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On this page there are some examples of directed sets. One of those cites:

"If $x_0$ is a real number, we can turn the set $\mathbb{R} − \{x_0\}$ into a directed set by writing $a \leq b$ if and only if $|a − x_0| \geq |b − x_0|$. We then say that the reals have been directed towards $x_0$. This is an example of a directed set that is not ordered (neither totally nor partially)."

Can you show me why such a set is not partially ordered?

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The relation is not antisymmetric: take $a=x_0-1$, $b=x_0+1$. Then $|a-x_0| = |b-x_0|=1$, so we have both $a\geq b$ and $b\geq a$, but we do not have $a=b$. Since the relation is not antisymmetric, it is not a partial order.

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Take, for example, $x_0=0$ take $a=-1$ and $b=1$. You have $a\neq b$ but $a\leq b$ and $b\leq a$. Anti-symmetry breaks down so it is not a partial order, therefore it is not a linear order either.

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Ordinarily a (partial or total) order relation is required to be antisymmetric, that is, $a\le b$ and $b\le a$ must not both hold unless $a$ and $b$ are the same element.

However, your example does not satisfy that criterion -- for example for $x_0=2$ we have $1\le 3$ and $3\le 1$.

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