Is my proof correct

Question

Just posted a similar question, but my proof had ambiguity in it so I rewrote it just so I can check if my reasoning is correct...

Proposition:

Let $f$ be a function defined as $f(x) = x$ if $x$ is rational and $f(x) = 0$ otherwise. The limit of $f(x)$ as $x$ approaches any number $a \neq 0$ does not exist.

My proof:

Suppose by contradiction that there is a limit $L$ for $f(x)$, as $x$ approaches $a \neq 0$. By definition of limit, there is a $\delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $\epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller. Let's take a irrational number $x$ in such an neighborhood of $a$. That would imply $|L| < \epsilon$. We know that we can find a $y$ in the same interval as $x$ such that $|y - a|<|x - a|$ implies that $|f(y) - L| < |L| < \epsilon$. Since there are infinitely many numbers like that who happen to be rational, we have that $|y - L| < |L|$. But we also have infinitely many irrational numbers $z$ for which $|z -a| < |y - a|$ such that $|f(z) - L| < |y - L| < |L| < \epsilon$, but this is absurd. We conclude then that there is no limit $L$. ■

Answer 1

It is still not correct. You wrote “there is a $\delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $\varepsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller”. Don't you see that it is rather strange, to say the least, that, after “there is a $\delta$-neighborhood”, $\delta$ is not mentioned again? And what does “gets smaller” mean in a formal proof?

Answer 2

By definition of limit, there is a $\delta$-neighborhood of $a$ in which, as $|x - a| > 0$ gets smaller, in a $\epsilon$-neighborhood of $L$, $|f(x) - L|$ also gets smaller.

By the definition of limit, for every $\epsilon>0$ there is a $\delta$-neighborhood of $a$ such that if $\delta>|x-a|>0$, then $|f(x)-L|<\epsilon$. You cannot necessarily say that the smaller $|x-a|$ is the smaller $|f(x)-L|$ is.

We know that we can find a $y$ in the same interval as $x$ such that $|y−a|<|x−a|$ implies that $|f(y)−L|<|L|<\epsilon$.

Again, this is not true because of what I said above. What you can say is that if $|y-a|<\delta$ then $|f(y)-L|<\epsilon$.

Also notice that you have not used the fact that $a \neq 0$. This will have to be used somewhere, because $f(x)$ is in fact continuous at $0$.

What you now have is an irrational number $x$ in a $\delta$-neighborhood of $a$, which tells you $|L|<\epsilon$. You also have a rational number $y$ in the same $\delta$-neighborhood of $a$, which tells you $|y-L|<\epsilon$. Intuitively, the first inequality tells you that $L$ is close to $0$, and the second tells you that $L$ is close to $y$ which is close to $a \neq 0$.

To make this intuition rigorous, notice that for every $\epsilon>0$, we can find such a $\delta$-neighborhood and an irrational $x$ in that neighborhood which tells us $|L|<\epsilon$. The fact that $|L|<\epsilon$ for every positive $\epsilon$ implies $L=0$.

Now we can use $L=0$ in the other inequality. For every $\epsilon>0$, we can find a $\delta$-neighborhood and a rational $y$ in that delta neighborhood which tells us $|y-L|<\epsilon$, i.e., $|y|<\epsilon$. Then $|a|=|a-y+y| \leq |a-y|+|y|<\delta+\epsilon$. I claim we can always choose $\delta$ so that it is smaller than $\epsilon$ (taking a smaller interval around $a$ will never hurt), so $|a|<2\epsilon$. But this is true for every positive $\epsilon$, so $a=0$. There is the contradiction.