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 !
I have a very limited exposure to this kind of topology. I am just a beginning high school student trying to learn a book called "Game Theory"-Michael Maschler, and this is one of the problems in the text that baffles me.
I THANK YOU VERY MUCH FOR YOUR ANSWER.
VECTORIAL SPACE :
Let say that the direction of $H^i$ is $D^i$. So $H^1 \subseteq H^2 \implies D^1 \subseteq D^2$
Let $(e_1;\dots;e_k)$ be a basis of $D^1$ so it's an independant vector family of $k$ vectors, (these vectors belong to $D^1$ which is in $D^2$ so there are in $D^2$) thus this is a basis of $D^2$.
Thus for all $v \in D^2$, $v = \Sigma_1^k (\lambda_i*e_i)$ so $v\in D^1$.
So $D^2 \subseteq D^1$
In conclusion $D^2=D^1=D \text{ (used after)}$
So with $h_0 \in H^1 \subseteq H^2$, $H^2=\{h_0+d;d\in D\}= H^1$
SIMPLEX :
Let take $S=<<x_0, \dots , x_k>>$ a simplex of $H^1$
But as $H^1 \subseteq H^2$ (we allready have the independance because S simplex in $H^1$), S is also as simplex of $H^2$
So if $H^1 \ne H^2$, there is $h \in H^2$\ $H^1$ such as $h = \sum_{l=0}^k \alpha^{l}x^{l}$ with $\sum_{l=0}^k \alpha^{l}=1$ (cause belong to $H^2$) but that also implies that $h \in H^1$
In conclusion $h \notin H^1$ and $h \in H^1$ so contradiction.
Thus $H^1 = H^2$