Fréchet Space (Functional Analysis) is Complete Metric Space

Theorem
Let $\struct {\R^\omega, d}$ be the Fréchet space on $\R^\omega$.

Then $\struct {\R^\omega, d}$ is a complete metric space.

Proof
From Fréchet Space (Functional Analysis) is Metric Space, $\struct {\R^\omega, d}$ is a metric space.

It remains to be demonstrated that $\struct {\R^\omega, d}$ is complete.

Let $ \family {x^n} _n $ be a Cauchy sequence in $\R^\omega$, where $x_n:=\family {x^n_i} _i$.

Lemma 1
For each $j\in\N$, the real sequence $ \family {x^n _j} _n $ is convergent.

Proof
Let $j\in\N$.

In view of Cauchy's convergence criterion, it suffices to show that $ \family {x^n _j} _n $ is a Cauchy sequence.

Let $\varepsilon>0$ be arbitrary.

Since $ \family {x^n} _n $ is a Cauchy sequence, there is a $N\in\N$ such that:
 * $\forall m, n \in \N: m, n \geqslant N:\, d(x^m,x^n)<\frac{\varepsilon}{2^j}$

Thus:
 * $ \dfrac {2^{-j} \size {x^m _j - x^n _j} } {1 + \size {x^m _j - x^n _j} } \leqslant \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x^m _i - x^n _i} } {1 + \size {x^m _i - x^n _i} } = d(x_m, x_n) < \frac{\varepsilon}{2^j}$

which implies:
 * $ \size {x^m _j - x^n _j} < \varepsilon$

Let $x^\infty := \family {x^\infty _i}_i \in \R^\omega$, where $x^\infty _i :=\lim _{n\to\infty} x^n _i$ for each $i$.

It remains to show that $ x^\infty $ is the limit of $ \family {x^n} _n $.