Well consider the Law of Iterated expectation $$\mathbb E[ \mathbb E[Y\lvert W]]= \mathbb E[W]$$

and apply it with $Y=\mathbf u$ and $W = \mathbf X$, to get $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]]= \mathbb E[\mathbf u]$ and $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]] = \mathbb E[0] = 0$, hence $\mathbb E[\mathbf u]=0$.

Consider a generalized version of Law of Iterated Expectation mentioned in the post A generalization of the Law of Iterated Expectations which states that

$$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$

and let $W_1 = \mathbf X\setminus \mathbf x_t$ (read $\mathbf X$ except the vector $\mathbf x_t$) and let $W_2=\mathbf x_t$ then

$$\mathbb E[\mathbb E[u_t\lvert \mathbf X]\lvert \mathbf x_t] = \mathbb E[\mathbb E[u_t\lvert \mathbf X\setminus \mathbf x_t,\mathbf x_t]\lvert \mathbf x_t] = \mathbb E[u_t \lvert \mathbf x_t]$$ and as before it must be 0 because $\mathbb E[u_t\lvert \mathbf X]=0$.

Use the same argument to get $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ and use this to conclude that

$$\mathbb E[u_t \mathbf x_s] = \mathbb E[\mathbb E[u_t \mathbf x_s\lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \mathbb E[u_t \lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \cdot 0 ] = 0$$

So the answer must be yes they are all valid statements and further more $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ as was used.

An intuitive statement of the statement $$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$ is that the least information set always dominates so you also have $$\mathbb E[\mathbb E[Y\lvert W_2] \lvert W_1, W_2] = \mathbb E[Y\lvert W_2]$$ again this is mentioned in the post referred to above.

Well consider the Law of Iterated expectation $$\mathbb E[ \mathbb E[Y\lvert W]]= \mathbb E[Y]$$

and apply it with $Y=\mathbf u$ and $W = \mathbf X$, to get $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]]= \mathbb E[\mathbf u]$ and $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]] = \mathbb E[0] = 0$, hence $\mathbb E[\mathbf u]=0$.

Consider a generalized version of Law of Iterated Expectation mentioned in the post A generalization of the Law of Iterated Expectations which states that

$$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$

and let $W_1 = \mathbf X\setminus \mathbf x_t$ (read $\mathbf X$ except the vector $\mathbf x_t$) and let $W_2=\mathbf x_t$ then

$$\mathbb E[\mathbb E[u_t\lvert \mathbf X]\lvert \mathbf x_t] = \mathbb E[\mathbb E[u_t\lvert \mathbf X\setminus \mathbf x_t,\mathbf x_t]\lvert \mathbf x_t] = \mathbb E[u_t \lvert \mathbf x_t]$$ and as before it must be 0 because $\mathbb E[u_t\lvert \mathbf X]=0$.

Use the same argument to get $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ and use this to conclude that

$$\mathbb E[u_t \mathbf x_s] = \mathbb E[\mathbb E[u_t \mathbf x_s\lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \mathbb E[u_t \lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \cdot 0 ] = 0$$

So the answer must be yes they are all valid statements and further more $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ as was used.

An intuitive statement of the statement $$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$ is that the least information set always dominates so you also have $$\mathbb E[\mathbb E[Y\lvert W_2] \lvert W_1, W_2] = \mathbb E[Y\lvert W_2]$$ again this is mentioned in the post referred to above.

Well consider the Law of Iterated expectation $$\mathbb E[ \mathbb E[Y\lvert W]]= \mathbb E[W]$$

and apply it with $Y=\mathbf u$ and $X = \mathbf X$, to get $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]]= \mathbb E[\mathbf u]$ and $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]] = \mathbb E[0] = 0$, hence $\mathbb E[\mathbf u]=0$.

Consider a generalized version of Law of Iterated Expectation mentioned in the post A generalization of the Law of Iterated Expectations which states that

$$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$

and let $W_1 = \mathbf X\setminus \mathbf x_t$ (read $\mathbf X$ except the vector $\mathbf x_t$) and let $W_2=\mathbf x_t$ then

$$\mathbb E[\mathbb E[u_t\lvert \mathbf X]\lvert \mathbf x_t] = \mathbb E[\mathbb E[u_t\lvert \mathbf X\setminus \mathbf x_t,\mathbf x_t]\lvert \mathbf x_t] = \mathbb E[u_t \lvert \mathbf x_t]$$ and as before it must be 0 because $\mathbb E[u_t\lvert \mathbf X]=0$.

Use the same argument to get $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ and use this to conclude that

$$\mathbb E[u_t \mathbf x_s] = \mathbb E[\mathbb E[u_t \mathbf x_s\lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \mathbb E[u_t \lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \cdot 0 ] = 0$$

So the answer must be yes they are all valid statements and further more $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ as was used.

An intuitive statement of the statement $$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$ is that the least information set always dominates so you also have $$\mathbb E[\mathbb E[Y\lvert W_2] \lvert W_1, W_2] = \mathbb E[Y\lvert W_2]$$ again this is mentioned in the post referred to above.

Well consider the Law of Iterated expectation $$\mathbb E[ \mathbb E[Y\lvert W]]= \mathbb E[W]$$

and apply it with $Y=\mathbf u$ and $W = \mathbf X$, to get $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]]= \mathbb E[\mathbf u]$ and $\mathbb E[ \mathbb E[\mathbf u \lvert \mathbf X]] = \mathbb E[0] = 0$, hence $\mathbb E[\mathbf u]=0$.

Consider a generalized version of Law of Iterated Expectation mentioned in the post A generalization of the Law of Iterated Expectations which states that

$$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$

and let $W_1 = \mathbf X\setminus \mathbf x_t$ (read $\mathbf X$ except the vector $\mathbf x_t$) and let $W_2=\mathbf x_t$ then

$$\mathbb E[\mathbb E[u_t\lvert \mathbf X]\lvert \mathbf x_t] = \mathbb E[\mathbb E[u_t\lvert \mathbf X\setminus \mathbf x_t,\mathbf x_t]\lvert \mathbf x_t] = \mathbb E[u_t \lvert \mathbf x_t]$$ and as before it must be 0 because $\mathbb E[u_t\lvert \mathbf X]=0$.

Use the same argument to get $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ and use this to conclude that

$$\mathbb E[u_t \mathbf x_s] = \mathbb E[\mathbb E[u_t \mathbf x_s\lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \mathbb E[u_t \lvert \mathbf x_s] ] = \mathbb E[ \mathbf x_s \cdot 0 ] = 0$$

So the answer must be yes they are all valid statements and further more $\mathbb E[u_t \lvert \mathbf x_s] = 0$ for all $s,t = 1,...,T$ as was used.

An intuitive statement of the statement $$\mathbb E[\mathbb E[Y\lvert W_1,W_2] \lvert W_2] = \mathbb E[Y\lvert W_2]$$ is that the least information set always dominates so you also have $$\mathbb E[\mathbb E[Y\lvert W_2] \lvert W_1, W_2] = \mathbb E[Y\lvert W_2]$$ again this is mentioned in the post referred to above.