I am asked to prove that the set $\succeq(x_o) = \{x \in X \subset \mathbb{R} : x \succeq x_o\}$ is convex for any $x_o$ given the following Axiom (1):
If $x_1 \succeq x_o$, then $tx_1 + (1-t)x_o \succeq x_o, \forall t \in [0,1]$.
The proof seems trivial if that Axiom is true, which makes me believe that I am missing something or making some silly assumptions. Can anyone give me some insight/help here? Here is what I would give as proof.
Consider $x_1, x_2 \in \ \succeq(x_o)$ then by the previous Axiom $tx_1+(1-t)x_o \succeq x_o$ and $tx_2+(1-t)x_o \succeq x_o$.
I will also use the following Axiom (2): For any $x_1, x_2 \in X \subset \mathbb{R}$ either $x_1 \succeq x_2$ or $x_2 \succeq x_1$ or both.
Suppose $x_1 \succeq x_2$ then by the Axiom 1 and then Axiom 2: $tx_1 + (1-t)x_2 \succeq x_2 \succeq x_o, \forall t \in [0,1]$ so clearly $tx_1 + (1-t)x_2 \in \ \succeq(x_o)$ thus $\succeq(x_o)$ is convex.
Suppose $x_2 \succeq x_1$, the proof is the same as above.
Suppose both $x_1 \succeq x_2$ and $x_2 \succeq x_1$, then use one of the relations and the proof is the same.
Is this sufficient, or allowed?