Proving partial order for a set

Question

My question is:

Prove that if $R$ is a partial order for a set $A$ and if $zRx_1,x_1Rx_2,x_2Rx_3,\cdots,x_{n-1}Rx_n$ and $x_nRz$, then $z=x_1=x_2=x_3=\cdots=x_n$.

I am not too sure how to go about this but I am guessing I need to first show each part of the partial condition. Or should I just use the properties since it's assumed $R$ is a partial order? I am not too sure how to start off the proof. Can I only use the fact that it is antisymmetric to get z=x_1 and so forth.

Answer 1

Since $ R $ is a partial order, $R $ is transitive and antisymmetric:

Transitive: If $xRy$ and $yRz$, then $xRz$ for any $x,y,z$

Antisymmetric: If $xRy$ and $yRz$, then $x=y$ for any $x,y$

OK, since $zRx_1$ and $x_1Rx_2$, by transitivity, $zRx_2$. Then, since $x_2Rx_3$, again by transitivity $zRx_3$, though notice we also get $x_1Rx_3$. Etc etc, so eventually you get $zRx_n$ ... But also $x_1Rx_n$, $x_2Rx_n$, etc etc., Finally, since $x_nRz$ we get $zRz$, ... but also $x_1Rz$, $x_2Rz$ etc. In sum, we have $zRx_i$ as well as $x_iRz$ for any $1 \leq i \leq n$. So, by antisymmetry, we have $z = x_i$ for all $1 \leq i \leq n$, meaning they are all the same.

Answer 2

You can straight-up apply the properties of $R$ being a partial order because that is given to you on the statement. In particular you can use the transitivity of $R$ to show what you want.

Start by using transitivity to show that $zRx_k$ and $x_kRz $ and then use anti-simmetry to show they are equal.

Answer 3

We have: $z \le x_1 \le x_2 \le ...\le x_{n-1} \le x_n \le z$. Thus $z \le x_k, x_k \le z\implies z = x_k, k = 1, .., n$. This means:$x_1 = x_2 = ...= x_n = z$.