Completeness of the set of convergent sequences

Solution 1:

1)First notice that a closed subset of a complete space is complete. Another way of understanding closed sets is that a closed set contains all its limit points. Equivalently, a subset $S$ is closed, if every convergent sequence in $S$ has its limit in $S$.

2. An element $(x_n)$ is a limit point of a set $S$, if every neighborhood of $(x_n)$ intersects $S$. It is not always the case that if $(x_n)$ is a limit point, that there exists a sequence converging to $(x_n)$; this is true for 1st-countable spaces (which includes metric spaces).

3)Now, try to show c is closed by showing that if $(x_n)$ is a limit point of c, then $(x_n)$ is in $S$, or that, if you have a convergent sequence in c, then its limit is also in c. The triangle inequality should help you there.

A general observation is that you can show the subspace $c$ is closed, by showing that its complement is open: consider a point $(x_n)$ in $\mathbb R^{\mathbb R} -c$ , i.e., a sequence of Reals that does not converge . Then show that there is an $\epsilon>0$ , so that the ball $B((x_n), \epsilon)$ (ball given in the $||.||_{\infty}$ metric ), so that no sequence in that ball converges. This should not be too hard; if a sequence $(x_n)$does not converge and $Sup|x_n-y_n| < \epsilon$ , then I think it is not too hard to show that $(y_n)$ does not converge either.