I'm currently reading Elements of Algebraic Topology by Munkres and I'm unable to proof exercises 3-5 page 7. So, as follows:
Let $\sigma$ be a n-simplex.
Show that Int $\sigma$ is convex and open in plane P (all definitions can be found in book pages 2-6) and its closure is $\sigma$. Show that Int $\sigma$ equals the union of all open line segments joining $a_0$ to points of Int $s$ where $s$ is the face of $\sigma$ opposite to $a_0$.
Let $\sigma$ be spanned by $a_0,\ldots,a_n$, let $s$ be the face of $\sigma$ spanned by $a_0,\ldots,a_p$ (p < n) and let $t$ be the face of $\sigma$ spanned by $a_{p+1},\ldots,a_n$: Show that $\sigma$ is the union of all line segments joining points of $s$ to the points of $t$ and two of these line segments intersect in at most a common end line. Show that Int $s$ is the union of all line segments joining points of Int $s$ to the points of Int $t.$
Let $U$ be bounded and open, $U \subset \mathbb{R}^n$. Suppose $U$ is star-corvex to the point $\theta$ (for all $x \in U$, the lines segments from $\theta$ to x lines in $U$). Show that a ray from $\theta$ may intersect Bd $U$ on more than one point. Show (by example) that $\overline{U}$ need not be homeomorphic to $B^n = \{x \in \mathbb{R}^n : ||x|| \leq 1\}$.
$\textbf{Edit 1}$ (26.10.2017) :
Sorry I had no time to answer earlier. So I'll post here what I had done so far:
Ex. 1:
convex:
Let $x,y \in $ Int $\sigma$.
$x=t_0a_0+\cdots+t_na_n$, where $\sum_{i=0}^n t_i=1$ and $t_i>0$ for $i=0,1,\ldots,n$.
$y=s_0a_0+\cdots+s_na_n$, where $\sum_{i=0}^n s_i=1$ and $s_i>0$ for $i=0,1,\ldots,n$.
Let $z=\lambda x+(1-\lambda)y$, $\lambda \in [0,1]$. We'll show that $z \in$ Int $\sigma$.
$$z=\lambda(t_0a_0+\cdots+t_na_n)+(1-\lambda)(s_0a_0+\cdots+s_na_n)=$$
$$=a_0(\lambda t_0+(1-\lambda)s_0)+\cdots + a_n(\lambda t_n+(1-\lambda)s_n).$$ Notice, that $\lambda t_i>0$ and $(1-\lambda)s_i>0$, so $\lambda t_i+(1-\lambda)s_i>0$ for $i=0,\ldots,n$.
$$\lambda t_0+(1-\lambda)s_0 + \cdots + \lambda t_n+(1-\lambda)s_n=$$
$$=\lambda (t_0+\cdots+t_n)+(1-\lambda)(s_0+\cdots+s_n)=\lambda +1-\lambda =1.$$ So $z \in$ Int $\sigma$.
- closure:
$\sigma$ is closed, so closure of Int $\sigma \subset \sigma$ and $\sigma = $ Int $\sigma$ $\cup$ Bd $\sigma$. If a point $x \in$ Bd $\sigma$ is not in the closure of Int $\sigma$, we can construct a sequence of points in Int $\sigma$ which converges to $x$.
Let $x \in$ Bd $\sigma$. Without loss of generality, assume that $x=\sum_{i=0}^k t_ia_i$, where $t_i(x)>0$ for $i=0,1,\ldots,k$ and $t_{k+1}(x)=\cdots=t_n(x)=0$. I've found this construction:
$$r=\frac{min\{t_i(x) : 0 \leq i \leq k \}}{2}$$ $$x_m=\sum_{i=0}^k \left(t_i(x)-\frac{r}{m(k+1)}\right)a_i+\sum_{i=k+1}^n \frac{r}{m(n-k)}.$$ I don't really have any idea how to show that $x_m \in$ Int $\sigma$. It's obvious that $x_m$ converges to $x$. (maybe there is another way to proof this?)
open: (no idea)
Int $\sigma$ $=$ union of all open line segments:
$(\subset)$ Let $x \in $ Int $\sigma$.
$x=t_0a_0+\cdots+t_na_n$, where $\sum_{i=0}^n t_i=1$ and $t_i>0$ for $i=0,1,\ldots,n$. $$x=t_0a_0+\lambda\sum_{i=1}^n \frac{t_i}{\lambda}a_i=t_0a_0+\lambda W,$$ where $\lambda=1-t_0=t_1+\cdots+t_n$.
$t_0+\lambda=1$, so $x \in$ line connecting $a_0$ with $W$. $\frac{t_1}{\lambda}+\cdots+\frac{t_n}{\lambda}=1$ and $\frac{t_i}{\lambda}>0,$ so $W \in$ Int $s$.
$(\supset)$ Let $x \in$ union of all open lines.
$x=\lambda a_0+(1-\lambda)v$< where $v \in$ Int $s$, $\lambda \in (0;1)$.
$v=t_1a_1+\cdots+t_na_n$, where $t_1+\cdots+t_n=1$, $t_i>0$, $i=1,\ldots,n$.
$x=\lambda a_0+(1-\lambda)t_1a_1+\cdots+(1-\lambda)t_na_n.$ Notice, that $(1-\lambda)t_i>0$.
$\lambda+(1-\lambda)t_1+\cdots+(1-\lambda)t_n=\lambda+(t_1+\cdots+t_n)-\lambda(t_1+\cdots+t_n)=\lambda+1-\lambda=1$, so $x \in$ Int $\sigma$.