Show that P is linearly ordered iff every maximal antichain in P has only one element. If P is a partially ordered set, a subset A of P is called an antichain in P provided:
whenever $x,y \in A$ with $x \le y$ we have $x=y$. 
(In other words, every pair of elements in an antichain are incomparable.)
GOAL: Show that P is linearly ordered if and only if every maximal antichain in P has a single element.


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*If we let M be a maximal antichain of P, then M is not a proper subset of any other antichain.

*First, suppose every maximal antichain in P has only a single element. So let $a \in M: a \le a\le \cdots$ where a is the single element of the maximal antichain M in P. (I get stuck trying to get P linearly ordered probably because I'm not noticing some obvious fact.)

*Now, suppose P is a linearly ordered set. Then any two elements are comparable and P looks like: $p_1 \le p_2 \le p_3 \le \cdots$ where each p element is in P.


I appreciate the help!

Here is a hint in each direction:
Recall the defining feature of a linear order is that every pair of elements is comparable.
First, if $P$ is linearly ordered, then take any two elements $x$ and $y$. In an antichain, we must have $x \not \leq y$ and $y \not \leq x$. But what is the one thing we know about linear orders?
Conversely, if the largest antichain has size 1, then take two elements $x$ and $y$. Can you show they must be comparable (using an argument similar to the above)? 

I hope this helps ^_^
