For $x \in \Bbb R$, for all $y (|y|<x \implies y<5)$

Question

Given that $x$ is a free variable beloging to the set of real numbers, find the set of values for $x$ for which the following statement is true:

$\forall y (|y|<x \implies y<5)$

My attempt:

• If $x < 0$ then the hypothesis $|y|<x$ is False, so the statement is True.
• If $x \geq 5$ then if $|y|$ is between $5$ and $x$, the hypothesis is True but the conclusion is False, so the statement is False.
• If $0 \leq x < 5$ then the hypothesis is True for all $|y| < x$ and the conclusion $y < 5$ is also True.

So my answer would be the set $x < 5$. However the answer in the back of the book says the set $(-\infty,0] \cup [5,\infty)$. Where am I wrong?

Answer 1

Consider the contrapositive which is logically equivalent and easier to work with:

If $\forall y( y \geq 5 \implies (|y| \geq x))\Longleftrightarrow \forall y $$( y \geq 5,\implies (y \geq x))$ since $y \geq 0$.

If $x > 5$, then choose $y = 5$ which contradicts the statement.

Hence we must have all $x \leq 5$, specifically $(-\infty,5]$ as BrianO commented.

You missed $x = 5$, here is why it works for your original statement:

Note that $y \leq |y|$. If $x = 5$, then the statement becomes $\forall y (y\leq|y|<5 \implies y < 5)$, which is seen to be true immediately.