Definition and properties of a half-space How does a half-space in $\mathbb{R}^m$ defined and which are its properties? How does the positive/negative half-space defined?
I would appreciate it if someone could provide some example as well.
About the properties of half-spaces I would also like to ask the following:
Are half-spaces convex and closed sets in general (meaning by definition)? If not, in which cases this happens?
Does convexity and the boundary condition (i.e. closed sets) in $\mathbb{R}^m$ suffice to assume the compactivity as a property of half-spaces?
 A: Let $a \in \mathbb{R}^m\setminus \{0\}$, $b \in \mathbb{R}$ and let $\langle\cdot,\cdot\rangle$ denote the standard dot product, i.e. $\langle x,y\rangle = x_1y_1+\ldots + x_my_m$, for $x,y \in \mathbb{R}^m$. Then, one defines a closed half-space as
$$
H = \{ x \in \mathbb{R}^m \mid \langle a,x \rangle \geq b \},
$$
and an open half space as
$$
H^\circ = \{ x \in \mathbb{R}^m \mid \langle a,x \rangle > b \}.
$$
Of course we could have used $\leq$ and $<$ as well.
The positive half-space is then defined by requiring that $x_m \geq 0$ (or $>0$). The negative half-space is defined similarly.
As an example, take $m=3$, $a=(3,1,5)$ and $b=0$. Then, we get a half-space
$$
H^\circ = \{ x \in \mathbb{R}^3 \mid 3x_1+x_2+5x_3 > 0 \}.
$$
You can plot the hyperplane $3x_1+x_2+5x_3 = 0$ with a $3d$ calculator such as this one, to get a better visual picture.
Another good example is the following: take $m=2$. Then, the positive half-space is given by
$$
\mathbb{H} = \{ (x_1,x_2) \in \mathbb{R}^2 \mid x_2 >0 \}.
$$
This is an important space in complex analysis, if we consider $\mathbb{C} \cong \mathbb{R}^2$, when one studies conformal maps for instance.
I recommend you make up some more examples of your own, playing with a graphing calculator, and experimenting mostly with the case $m=1,2,3$.
I hope this was helpful!
Edit: Not all half-spaces are closed. Take for instance $m=1$, $a=1$ and $b=0$. Then you get $H^\circ = \{x \in \mathbb{R} \mid x > 0\} = (0,\infty)$, and this is certainly not closed. So $H$ is always closed, but $H^\circ$ is not.
However, it is true that all half-spaces are convex. I'll sketch a proof for the case $m=2$, $a = (a_1,a_2) \in \mathbb{R}^2$ and $b=0$: Define a half-space in $\mathbb{R}^2$ by
$$
H := \{ x \in \mathbb{R}^2 \mid \langle (a_1,a_2),(x_1,x_2)\rangle \geq 0 \}.
$$
Let $t \in [0,1]$ and fix points $x,y \in H$, meaning we have
$$
a_1x_1+a_2x_2\geq0 \quad\text{and}\quad a_1y_1+a_2y_2\geq0.
$$
Define $p(t) := (1-t)x+ty$. We want to show that $\langle a,p(t)\rangle \geq 0$. Indeed, we see that
\begin{align}
\langle a,p(t)\rangle &= a_1((1-t)x_1+ty_1)+a_2((1-t)x_2+ty_2)\\
&= a_1(1-t)x_1+a_1ty_1+a_2(1-t)x_2+a_2ty_2\\
&= (1-t)(\underbrace{a_1x_1+a_2x_2}_{\geq0})+t(\underbrace{a_1y_1+a_2y_2}_{\geq0}) \geq 0,
\end{align}
since $1-t$ and $t$ are $\geq0$. So we have shown that $H$ is convex. I leave it up to you to show this for general $m \in \mathbb{N}$ and $b \in \mathbb{R}$.
If $H$ is closed (it is convex anyway, as we have seen) it is not necessarily compact. Take $H = [0,\infty) \subset \mathbb{R}$. $H$ is closed, but certainly not compact, since
$$
[0,2) \cup \bigcup_{i=1}^\infty (i,i+2)
$$
is an open covering of $H$ with the subspace topology, but it does not admit a finite sub-covering of $H$.
