Help with this convex set proof

Question

Take $C ⊂ \mathbb{R}^n$ a convex set. Fix $x_0 ∈ C$ and a nonzero vector $v ∈ \mathbb{R}^n$ . Define the set $I(x_0,v) := \{t ∈ R : x_0 + tv ∈ C \}$. Prove that $I_(x_0,v)$ is a convex subset of $\mathbb{R}$

My confusion arises from the fact that $I(x_0,v)$ is simply a line with direction $v$ passing through $x_0$. Since it's a subset of $C$ which is contained in $\mathbb{R}^n$ isn't it quite obvious that $I(x_0,v)$ is a convex subset of $\mathbb{R}$ ?

How would one prove this?

edit: My attempt so I need to show that $I(x_0,v)$ is an interval. We do this by showing that for some $$t \in (a,b), x_0+tv \in C$$

Assume that $$x:=x_0 + av \in C$$ $$y:=x_0 + bv \in C$$

Because $C \subset \mathbb{R}^n $ is convex, $\forall x,y \in C, t \in [0,1]$ $$ tx + (1-t)y \in C$$

Is this on the right track?

Answer 1

Note that $I(x_0, v)$ actually is an interval. Intervals are convex subsets of $\mathbb R$. The only thing you really need to show is that $I(x_0, v)$ is an interval, i.e. if $\{a,b\}\subset I(x_0, v)$, so is $(a,b) \subset I(x_0,v)$. Do you have an idea how to do that?

---

To prove the statement you proceed in two steps:

• Let $I:= I(x_0, v)$. Assume $\{a,b\}\in I$, i.e. $$x := x_0 + av \in C\\ y := x_0 + bv \in C$$
• Let $\alpha \in (a,b)$ be arbitrary. We must show $\alpha\in I$ or $$w := x_0 + \alpha v \in C$$ To show this, write $w$ as a convex combination of $x$ and $y$, i.e. find a $\lambda\in (0,1)$ such that $$w = \lambda x + (1-\lambda) y$$