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This exercise is from Bazaraa Linear programming book.
And I don't see the point of the exercise since the definition of direction is already given by hypothesis.
Am I missing something or the exercise it's not well stated?
Let $S$ be a closed convex set in $\mathbb R^n$ and let $x\in S$ . Suppose that $d$ is a nonzero vector in $\mathbb R^n$ and that $x +\lambda d\in S$ for all $\lambda\ge 0$. Show that $d$ is a direction of $S.$
You need to show that if $y \in S$ then $y+\lambda d \in S$ for all $\lambda >0$.
Note that for $t \in [0,1]$ we have $(1-t) y + t(x+\lambda d) \in S$.
Pick $\mu >0$, and let $t_n = {1 \over n} \mu$ and let $\lambda = n$ in the above equation (pick $n$ large enough so that $t_n \in (0,1]$).
This gives $(1-t_n) y + t_n x + \mu d \in S$ for all $n$.
Let $n \to \infty$ and use the fact that $S$ is closed to get the desired result.
Excuse the crude diagram:
To illustrate why closure is specified, consider the set $S=\mathbb{R} \times (0,\infty) \cup (0,0)$. Then with $d=(1,0)$ we see that if $x \in \mathbb{R} \times (0,\infty) $ then $x+ \lambda d \in S$ for all $\lambda$ but $(0,0)+ \lambda d \notin S$ for any $\lambda \neq 0$.
