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Inequality on open interval

  1. Mar 9, 2010 #1
    Given the fact that the following inequality must hold;

    x > y-1 For all y[tex] \in[/tex] ]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?
  2. jcsd
  3. Mar 9, 2010 #2


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