# First number in an open interval?

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?

• No, it doesn't make sense. There is no lowest number. – rae306 Oct 19 '14 at 17:54
• 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. – mfl Oct 19 '14 at 17:55
• If you use the classic ordering on reals, then no. If not, and you assume Zorn valid, then any set has a well-ordering. – Milly Oct 19 '14 at 17:58
• 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). – posilon Oct 19 '14 at 18:07
• 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 – Kelvin Soh Oct 19 '14 at 18:39

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 $0$.

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 $(0,1)$.