There are already good answers. Here we merely use different words and stress different things.

1. The $n$'th Fock space ${\cal H}_n$ is the vector space spanned by $n$ particle states. In other words, if we omit the zero-vector, then ${\cal H}_n\backslash\{0\}$ is the space of $n$ particle states. Be aware that we will often casually refer to ${\cal H}_n$ as the space of $n$ particle states, although the zero-vector $0$ is not an $n$ particle state.

2. For negative $n<0$, let us define ${\cal H}_{n<0}:=\{0\}$ to be the zero vector space.

3. The Fock space $${\cal H}_0~=~ \mathbb{C} |\Omega \rangle$$ of $0$ particles is spanned by the vacuum state$^1$ $|\Omega\rangle$. Be aware that the vacuum state $|\Omega\rangle$ is often written as $|~ \rangle$ or $|0\rangle$. The latter should not be confused with the 1 particle state $|\vec{p}\!=\!\vec{0}\rangle$ nor the zero-vector $0$.

4. The annihilation operator $$ a_{\vec{p}}:{\cal H}_n\to {\cal H}_{n-1}$$ goes from the Fock space ${\cal H}_n$ of $n$ particles to the Fock space ${\cal H}_{n-1}$ of $n\!-\!1$ particles. It removes a particle with momentum $\vec{p}$. If none of the $n$ particles have momentum $\vec{p}$, then the result will be the zero-vector $0$; not an $n\!-\!1$ particle state.

5. OP's three equations can alternatively be written as $$a^{\dagger}_{\vec{p}} | \Omega \rangle ~=~ |\vec{p}\rangle, \qquad a_{\vec{p}} | \vec{p} \rangle ~=~ |\Omega\rangle, \qquad a_{\vec{p}}|\Omega \rangle~=~0, $$ respectively.

--

$^1$ Here we will for simplicity only consider the perturbative vacuum.

There are already good answers. Here we merely use different words and stress different things.

1. The $n$'th Fock space ${\cal H}_n$ is the vector space spanned by $n$ particle states. In other words, if we omit the zero-vector, then ${\cal H}_n\backslash\{0\}$ is the space of $n$ particle states. Be aware that we will often casually refer to ${\cal H}_n$ as the space of $n$ particle states, although the zero-vector $0$ is not an $n$ particle state.

2. For negative $n<0$, let us define ${\cal H}_{n<0}:=\{0\}$ to be the zero vector space.

3. The Fock space $${\cal H}_0~=~ \mathbb{C} |\Omega \rangle$$ of $0$ particles is spanned by the vacuum state$^1$ $|\Omega\rangle$. Be aware that the vacuum state $|\Omega\rangle$ is often written as $|~ \rangle$ or $|0\rangle$. The latter should not be confused with the 1 particle state $|\vec{p}\!=\!\vec{0}\rangle$ nor the zero-vector $0$.

4. The annihilation operator $$ a_{\vec{p}}:{\cal H}_n\to {\cal H}_{n-1}$$ goes from the Fock space ${\cal H}_n$ of $n$ particles to the Fock space ${\cal H}_{n-1}$ of $n\!-\!1$ particles. It removes a particle with momentum $\vec{p}$. If none of the $n$ particles have momentum $\vec{p}$, then the result will be the zero-vector $0$; not an $n\!-\!1$ particle state.

5. OP's three equations can alternatively be written as $$a^{\dagger}_{\vec{p}} | \Omega \rangle ~=~ |\vec{p}\rangle, \qquad a_{\vec{p}} | \vec{p} \rangle ~=~ |\Omega\rangle, \qquad a_{\vec{p}}|\Omega \rangle~=~0, $$ respectively.

--

$^1$ Note that the notions of vacuum $|\Omega\rangle$ and particle number depend on the choice of creation and annihilation operators. They may not be invariant under e.g. Bogoliubov transformations, and non-perturbative effects. See also the notion of ground state.

There are already good answers. Here we merely use different words and stress different things.

1. The $n$'th Fock space ${\cal H}_n$ is the vector space spanned by $n$ particle states. In other words, if we omit the zero-vector, then ${\cal H}_n\backslash\{0\}$ is the space of $n$ particle states. Be aware that we will often casually refer to ${\cal H}_n$ as the space of $n$ particle states, although the zero-vector $0$ is not an $n$ particle state.

2. For negative $n<0$, let us define ${\cal H}_{n<0}:=\{0\}$ to be the zero vector space.

3. The Fock space $${\cal H}_0~=~ \mathbb{C} |\Omega \rangle$$ of $0$ particles is spanned by the vacuum state $|\Omega\rangle$. Be aware that the vacuum state $|\Omega\rangle$ is often written as $|0\rangle$, which should not be confused with the 1 particle state $|\vec{p}\!=\!\vec{0}\rangle$ nor the zero-vector $0$.

4. The annihilation operator $$ a_{\vec{p}}:{\cal H}_n\to {\cal H}_{n-1}$$ goes from the Fock space ${\cal H}_n$ of $n$ particles to the Fock space ${\cal H}_{n-1}$ of $n\!-\!1$ particles. It removes a particle with momentum $\vec{p}$. If none of the $n$ particles have momentum $\vec{p}$, then the result will be the zero-vector $0$; not an $n\!-\!1$ particle state.

5. OP's three equations can alternatively be written as $$a^{\dagger}_{\vec{p}} | \Omega \rangle ~=~ |\vec{p}\rangle, \qquad a_{\vec{p}} | \vec{p} \rangle ~=~ |\Omega\rangle, \qquad a_{\vec{p}}|\Omega \rangle~=~0, $$ respectively.

There are already good answers. Here we merely use different words and stress different things.

1. The $n$'th Fock space ${\cal H}_n$ is the vector space spanned by $n$ particle states. In other words, if we omit the zero-vector, then ${\cal H}_n\backslash\{0\}$ is the space of $n$ particle states. Be aware that we will often casually refer to ${\cal H}_n$ as the space of $n$ particle states, although the zero-vector $0$ is not an $n$ particle state.

2. For negative $n<0$, let us define ${\cal H}_{n<0}:=\{0\}$ to be the zero vector space.

3. The Fock space $${\cal H}_0~=~ \mathbb{C} |\Omega \rangle$$ of $0$ particles is spanned by the vacuum state$^1$ $|\Omega\rangle$. Be aware that the vacuum state $|\Omega\rangle$ is often written as $|~ \rangle$ or $|0\rangle$. The latter should not be confused with the 1 particle state $|\vec{p}\!=\!\vec{0}\rangle$ nor the zero-vector $0$.

4. The annihilation operator $$ a_{\vec{p}}:{\cal H}_n\to {\cal H}_{n-1}$$ goes from the Fock space ${\cal H}_n$ of $n$ particles to the Fock space ${\cal H}_{n-1}$ of $n\!-\!1$ particles. It removes a particle with momentum $\vec{p}$. If none of the $n$ particles have momentum $\vec{p}$, then the result will be the zero-vector $0$; not an $n\!-\!1$ particle state.

5. OP's three equations can alternatively be written as $$a^{\dagger}_{\vec{p}} | \Omega \rangle ~=~ |\vec{p}\rangle, \qquad a_{\vec{p}} | \vec{p} \rangle ~=~ |\Omega\rangle, \qquad a_{\vec{p}}|\Omega \rangle~=~0, $$ respectively.

--

$^1$ Here we will for simplicity only consider the perturbative vacuum.