Prove x < y + ϵ for every ϵ > 0 is equivalent to x <= y + ϵ for every e > 0.
I thought it may be trivial, but I am not sure. HELP!!
To prove that the inequality x < y + ε for every ε > 0 is equivalent to x ≤ y + ε for every ε > 0, we need to show that these two statements are logically equivalent.
For the forward direction, assume x < y + ε for every ε > 0. To prove that x ≤ y + ε for every ε > 0 holds as well, we need to show that x ≤ y + ε. Since the inequality x < y + ε holds for every ε > 0, it also holds for ε = 0. Therefore, x ≤ y.
For the reverse direction, assume x ≤ y + ε for every ε > 0. To prove that x < y + ε for every ε > 0 holds as well, we need to show that x < y + ε. Since x ≤ y + ε for every ε > 0, it also holds for ε = 0. Therefore, x < y.
By proving both directions, we have shown that x < y + ε for every ε > 0 is equivalent to x ≤ y + ε for every ε > 0.