Let $x∈ℝ$ be given. By using a contradiction argument, show the equivalence of $x≤0$ $⇔$ $x<ε$, for all $ε>0$

Question

Let $x∈ℝ$ be given. By using a contradiction argument, show the equivalence of $x≤0$ $⇔$ $x<ε$, for all $ε>0$.

Since this is an equivalence question, I know that I must prove both sides of the argument, i.e. $⇐$ and $⇒$.

For $⇒$ I started by supposing that $x≥ε$, and we know this must satisfy for all $ε>0$, thus implying that $x≥0$ ... which I know is a contradiction to what we are given on the LHS, but I'm unsure how to express this from the stage I got to, up to the end result.

For $⇐$ I started by supposing that $x>0$ but I was a bit stuck on how to progress from this point, as I don't know which other information I should use to continue.

Any help/pointing out any of my errors would be appreciated. Thank you.

Answer 1

About $\implies$:

"For ⇒ I started by supposing that x≥ε, and we know this must satisfy for all ε>0, thus implying that x≥0 ... which...."

That's kind of garbled but you have the idea.

The opposite of "$x < \epsilon$ for all $\epsilon > 0$" is not "$x\ge \epsilon$ for all $\epsilon > 0$". It is "there exists at least one $\epsilon > 0$ so that $x \ge \epsilon$. And from that we can conclude that $x \ge \epsilon > 0$ so $x > 0$. (That is a stronger result than $x \ge 0$). And that contradicts $x \le 0$. ( $x \ge 0$ and $x \le 0$ are NOT contradictions as it would be possible for $x = 0$ and both those statements would be true.)

So that's that

.....

Frankly I would not do the $\implies$ direction by a contradiction and I don't think the excercise was intending to use a contradiction for this dirction.

If $x \le 0$ then for all $\epsilon > 0$ we have by transitivity, $x \le 0 < \epsilon$ or $x < \epsilon$.

That's it. I think the exercise intended the $\implies$ direction to be too easy and trivial to require much proof.

I think the contradiction was for proving the $\Leftarrow$ dirction:

About $\Leftarrow$

Asumme $x > 0$ and get a contradiction:

For $x > 0$ the remember that the opposite of "$x < \epsilon$ for all $\epsilon > 0$" is ""there exists at least one $\epsilon > 0$ so that $x \ge \epsilon$". Can you find such an $\epsilon$? What if you picked an $\epsilon $ so that $0 < \epsilon < x$?

Are there any such numbers between $0$ and $x$? If there were, would that be a contradiction to $x \ge \epsilon$?

1:

Well, $\epsilon = \frac x 2$ is such a number $0 < \frac x2=\epsilon < x$. THere's infinitely many others, but this one is ... right there.

2:

Obviously if $0 < \epsilon = \frac x2 < x$ then that is an $\epsilon > 0$ where $x < \epsilon $ is not true.