I need to check my homework. I'm not good at discrete mathematics and I need to have this homework correct.
Let $S=\{1,2,\ldots,100\}^2$. The partial order $\le_S$ is defined as follows: for $\langle a,b\rangle,\langle x,y\rangle\in S$, $\langle a,b\rangle\le_S\langle x,y\rangle\iff a\le x\land b\le y$.
And I have to find the longest chain and the longest antichain. The minimal element is ordered pair (1,1) and the maximal element is (100,100). I suppose the longest antichain are ordered pairs (1,100), (2,99), ..., (100,1). Am I right? Can I write this result using following formula? $$\alpha(\{1, 2, ..., 100\}^2,\le S)=\bigcup_{i=1}^{100}(i,101-i)$$ The longest chain goes from minimal element (1,1) to maximal element (100,100). I think there is a lot of longest chains. As an example I chose this one - (1,1), (1,2), …, (1,99), (1,100), (2,100), …, (99,100), (100,100). I think it has length of 199 ordered pairs. Am I right? Can I write this result using following formula? $$\omega(\{1, 2, ..., 100\}^2,\le S)=\bigcup_{i=1}^{100}(1,i) \cup \bigcup_{i=2}^{100}(i,100)$$ Thank you very much for your help.
You have correctly identified the longest antichain in $S$, but your notation for it is incorrect. It is not the union of the ordered pairs $\langle i,101-i\rangle; it is the set of those ordered pairs,
$$\big\{\langle i,101-i\rangle:i\in\{1,2,\ldots,100\}\big\}\;.$$
You have also correctly identified one of the longest chains, all of which do indeed have $199$ elements, but you’ve made the same mistake in your notation; it should be
$$\big\{\langle 1,i\rangle:i\in\{1,2,\ldots,100\}\big\}\cup\big\{\langle i,100\rangle:i\in\{2,3,\ldots,100\}\big\}\;.$$
As Did notes in the comments, your description of the partial order is very badly written. I suspect that it is supposed to be something like this: