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Given a poset $P = (X, \le)$, I am trying to create a new poset by adding a new element $x^*$ to $X$ and the ordered pair $(x^*,x^*)$ to the relation.

How does this effect the height, width, the set of maximal and minimal elements, the existence of a maximum and minimum?

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  • $\begingroup$ Do you mean the pair $(x^*,x^*)$? $\endgroup$ – Thomas Andrews Apr 24 '13 at 16:44
  • $\begingroup$ Whoops! Yeah I meant to put that but for some reason it isn't showing up. $\endgroup$ – Atom Apr 24 '13 at 16:46
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Let us assume that $X$ was nonempty to begin with. Denote $X' = X \cup \{x^*\}$, and $P' = (X', \le)$, with $\le$ extended so as to include $x^* \le x^*$.

Observe that for any $x \in X'$, $x \le x^*$ or $x^* \le x$ implies that $x = x^*$.

We deduce that the longest chain involving $x^*$ has length $1$. Because any element on its own defines a chain of length $1$, the height of $X'$ is the same as that of $X$.

Suppose $A$ is a maximal antichain of $P$. Then $A \cup \{x^*\}$ is easily seen to be an antichain as well; its maximality as an antichain of $P'$ is immediate from $A$ being one of $P$. Thus $\operatorname{width}(P') = \operatorname{width}(P)+1$ (using cardinal arithmetic in case these are infinite).

Also, $x^*$ is easily seen to be both a maximal and a minimal element, but neither a greatest nor a smallest element (which I suppose is the meaning of maximum and minimum).

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