Hilbert function for points in projective space In Lecture 13 of Harris's Algebraic Geometry, we have the Hilbert functions, $h_X$, for the following two cases:


*

*If $X=\{p_1,p_2,p_3\}\subseteq \mathbb{P}^2$ then there are two possible Hilbert functions:


$$h_X(m) = \begin{cases}2 & m=1\\ 3& m\geq 2\end{cases} \quad \text{and} \quad h_X(m)=3 \ \forall m\in \mathbb{N}$$ 
depending on whether the points are collinear or not, respectively [proof].


*

*If $X=\{p_1,p_2,p_3, p_4\}\subseteq \mathbb{P}^2$ then there are two possible Hilbert functions:


$$h_X(m) = \begin{cases}2 & m=1\\ 3& m= 2\\ 4 & m\geq 3\end{cases} \quad \text{and} \quad h_X(m)=\begin{cases}3 & m=1\\ 4& m \geq 2\end{cases}$$ 
depending on whether the points are collinear or not, respectively [proof].
Then it leaves the following generalization as an exercise:

Let $X\subseteq \mathbb{P}^n$ be a set of $d$ points. Show that for $m\geq d-1$, the Hilbert function $h_X(m)=d$.

I have no idea about how to approach. How can we generalize the specific arguments given in the proofs (like the case $m=2$)?  To begin with, how to approach the following problem:

If $X=\{p\}\in\mathbb{P}^n$ be a point then $h_X(m)=1$ for all $m\in \mathbb{N}$.

Thank you for your time, please let me know if you have any hints/solution for this problem.
 A: I asked this question when I was an undergraduate student and now I am a graduate student. Though my understanding hasn't improved much, here is an answer from Gathmann's notes (Lemma 6.1.4):
Note that $X$ is a zero-dimensional affine subscheme of $\mathbb{P}_k^n$. That is, $X=\mathrm{Spec}(R)$ for some $k$-algebra $R$.

Claim: $h_X(m)=d$ for large values of $m$, where $d$ is the number of points in $X$ counted with scheme-theoretic multiplicities.

Without loss of generality, we can assume that $X$ is irreducible, i. e. consists of only one point (but may have a non-trivial scheme structure) since in the reducible case $X = X_1 \sqcup \cdots \sqcup X_r$ with
$X_i = \mathrm{Spec}(R_i)$ for $i = 1, \ldots , r$ we have $R = R_1 \times \cdots \times R_r$. Moreover, by a change of coordinates we can assume that this point is the origin in $\mathbb{A}_k^n$. If $X = \mathrm{Spec}( k[x_1,\ldots, x_n]/I)$ we then must have $\langle x_1,\ldots , x_n\rangle = \sqrt{I}$ by the Nullstellensatz. That is, $x_i^m \in I$ for some $m$ and all $i$. Therefore, every monomial of degree at least $D = m\cdot n$
lies in $I$. In other words, $k[x_1 ,\ldots , x_n]/I$ has a $k$-basis of polynomials of degree less than $D$. But the space of such polynomials is finite-dimensional and equals $\dim_k\left(k[x_1,\ldots, x_n]/I\right)=d$ if $I(X)= \langle x_0^d\rangle$ where $I(X)$ be the homogenization of $I$, i.e. $I=I(X)|_{x_0=1}$.
Next, let $S(X)$ be the homogeneous coordinate ring (graded). Hence, for $m\geq m\cdot n$, there is an isomorphism $S(X)^{(m)} \to R$ as vector spaces over $k$ given by
\begin{align*}
\left(k[x_0,\ldots, x_n]/I(X)\right)^{(m)} & \to k[x_1,\ldots, x_n]/I\\
f & \mapsto f|_{x_0=1}
\end{align*}
and the inverse
\begin{align*}
k[x_1,\ldots, x_n]/I &\to \left(k[x_0,\ldots, x_n]/I(X)\right)^{(m)} \\
f &\mapsto f^h \cdot x_0^{m-\deg(f)}
\end{align*}
where $f^h$ denotes the homogenization of a polynomial. Note that the inverse map is well-defined as $k[x_1, \ldots,x_n]/I$ has a basis of polynomials of degree less than $D=m\cdot n$. That is,
$$h_X(m) = \dim_kS(X)^{(m)} = d$$
for large values of $m$.
