Question about notation of sequences and equivalence classes.

Question

In these notes (see pg3 second-to-last paragraph), what does $d(x_k,x^\ast_{N_k})$ mean? The term $x_k$ lies in $X$, but $x^\ast_{N_k}$ is a class of Cauchy sequences in $X$. Should I take $d(x_k,x_{N_k}^\ast)$ to mean the distance function on $X^\ast$ and identify $x_k$ with the constant Cauchy sequence whose terms are all $x_k$? If so, I can see how one deduces that $\{x_k\}$ is at distance zero from $\{x_n^\ast\}$, but I don't see how to show that $\{x_k\}$ is Cauchy.

Answer 1

Yes, you’re to identify $x_k$ with the equivalence class of the constant sequence $\langle x_k,x_k,x_k,\ldots\rangle$ and take the distance in $X^*$. Given $k\in\Bbb Z^+$, there is an $m_k\in\Bbb N$ such that $d(x_\ell^*,x_n^*)<\frac1k$ whenever $\ell,n\ge m_k$. Without loss of generality we may assume that $m_k\ge k$ for each $k\in\Bbb Z^+$ and that the sequence $\langle m_k:k\in\Bbb Z^+\rangle$ is increasing. But then for $\ell,n\ge m_k$ we must have

$$d(x_\ell,x_n)\le d(x_\ell,x_{m_\ell}^*)+d(x_{m_\ell}^*,x_{m_n}^*)+d(x_{m_n}^*,x_n)<\frac1\ell+\frac1k+\frac1n\le\frac3k\;,$$

and $\langle x_n:n\in\Bbb Z^+\rangle$ is a Cauchy sequence in $X$.