Understanding proof that multiplicative inverses exist in $\mathbb{R}$

Question

Context: For this course, the real numbers were constructed using Cauchy sequences, i.e., each real number is the equivalence class of a Cauchy sequence of rational numbers.

I'll only post the relevant part of the proof here, as its not used again and the rest is clear to me.

The proof: Suppose $\left[\left(x_i\right)\right]\neq \left[\left(0\right)\right]$ where $(0)=(0,0,0,\ldots)$. This implies that $\exists \varepsilon>0$ such that $\forall N$ $\exists n>N$ such that $\left|x_n\right|>\varepsilon$. But $x_n$ is Cauchy, so $\exists M$ such that $\forall k,\ell > M$, $\left|x_k - x_{\ell}\right|<\varepsilon/2$.

Now take $m>\text{max}\{M,N\}$. Then $\left|x_m\right|\geq \varepsilon/2$.

For context again, he's doing all this to show that $\left(x_i\right)$ is eventually non-zero.

My problem: For one, I don't understand how $\left|x_m\right|\geq \varepsilon/2$ follows from what he stated beforehand. I tried rearranging and trying to get it from the Cauchy criterion but I don't see it.

Second, I don't understand why this even needs to be shown. If $\left[\left(x_i\right)\right]\neq \left[\left(0\right)\right]$, then doesn't it follow that $\left(x_i\right)$ is eventually non-zero? Since the statement isn't that $\left(x_i\right)\neq (0)$, but rather that none of the sequences equivalent to $\left(x_i\right)$ equal any of the sequences equivalent to $(0)$. It should follow naturally that $\left(x_i\right)$ must be 'eventually non-zero', no?

Note: I can't ask the professor because this course isn't actually running at the moment. I'm studying ahead for it.

Any help would be much appreciated.

Answer 1

Let $m>M$. As described, there exists $n>M$ with $|x_n|>\epsilon$. As $m$ and $ n$ are $>M$, we have $|x_m-x_n|<\epsilon/2$. Then $|x_m|\ge |x_n|-|x_m-x_n|>\epsilon/2$ by triangle inequality.

Note that “sequence is eventually non-zero” ($\exists N, \forall n>N, x_n\ne0$) is not enough; we need “does not converge to $0$” aka. “Is not a zero sequence”. For example, the sequence given by $x_n=\frac 1n$ is eventually (in fact, immediately) non-zero, but $[(x_k)]=[(0)]$.

Answer 2

First, there appears to be an error in what you wrote. You wrote the expression $\max(M, N)$, but you did not define $N$ in this context. $N$ does occur earlier on the page, but it is quantified over, and $\max(M, N)$ occurs outside the scope of the $\forall N$. I will instead assume instead that $m > M$.

For one, I don't understand how $\left|x_m\right|\geq \varepsilon/2$ follows from what he stated beforehand.

Take some $n > m$ such that $|x_n| > \epsilon$. Then $|x_n - x_m| \leq \frac{\epsilon}{2}$. Now $x_n = (x_n - x_m) + x_m$, so $\epsilon < |x_n| = |(x_n - x_m) + x_m| \leq |x_n - x_m| + |x_m| \leq \frac{\epsilon}{2} + |x_m|$. We used the triangle inequality for one of these steps.

Then $\epsilon < \frac{\epsilon}{2} + |x_m|$, and therefore $\frac{\epsilon}{2} < |x_m|$, as required.

Second, I don't understand why this even needs to be shown. If $\left[\left(x_i\right)\right]\neq \left[\left(0\right)\right]$, then doesn't it follow that $\left(x_i\right)$ is eventually non-zero?

It does indeed follow. However, a sequence can have all its elements nonzero and still converge to $0$. So the fact that all but finitely many elements of $x$ are nonzero is actually not what we’re trying to show. We’re trying to show there is some $w > 0$ such that all but finitely many elements of $x$ have absolute value at least $w$. This is the result we really need to show that the inverse of $[x]$ exists and can (almost) be constructed pointwise.