About the implication of $\epsilon>0$ being arbitrary.

Question

It is indeed a very basic question. I always read proofs stating that since $\epsilon>0$ is arbitrary, then some inequality holds. Also, I've been accepting it without problems. Now, I just figured out that this point is not completely clear to me.

For example: as $\epsilon>0$ is arbitrary, we have that

1. $f(x)>z-\epsilon \implies f(x)\geq z$

2. (I will be more specific here) Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a function, $z\in\mathbb{R}$ a given point and assume $\forall \epsilon>0, [f^{-1}(z-\epsilon,z+\epsilon)]^c=\{x\in \mathbb{R}:\mid f(x)-z\mid \geq \epsilon \}:=A_{\epsilon}$ is countable (on a crazy topology on domain called co-countable). I want to show that as $\epsilon>0$ is arbitrary then actually all $\mathbb{R}$ without the point $x:f(x)=z$ is countable (again, $\mathbb{R}$ together with the crazy co-countable topology). Looking at $\{x\in \mathbb{R}:\mid f(x)-z\mid \geq \epsilon \}$ if $f(x)=z$ were possible, $0\geq \epsilon>0$ which is a contradiction.

Both assertions are acceptable to me, but I can't just put (intuitively) $\epsilon=0$ and infer the results. How would you properly argue it?

Answer 1

Well to be absolutely clear, we must specify the conditions are true for all $\epsilon > 0$.

So for one. If $f(x) > z-\epsilon$ for all $\epsilon \ge 0$, then $f(x) < z$ is impossible. If $f(x) < z$ were possible then the we'd have $\Delta = z - f(x) > 0$. So $f(x) > z - \Delta = z-(z-f(x))= f(x)$ which is impossible. So as $f(x) < z$ is impossible, $f(x)\ge z$ must be true.

That's it. Furthermore $f(x)= z$ is possible. As $z> z-\epsilon$ for all $\epsilon$. And $f(x) > z$ is possible as $z > z-\epsilon$ so if $f(x) > z$ we'd also have $f(x) > z-\epsilon$.

Number two is actually different.

If $\epsilon$ actually is arbitrary we could take $\epsilon$ to be enormous. This is impossible. If $y$ is a value that must be at least as far away as any possible value of $\epsilon$ then $|f(y)-z|\ge |f(y)-z| + 1$ which is impossible.

So... I actually have to see exactly what was being asked and when was $\epsilon$ introduced for that to make any sense.

.....

However if the question was "We can find some $\epsilon > 0$ so that $y\in \{x||f(x)-z|\ge \epsilon\}$" then in this case $\epsilon$ was NOT arbtrary. But this would mean $|f(y) - z| \ge \epsilon >0$ so $|f(y)-z|\ne 0$ so $f(y) - z \ne 0$ and $f(y) \ne z$.

==== with more info of the second example ===

Well, you can't show that all $\mathbb R$ without the point $x$ is countable because that would mean all $\mathbb R$ is countable.

Now heres a thing. For any $\epsilon > 0$ we can find an $n \in \mathbb N$ so that $n > \frac 1\epsilon$ and $\epsilon > \frac 1n >0$.

We can take the countable union $\cup_{n\in \mathbb N} A_{\frac 1n}$ is countable. Now for any $y \ne x$ we have $|f(y) -f(x)|_{\mathscr T}> 0$ so there is an $k\in \mathbb N: |f(y)-f(x)|_{\mathscr T} > \frac 1k > 0$. So $y \in A_{\frac 1k}\subset \cup_{n\in \mathbb N} A_{\frac 1n}$.

So $\mathbb R\setminus\{x\}\subset \cup_{n\in \mathbb N} A_{\frac 1n}$ so $\mathbb R\setminus\{x\}$ is countable. This is a contradiction.

So such a metric $|a - b|_{\mathscr T} = d(a,b)$ simply can not exist on $\mathbb R$.

(I'm not sure in what context this question is in but... such a Topology can not exist on the real numbers.)

(But I feel we hae really gotten off on a tangent.)

Answer 2

In many proofs you do not know that the inequality holds for $\epsilon =0$. You only know it for every $\epsilon >0$ and then you take the limit.

Proof is by contradiction. For example, in 1) suppose $f(x) <z$. Take $\epsilon =\frac {z-f(x)} 2$ and check that the hypothesis fails.

In 2) if the hypothesis holds for one value of $\epsilon >0$ we can say that $f(y) \neq z$.

Answer 3

I would say that $$ f(x)>z-\epsilon \implies f(x)\geq z $$ is false.
Instead, $$ (\forall \epsilon > 0)(f(x)>z-\epsilon) \implies f(x)\geq z $$ is what should be stated.