Inequality on open interval

[azay]

Given the fact that the following inequality must hold;

x > y-1 For all y$$\in$$ ]0,1[ (an open interval)

and given the fact that one can choose y After one chooses x, can one then state that x > 0 holds?

My idea was to say that at least x >= 0 holds because:

1) Someone picks a negative x that is arbitrarily close to 0, say -0.000...001.
2) I can now choose a y from the interval ]0,1[, say 0.999999... so that y-1 > x
3) Therefore nobody can pick a negative x so that the inequality holds

However, I am even more unsure about the strict inequality x > 0. It seems unlikely to me that it holds.

How do you properly reason about these kind of things?

 

[vela]

If $y \in (0,1)$, then $y-1 \in (-1,0)$, so $x \in [0,\infty)$, that is, $x \ge 0$, satisfies the inequality. Note that this is a weaker condition than x>0, so both conditions hold.