)$ we have $0<x$, and therefore $0<{x\over 2}$. But $x/2 < x$, hence $x$ can never be the lowest number in the open interval. This is essentially identical to the proof that there are no real "infinitesimals"; if you apply this logic to the interval $(0,\infty)$ you see that if a positive real number is smaller than each $\varepsilon>0$ then said number is equal to $0$.

$\endgroup$
2
$\begingroup$

There isn´t the first number in interval $(0,1)$. Suppose, by contratiction, that $0<\alpha<1$ is the lowest number in interval $(0,1)$. Note that $0<\frac{\alpha}{2}<\alpha<1$, i.e., $\frac{\alpha}{2} \in (0,1)$ it´s less than $\alpha$, it´s contradiction. Hence there isn´t que lowest number in interval $(0,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.