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

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.

• Actually it's more than just showing that $(x_i)$ is eventually non zero. The sequence $x_i=1/i$ is eventually non zero... in fact... it's never zero. And yet, in the construction of the real numbers, the sequence $x_i=1/i$ represents zero. Commented Jan 28, 2023 at 1:04
• @Lee Mosher That's true, but since $x_i =1/i$ represents zero, it is contained in $[(0)]$, isn't it? So my point still stands. Commented Jan 28, 2023 at 1:06
• I don't know what the point is that you want to stand... but... the point you should pay attention to is this: What distinguishes the Cauchy sequences that do represent zero from the Cauchy sequences that do not represent zero? Commented Jan 28, 2023 at 1:09
• Ah, I think I understand now. Just to check, he's saying that while a sequence such as $x_i=1/i$ is also eventually non-zero, we want our sequence to be non-zero in the sense that every element past a certain 'point' is greater than a fixed number (here $\varepsilon/2$), which is not true for the aforementioned $1/i$. Is that correct? Commented Jan 28, 2023 at 1:15
• Yes, that's a good way of putting it. Commented Jan 28, 2023 at 15:31

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 $$[(x_k)]=[(0)]$$.

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.

• About the error, I copied this word for word from what he wrote, so he might've made a mistake there. As for your working, I understand every step except your very first one: why can we take $n>m$? I thought his statements only guarantee that $n>N$, not that $n>m>\text{max}(M,N)$. Could you explain please? Commented Jan 28, 2023 at 1:20
• Never mind, I see it now. Thank you for the answer. Commented Jan 28, 2023 at 1:21