Talk:Fréchet Space (Functional Analysis) is Complete Metric Space

Why is it not clear that $\sequence {x^n}_{n \mathop \in \N}$ is a sequence in $\R^\omega$?
Hi prime mover, thank you for reviewing. What is still not clear and what would be your suggestion?

It was already the best clearly stated that
 * Let $\sequence {x^n}_{n \mathop \in \N} := \tuple {x^0, x^1, x^2, \ldots}$ be a Cauchy sequence in $\R^\omega$.


 * So why use $x^n$ for it?

A Cauchy sequence is just a Cauchy sequence, not more or less. $x^n$ denotes the $n$-the sequence element, not a power of some undefined symbol $x$. In the space $\R ^\omega$, there is no multiplication, no power.

The expression $\sequence {x^n}_{n \mathop \in \N}$ has the same meaning as $\sequence {x_n}_{n \mathop \in \N}$. I am using the former instead of the latter, only since the latter is already used to express the elements in $\R ^\omega$.


 * That's where the problem is. Using $x^n$ when you really mean $x_n$ is ridiculous.

Now, I improved a Cauchy sequence to an arbitrary Cauchy sequence.

If this is still not satisfactory, please give me a suggestion for better notation.


 * Don't use $x^n$ for something that does not mean $x^n$, or if you do use it, don't explain it in terms of $x$ with the subscript of a sequence of numbers.

---Usagiop