Finding a bijection between $c$ (set of real sequences that converge) and $c_0$ (set of real sequences that converge to $0$) Actually I have to prove that there is an uniform homeomorphism between $c$ and $c_0$ (regarding the $sup$ metric), but I can't even find a bijection between them in the first place. I tried to construct some "rule" (I know not all functions are defined by a few rules, but I think an uniform homeomorphism is intuitive enough to be) by combining additive and multiplicative rules (multiplying each term of the sequence by a constant/function that depends on the sequence's index and summing it to another constant/function that depends on the sequence's index) but it doesn't work. The function is never bijective.
 A: Here's a simpler example: given a sequence $\alpha=(a_0, a_1, a_2, . . .)$ converging to $0$, let $\alpha'=(a_0+a_1, a_0+a_2, a_0+a_3, . . .)$. Then 


*

*The map $F:\alpha\mapsto\alpha'$ is a map from $c_0$ to $c$.

*$F$ is surjective: given any $\beta=(b_0, b_1, b_2, . . . )\in c$ converging to $z$, we have $\beta=F(z, b_0-z, b_1-z, b_2-z, . . . )$, and $(z, b_0-z, b_1-z, b_2-z, . . . )\in c_0$.

*$F$ is injective: Suppose $F(a_0, a_1, a_2, .. . )=F(a_0', a_1', a_2', . . . )$. Then first of all, we must have $a_0=a_0'$ since $F(a_0, a_1, a_2, .. . )$ and $F(a_0', a_1', a_2', . . . )$ converge to $a_0$ and $a_0'$ respectively. But now this implies that $a_1=a_1'$, since we must have $a_0+a_1=a_0'+a_1'$. Similarly, $a_2=a_2'$, etc.
Showing that $F$ is a homeomorphism is no harder.
A: As you're aware, every convergent sequence has a unique translation that converges to $0.$ Unfortunately, translations simply provide an infinite family of non-surjective injections $c_0\to c,$ so that isn't enough, on its own. However, these do allow us to construct the desired homeomorphism, with a bit of tweaking.
Given $r\in\Bbb R,$ let $c_r$ be the set of real sequences converging to $r.$ Clearly, then, $c_r$ is simply a translation of $c_0,$ and $c$ is the (disjoint) union of all the sets $c_r.$ Let $\Bbb{R^N}$ denote the set of all sequences of real numbers. Readily, $\Bbb{R^N}$ is a vector space over $\Bbb R,$ so it makes sense to think of such sequences as vectors with real entries and all natural indices, such as $$\vec a=\langle a_1,a_2,a_3,a_4,...\rangle.$$ Given $k\in\Bbb N,$ we define the sequence $\vec e^{(k)}$ by $$e^{(k)}_n=\begin{cases}1 & n=k\\0 & n\ne k.\end{cases}$$ We can readily show that $\left\{\vec e^{(k)}:k\in\Bbb N\right\}$ comprises a basis for $\Bbb{R^N},$ and that for any $\vec a\in\Bbb{R^N},$ we have $$\vec a=\sum_{k\in\Bbb N}a_k\vec e^{(k)}.$$
Consider the "right-shift operator" $R:\Bbb{R^N}\to\Bbb{R^N}$ defined as the unique linear operator on $\Bbb{R^N}$ such that $\vec e^{(k)}\mapsto\vec e^{(k+1)}$ for all $k\in\Bbb N.$ More explicitly, $$R(\vec a)=\langle0,a_1,a_2,a_3,a_4,...\rangle$$ for any $\vec a\in\Bbb{R^N}.$ It is fairly straightforward to show that if $\vec a\in c_0,$ then $R(\vec a)\in c_0,$ as well.
Next, consider the function $S:\Bbb{R\times R^N\to R^N}$ given by $$S(r,\vec a)=r\vec e^{(1)}+\vec a=\langle r+a_1,a_2,a_3,a_4,...\rangle.$$ Readily, if $\vec a\in c_0,$ then for any $r\in\Bbb R,$ we have $S(r,\vec a)\in c_0.$
Next, consider the function $T:\Bbb{R\times R^N\to R^N}$ given by $$T(r,\vec a)=\vec a-r\sum_{k\in\Bbb N}\vec e^{(k)}=\langle a_1-r,a_2-r,a_3-r,a_4-r,...\rangle.$$ Once again, it is straightforward to show that if $\vec a\in c_r,$ then $T(r,\vec a)\in c_0.$
Finally, we define a function $L:c\to\Bbb R$ by $$L(\vec a)=\lim_{n\to\infty} a_n.$$ Then for each $\vec a\in c,$ we have $\vec a\in c_{L(\vec a)}.$
Now, define a function $f$ on $c$ by $$f(\vec a)=S\biggl(L(\vec a),R\Bigl(T\bigl(L(\vec a),\vec a\bigl)\Bigr)\biggl).$$ The idea, here, is that $f$ sends $\vec a$ to its translated counterpart in $c_0,$ then right-shifts the result, then appends the limit of $\vec a$ as the new first component (to allow us to recover the original sequence). By the work above, we have $f:c\to c_0,$ and it's fairly straightforward to prove that $f$ is the desired uniform homeomorphism.
