For the sake of completeness, I would like to give you some concepts before asking the questions:
For every simplex $S=<<x^{0},x^{1},...,x^{k}>>$ in $\Bbb R^{n}$, denote by $H_s$, the affine space spanned by vectors in S,
$H_s:=\{\sum_{l=0}^k \alpha^{l}x^{l}:\sum_{l=0}^k \alpha^{l}=1\}$ where $x^{0},...,x^{k}$ are vectors in $\Bbb R^{n}$ and $\alpha^{0},...,\alpha^{k}$ are real numbers.
So, after spending half an hour trying to prove this theorem, I have given up on it:
Theorem: Prove that if $H^{1}$ and $H^{2}$ are two affine spaces of the same dimension $k$ in $\Bbb R^{n}$(that is, each one of them is spanned by a $k$-dimensional simplex), and if $H^{1} \subseteq H^{2}$, then $H^{1}=H^{2}$.
I try to prove it by first writing down the definition of affine space, that is, let: $H^{1}:=\{x^{0},...,x^{k-1},y: \sum_{l=0}^{k-1} \alpha^{l}x^{l} + \alpha^{k}y\}$ and $H^{2}:=\{x^{0},...,x^{k-1},x^{k}: \sum_{l=0}^{k-1} \alpha'^{l}x^{l} + \alpha'^{k}x^{k}\}$
then I give up !
Please help me with the question.
I THANK YOU VERY MUCH FOR YOUR ANSWER.
Because the space in a comment is too small I reproduce my text as an answer althoug it isn't one. First of all consider an simple example (without all these big $\sum$ signs) and work in $\Bbb R^2$. Take $3$ points (vectors) in $\Bbb R^2$: $x^0, x^1$ and $x^2$. Now we have three affine spaces defined by these points: one by the points $x^0$ and $x^1$, another by the points $x^0$ and $x^1$, and a third defined by $x^1$ and $x^2$. Let us consider the first space : $H^1$ is defined by the equation $\alpha x^0+\beta x^1$ with $\alpha+\beta=1$. Now take $\alpha=t$ for some $t$ and $\beta=1-t$, so we can get rid of the equation $\alpha+\beta=1$, and then the equation for $H^1$ becomes $tx^0+(1-t)x^1$. Now this should ring a bell since this is the parameter equation for a line. It is the line passing through $x^0$ for $t=1$ and through the point $x^1$ for the value $t=0$. Applying the theorem to this situation is equivalent to saying that if all the points of a line belong to another line then these two lines are equal.