Convexity in one dimension

Question

For which values of $x ∈ \mathbb R$ is the set $A=[0,2] ∪ [x,x+2]$ convex?

(a) $−2 < x ≤ 1 $
(b) $−1 ≤ x ≤ 2 $
(c) $−2 ≤ x < 2$
(d) $−2 ≤ x ≤ 2 $

My attempt: both $b$ and $d$ by using $(1 − t)x + ty$? But there should only be 1 correct answer.

Answer 1

We’ll first prove that any interval is convex. Consider an interval $S\subset\mathbb{R}$, and two of its elements $x$, $y$, with $x\leq y$ without loss of generality. We need to prove that for every $0\leq t\leq1$, $$tx+(1-t)y\in S.$$ However, since $$x=tx+(1-t)x\leq tx+(1-t)y\leq ty+(1-t)y=y,$$ this is immediate from the fact that $S$ is an interval, and therefore contains any number between any two others in itself.

Now, for $S=[0,2]\cup[x,x+2]$ to be convex, we can’t have $x>2$, since otherwise $S$ would contain $0$, $x+2$, but not $\frac{x+2}{2}$, an absurd. We also can’t have $x<2$, since in that case, $S$ would contain $x$, $2$, but not $\frac{x+2}{2}$. However, if $0\leq x\leq2$, $S=[0,x+2]$ will be convex, being an interval. Likewise, for $-2\leq x\leq0$, $S=[x,2]$ will also be convex. Therefore, the answer is d.