Smallest number less than a given number

Question

Consider the inequality $0<y<x$. How do I write this inequality using the sign "$\leq$" instead of "$<$" on the RHS? My idea is this: since $y$ will never quite be $x$, can I write $0<y\leq x^{1-\epsilon}$, where $\epsilon>0$ is arbitrarily small? It would make sense to me that I could do this only if $\epsilon$ is the smallest number greater than $0$. Is it better to write, instead, $0<y\leq \lim_{\epsilon \to 0} x^{1-\epsilon}$, where $\epsilon$ can be taken to $0$ from either the right or the left? This does not make as much sense to me because $\epsilon$ will still be $0$ eventually. What is the best, most accurate, way to do this?

Answer 1

There is no "proper" way to do this (*). You could "cheat", for example you could say that there is an $\epsilon > 0$ such that $\epsilon \leq y \leq x - \epsilon$, but you still have the $>$ in the condition on $\epsilon$, and it's no longer a simple condition, because now you have to deal with the existence of some $\epsilon$. Another way to "cheat" would be to say that both $0 \geq y$ and $y \geq x$ are false; essentially this means using $\ngeq$ as a synonym for $<$.

(*) This is an informal statement: it necessarily depends on what you mean by the word proper. If you want a more detailed answer, you'll need to think about why you want to rewrite this strict inequality in terms of non-strict inequalities.

Answer 2

Suppose $\ c\ $ and $\ d\ $ are constants such that:

$$0<y<c\quad (1)$$

and

$$0<y\leq d\quad (2)$$

are equivalent statements, that is, $(1)\implies (2)\ $ and $\ (2)\implies (1).$

Since $(1)\implies (2),\ d\geq c,\ $ for if $\ d<c,\ y=\frac{c+d}{2}\ $ would satisfy $\ (1)\ $ but not $\ (2),\ $ which would contradict the statement, "$\ (1)\implies (2)$. "

But since $\ d\geq c,\ y=c\ $ will satisfy $\ (2)\ $ and not $\ (1),\ $ contradicting the statement: $\ (2)\implies (1).\ $ Therefore we cannot have both: $(1)\implies (2)\ $ and $\ (2)\implies (1),\ $ i.e. statements $\ (1)\ $ and $\ (2)\ $ cannot be equivalent.

Answer 3

I think the premise of your question needs some clarification. When you ask for the "largest number less than a given x", what is your criterion for a number? Whenever we ask questions about how numbers relate to each other we have to clarify what set of numbers we are working over. If we are working over natural numbers or integers, then we can indeed define it in a clean inequality as you want:

$0 < y \leq x - 1$

If we are working over the rationals or the reals, there isn't a "largest number less than a given x". Suppose for contradiction that there was a y which is "right next to x". But then we could define $z = \frac{x+y}{2}$. This number of course lies between y and x so our original premise is wrong. z is closed to x than y is.

We can, however, define a rigorous definition of all numbers y that lie between 0 and x. This is a set that has the least upper bound, or supremum, of x. More rigorously, we can define a set $Y = \{y\in \mathbb{R} |\exists \epsilon>0 \text{ : } 0 < y \leq x - \epsilon\}$. In other words, we can talk about the set of all numbers y, where y lies between 0 and some possible positive $\epsilon$ subtracted from. This means we can imagine $\epsilon$ getting incredibly small, and thus pick y's that are closer and closer to x. However, as stated before, when we are talking about rationals or real numbers, there is no such valid y that is the closest to x.

Hope this helps!