# Very basic property of real numbers

I'm wondering if the following assertion is true.

Let $x, b \in \mathbb R$. If $x < b$, then there exists some $\delta > 0$ such that $x < b - \delta$.

• Choose $\delta = {1 \over 2} (b-x)$. If you have $x \ge b-\delta$ for all $\delta >0$ then you would have $x \ge b$. – copper.hat Jul 25 '14 at 17:13
• @copper.hat Thanks. So I believe the following would be true as well: If $$AX \ge b, \; A^tY \le c, \; X \ge 0,\; Y\ge 0, \; c^tX \le Y^tb.$$ is unsolvable, then so is: $$AX \ge b, \; A^tY \le c, \; X \ge 0,\; Y\ge 0, \; c^tX - \delta \le Y^tb.$$ for some positive $\delta$? – user7348 Jul 25 '14 at 17:53
• You need to elaborate a little. It is not clear what relationship this has with the question asked. – copper.hat Jul 25 '14 at 18:02
• @user7348 If the first system is not solvable, because $c^\top X \le Y^\top b \, (*)$ is violated, thus $Y^\top b < c^\top X$ then there is a positive $\delta$ with $Y^\top b < c^\top X - \delta$ which means that for this $\delta$ the modified condition $c^\top X - \delta \le Y^\top \, (**)$ is violated as well. Note that this is different from your version, which makes not sure, which conditions are violated. So there might be no solution, but $(*)$ might still be valid. Note in that case $c^\top X - \delta < c^\top X \le Y^\top b$ for all positive $\delta$. – mvw Jul 25 '14 at 19:25
• @user7348 However because one of the other conditions has to be violated in the second system, the system is still not solvable. So in a sense you can take any $\delta$, it won't matter. In both cases, (*) true or not, $\delta = \frac{c^\top X - Y^\top b}{2}$ should ensure an unsolvable system stays unsolvable in the modified case. So your statement is correct. (Note that both comments used the original question) – mvw Jul 25 '14 at 19:49

There is always an intermediate point $y = \frac{x+b}{2}$. Choose $\delta := y - x = \frac{b-x}{2}$.
In fact we can exploit a whole range of points between $x$ and $b$, all less than $b$: $$y(t) = x (1-t) + b t \in [x, b)$$
for $t \in [0, 1)$. So $$\delta(t) := y(t) - x = (b - x) t$$ will ensure $$b - \delta(t) = b - (b-x) t = b (1-t) + x t \in (x, b]$$ and $$x < b - \delta(t).$$