$F$ is a set, $A$ and $B$ and finite subsets of $F$, $k$ is a commutative ring, if $A$ is a subset of $B$, considering the canonical map $k[B] \to k[A]$, that is mapping all elements of $B-A$ to zero. for example $k[x_1,x_2]\to k[x_1]$, it means mapping $x_2$ to zero, and mapping $x_1$ to itself.
what is inverse limit $\varprojlim (k[B] \to k[A]) $ in Ring category? here, $A \subset B$ and both are finite sets of $F$.
what is inverse limit $\varprojlim (k[B] \to k[A]) $ in Graded Ring category? here, $A \subset B$ and both are finite sets of $F$.
I think when $F$ is a finite set, the answer to these two questions is $k[F]$. If $F$ is an infinite set, what are the answers and why?
I think for graded ring case, the answer should be $\bigoplus A_d$, here $A_d$ should be formal k-linear combinations of the degree $d$ monomials in the generators $x_a$, here $a\in F$. I do not know how to prove it is an inverse limit.
for ungraded ring case, I think the answer should be formal series of the form $\sum_{d\geq 0} x_d$, here $x_d \in A_d$, but I am not sure.
Please tell me which step you think is unclear, if you want to close my question.
I'm going to have $f_{AB}$ denote the map $k[B]\to k[A]$ that you defined above.
If $F$ is any set, even if it is infinite, we can still define the ring of polynomials $k[F]$ in the variables $x_\alpha$ for $\alpha\in F$. Similarly, we can define the ring of formal power series in these variables which I will denote $k[[F]]$ (which I think is the same as the ring you predicted for the ungraded case).
For the ungraded case, let $R$ denote the subring of $k[[F]]$ where elements satisfy that for each finite subset $A\subseteq F$, the number of monomials that contain only variables $x_\alpha$ for $\alpha\in A$ with a nonzero coefficient is finite. That is, there are finitely many monomials of the form $\prod_{\alpha\in A} x_{\alpha}^{c_\alpha}$ with non-zero coefficient. Let $\pi_A$ then be the projection from $R$ onto $k[A]$ (which is well defined by the construction of $R$).
To show that $R$ together with the projections is the inverse limit we do the following:
First, check that for each finite $A\subseteq B\subseteq F$ where $A$ and $B$ are finite, $\pi_A=f_{AB}\circ \pi_B$
Next, show that the universal property holds. Suppose $Y$ is another ring and for each finite $A\subseteq F$, $\psi_A:Y\to k[A]$ is a ring homomorphism such that for each $A\subseteq B\subseteq F$ ($A$,$B$ finite), $\psi_A=f_{AB}\circ \psi_B$. We want to show that for each such $(Y,\psi_A)$ there exists a unique $u:Y\to R$ such that $\psi_A=\pi_A\circ u$ for each $A$.
To show uniqueness, take some element $y\in Y$ and use the condition that for each $A$, $\psi_A=\pi_A\circ u$ to determine what the coefficients of $u(y)$ must be.
For existence, show that the map you defined above satisfies the condition (Make sure to show that the coefficients gives an element of R).
Notice that in the case where F is finite, $R=k[F]$
For the case of the graded ring category, I think your prediction (which is equivalent to the subring of $k[[F]]$ where each element has the degrees of the monomial terms uniformly bounded) works. To show this, you use the same method to prove it, only this time you will need to work with the gradings.