# A clarification regarding maximal linearly ordered subsets and the Hausdorff maximal principle

I am currently beginning to read Folland after finishing chapters 1 - 6 in baby Rudin. I am a little confused about the section on partially ordered sets. Firstly, I think I know the answer to this, but if $$x \leq y$$ is false, then that doesn't necessarily imply $$y < x$$ i.e. $$y < x$$ isn't necessarily the negation of $$x \leq y$$. Clearly, this will imply in the Real numbers, which I am used to.

Also, I am a bit confused about the Hausdorff maximal principle, which is "Every partially ordered set has a maximal linearly ordered subset".

When it's saying maximal linearly ordered subset does that mean that it has an element that is maximal to the entire set or specifically for the subset?

You are correct that in a partially ordered set $$x \not\leq y$$ does not imply $$y < x$$. This is essentially the "partial" in the name - we may have neither $$x \leq y$$ nor $$y \leq x$$ for some elements.
The "maximal" in the Hausdorff maximal principle means the following: $$A$$ is a maximal linearly ordered subset of a partially ordered set $$X$$ if, whenever $$B$$ is another linearly ordered subset of $$X$$ with $$A \subseteq B$$, then $$A=B$$. In other words, a maximal linearly ordered subset is one where you cannot find any linearly ordered subset of $$X$$ that strictly contains it. The Hausdorff principle then says that such a maximal linearly ordered subset of $$X$$ exists. Note that this is not (directly) a claim about any particular element being maximal in any sense, it's about the subset itself being maximal.
• Oh, I see, it's referring to a maximal element of $P(X)$ provided that it's linearly ordered. Aug 13 at 20:50
• Yes, maximal with respect to $\subseteq$ among the subsets of $X$ linearly ordered by the partial ordering $\le$ . Aug 14 at 3:06