Space of convergent sequences

Example

Let $c$ be the set of convergent sequences $x = (x_n)_{n \ge 1}$ of real (or complex) numbers, with the norm $\norm{x}_\infty = \sup_{n} \abs{x_n}$. Then $c$ is a sequence space and a Banach space.

Check that $c$ is a sequence space: the sum of two convergent sequences converges, and so does a scalar multiple of a convergent sequence.

Check that $c$ is complete: convergent sequences are bounded, so $c$ is a subspace of the Banach space $\ell^\infty$ of bounded sequences. So it suffices to show that $c$ is closed in $\ell^\infty$. Let $x^{(j)} \in c$ with $\norm{x^{(j)} - x}_\infty \to 0$ for some $x \in \ell^\infty$. For all $m, n$ and $j$, $$\abs{x_m - x_n} \le 2 \norm{x - x^{(j)}}_\infty + \abs{x^{(j)}_m - x^{(j)}_n}.$$ Choose $j$ to make the first term small, then use that $x^{(j)}$ is a Cauchy sequence. So $x$ is a Cauchy sequence of numbers, hence convergent, and $x \in c$.