A set of algebra homomorphisms from the real sequence space

Question

I was myself wondering the following question. Consider the sequence space $\mathbb{R}^{\mathbb{N}}$, that is

$\mathbb{R}^{\mathbb{N}} := \{ (x_{k})_{k=1}^{\infty} \; | \; x_{k} \in \mathbb{R} \}$.

It has a structure of a (real) commutative associative algebra with the unit, with the multiplication given by $(x_{k})_{k=1}^{\infty} \cdot (y_{k})_{k=1}^{\infty} = (x_{k} \cdot y_{k})_{k=1}^{\infty}$. Equivalently, $\mathbb{R}^{\mathbb{N}}$ is a direct product algebra $\prod_{k=1}^{\infty}\mathbb{R}$ or an algebra of maps $f: \mathbb{N} \rightarrow \mathbb{R}$.

Question: I am interested in the set of all algebra homomorphisms (preserving the unit) from $\mathbb{R}^{\mathbb{N}}$ to the algebra $\mathbb{R}$.

I have the following remarks/observations:

1. This set certainly contains all projections $\pi_{j}(x_{k})_{k=1}^{\infty} = x_{j}$, so this set has to be at least as big as $\mathbb{N}$.
2. If one replaces $\mathbb{R}$ with $\mathbb{Z}$, it contains only these projections. This is a well-known statement, see e.g. What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?. This means that in this case, the set in question is precisely $\mathbb{N}$. In the same question they claim that the result holds for any ring $R$ which is UFD with at least two distinct prime elements. But $\mathbb{R}$ has no prime elements.
3. Any homomorphism $f: \mathbb{R}^{\mathbb{N}} \rightarrow \mathbb{R}$ must act as a scalar multiple of a projection $\pi_{j}$ on sequences with finitely many non-zero elements. This is easy to see - if $\mathbf{e}_{a}$ is the sequence with $1$ on the $a$-th place and zeros everywhere else, one can show that $f(\mathbf{e}_{a}) \neq 0$ only for a single $a \in \mathbb{N}$, which follows from the equation $f(\mathbf{e}_{a} \cdot \mathbf{e}_{b}) = f(\mathbf{e}_{a}) \cdot f(\mathbf{e}_{b})$ and the fact that $\mathbf{e}_{a} \cdot \mathbf{e}_{b} = 0$ for $a \neq b$. Hence $f(x)_{k=1}^{n} = \lambda x_{a}$ for some $a \in \mathbb{N}$ and $\lambda \in \mathbb{R}$ for any sequence $(x_{k})_{k=1}^{n}$ with finitely many non-zero elements. However, one cannot repeat the argument from the reference in 2. to show that $f$ acts in this way on any sequence.

4. Clearly, the set in question is a subset of the dual space $(\mathbb{R}^{\mathbb{N}})^{\ast}$. Since $\mathbb{R}^{\mathbb{N}} = (\mathbb{R}^{\infty})^{\ast}$, where $\mathbb{R}^{\infty}$ is a set of all sequences with finitely many non-zero terms (that is $\mathbb{R}^{\infty} = \bigoplus_{k=1}^{\infty} \mathbb{R}$), we are looking for a subset of the double dual $(\mathbb{R}^{\infty})^{\ast \ast}$ which is supposedly an ugly space. It known that $\mathbb{R}^{\infty}$ embeds as a proper subspace into $(\mathbb{R}^{\infty})^{\ast \ast}$. Is there some explicit or useful description (or at least some nice examples) of elements of the double dual (that is of $(\mathbb{R}^{\mathbb{N}})^{\ast}$) which are not in this subspace? I was thinking that maybe some of these examples would be also an algebra homomorphism, proving that the set of all homomorphisms from $\mathbb{R}^{\mathbb{N}}$ to $\mathbb{R}$ is strictly bigger then $\mathbb{N}$.

Bonus Question: Instead of $\mathbb{N}$, I would like to know the result for general set $M$, that is consider a direct product algebra $\prod_{m \in M} \mathbb{R}$, or equivalently the algebra of all functions $f: M \rightarrow \mathbb{R}$, and ask the same question - what is the set of all homomorphisms from this algebra to $\mathbb{R}$? It is easy to show that for a finite $M$, it is isomorphic to $M$. Is there some answer for a general $M$?