Prove that for any two points $A$ and $B$ $\overrightarrow{AB} \cup \overrightarrow{BA} = \overleftrightarrow{[AB]}$

Question

Question: Prove that for any two points $A$ and $B$ $\overrightarrow{AB} \cup \overrightarrow{BA} = \overleftrightarrow{[AB]}$

The right hand side of the statement that I am trying to prove is a line AB in a set. It's just a single line AB in the set meaning that there is only one element in line AB. Attempt:

By definition of ray, $\overrightarrow{AB}$ is the set of all points on the segment $A$ together with all points $C$ such that $A*B*C$

Also, by definition of segment, $AB$ is defined as the set of all points between A and B together with endpoints A and B.

Moreover, by definition of union definition

$\overrightarrow{AB} \cup \overrightarrow{BA} $

${ x: x \in AB \lor x \in BA}$

$\overrightarrow{AB} \cup \overrightarrow{BA}$

$ { AB: AB \in AB \lor AB \in BA}$

Since we have an or statement, consider $\overrightarrow{AB}$. Then $C$ must be the only element in the set for Line AB.

Let Point C belong to the union of $\overrightarrow{AB}$. If $C=A$ or $C=B$ , then $C$ is a set of points lying on $\overleftrightarrow{AB}$

Really? Seems to be going fast after I did that.

I get stuck afterwards.. I mean I know I need to use set union definition and I have two rays. So I need to somehow prove along with definitions that the outcome will be a line AB in a set which means that there is a single element. I'm thinking that the single element must be C since by the ray definition that there is a set of all points with all points C. However, B must be between A and C. We can also have B must be between C and A.

Answer 1

$AB = \{P \in \overleftrightarrow{AB} . A - P - B\} \cup \{A, B\}$

$\overrightarrow{AB} = AB \cup \{P \in \overleftrightarrow{AB} . A - B - P\}$

We want to prove that $\overrightarrow{AB} \cup \overrightarrow{BA} = \overleftrightarrow{AB}$:

$P \in \overrightarrow{AB} \cup \overrightarrow{BA}$ iff (logic)

$P \in \overrightarrow{AB} \vee P \in \overrightarrow{BA}$ iff (def. of ray)

$(P \in AB \vee A - B - P) \vee (P \in AB \vee B - A - P)$ iff (logic)

$P \in AB \vee A - B - P \vee B - A - P$ iff (def. of segment)

$P \in \{A, B\} \vee A - P - B \vee A - B - P \vee B - A - P$ iff (third axiom of order)

$P \in \overleftrightarrow{AB}$