Understanding proof that Cauchy sequence of linear maps converges

Question

This is basically just saying that the space of linear map is a complete space as all Cauchy sequence converges.

The part that I don't understand is the second paragraph starting with "consider any $x \in R^n$. We are given that the sequence of linear maps are Cauchy, but why would the sequence of vectors in $R^m$ also be cauchy if we evaluate the sequence of maps at any random point.

Answer 1

As the proof says, $$\|A_p(x) - A_q(x)\| = \|(A_p - A_q)(x)\| \le \|A_p - A_q\|\|x\|.$$ This is sufficient to show $(A_p(x))$ is Cauchy, as I will demonstrate.

If $x = 0$, then $A_p(x) = 0$ for all $p$, so the sequence is clearly Cauchy in this case.

If $x \neq 0$, then $\|x\| > 0$. Fix $\varepsilon > 0$, and note that $\varepsilon / \|x\| > 0$ too. Since $(A_p)$ is Cauchy, there must exist some $N$ such that \begin{align*} m, n > N &\implies \|A_p - A_q\| < \frac{\varepsilon}{\|x\|} \\ &\implies \|A_p - A_q\|\|x\| < \varepsilon \\ &\implies \|A_p(x) - A_q(x)\| < \varepsilon, \end{align*} proving $(A_p(x))$ is Cauchy, as claimed.

Answer 2

Because $\mathbb{R}^m$ is complete, and in this case we do not have a sequence of linear functions, but a sequence in $\mathbb{R}^m$. We exploit the completeness of $\mathbb{R}^m$ and then we go back proving completeness of the space of linear maps.