Is this proof of countinuity correct ($\epsilon \delta$)

Question

$c$ = space of all convergent sequences, $c_0$ = space of sequences convergent to zero

We consider both spaces with supremum norm.

$F: c \ni (a_n)_{n \in \mathbb{N}} \rightarrow (\lim a_n, \ a_1 - \lim a_n, \ a_2 - \lim a_n, ...) \in c_0$

$G: c_0 \ni (a_n)_{n \in \mathbb{N}} \rightarrow (a_1+a_2, \ a_1 +a_3, \ a_1 + a_4, ...) \in c$

I need to prove that both maps are continuous:

$x,y \in c$ or $x,y \in c_0$, $\ \ ||x-y|| = sup \{|x_n - y_n|, n \in \mathbb{N} \} < \delta$

$||F(x)-F(y)||= ||(\lim x_n - \lim y_n, \ x_1 - \lim x_n - (y_1 - \lim y_n), ...)|| = ||\lim (x_n-y_n), x_1-y_1- \lim (x_n-y_n), ..._)|| < \sup \{ |x_n - y_n|, n\in \mathbb{N} \}$, because $\lim (x_n-y_n)< \sup \{ |x_n - y_n|, n \in \mathbb{N} \}$ and obviously $x_n - y_n -\lim (x_n-y_n)< \sup \{ |x_n - y_n|, n \in \mathbb{N} \}$.

$||G(x)-G(y) || = ||(x_1 - y_1 + x_2 - y_2, x_1 - y_1 + x_3 - y_3, ...)||= \sup \{|x_1 - y_1 + x_n - y_n|, n \in \mathbb{N} \} \le \sup \{|x_1 - y_1| + |x_n - y_n|, n \in \mathbb{N} \} < \delta + |x_1 - y_1|$.

But delta cannot be dependent on y. How can I fix it?

Answer 1

So in the first functions all the limits go away, because they're converging to $0$. So you get $$sup ||F(x) - F(y)|| = sup_n |x_n - y_n| < \delta$$ (with maybe an added $0$ when $n =1$)

On the second function when you ask $$sup_n |x_n - y_n| < \delta$$ you get $$||G(x) - G(y)|| \leq |x_n - y_n| + |x_1 - y_1| < 2\delta$$ Remember that if $sup_n|x_n - y_n| < \delta$ then $|x_1 - y_1| < \delta$ !