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Prove a squeeze-like property for reals.

by RoRi

If x \in \mathbb{R} and 0 \leq x < h for all h \in \mathbb{R}^+, then x = 0.

Proof. Suppose otherwise, that 0 \leq x < h for all h \in \mathbb{R}^+, but x \neq 0. Then, we have 0 \leq x and 0 \neq x implies 0 < x, so x is a positive real number. Then let h = x (since x \in \mathbb{R}^+ this is a valid choice of h). By assumption we then have x < h = x (since x < h for all h \in \mathbb{R}^+), but this is a contradiction since x = x (equality is symmetric) implies x \not< x. Therefore, we must have x = 0. \qquad \blacksquare

2 comments

  1. Camilo says:

    Even when is well known that “equality” is an equivalence relation (i.e symmetric) we can prove that x<x cannot hold since if it happens, then x-x is in R+, which means, 0 is in R+, leading to a contradiction with the first Axiom of order.

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