Inconsistency related to the definition of net/directed set in Kelley's General Topology

Question

I've been confused about the definition of directed set in John L. Kelley's book "General Topology" (1), and feel inconsistency in multiple relevant places.

Overall, I feel that in the following sentence on p.65

A binary relation $\geq$ directs a set D if D is non-void and ...

the symbol "$\geq$" should have been "$\leq$". Such change would make it consistent with other textbooks, such as Munkres (2) and Willard (3).

More importantly, Kelley seems to be inconsistent with himself, e.g.,

1. in the paragraph right below the definition of directs on p.65, it says

We say that m follows n in the order "$\geq$" and ... iff $m \geq n$

while earlier on p.13 he said

If $<$ is an ordering and $x<y$, then ... $y$ follows $x$

which seems more natural to me.

The inconsistency between the two places is that if we treat $<$ and $\geq$ as a more abstract relation $R$, then on p.65 he says $m$ follows $n$ if $\mathbf{mRn}$, while on p.13 he says $m$ follows $n$ if $\mathbf{nRm}$ (I made substitution $x\rightarrow n$ and $y\rightarrow m$ to make the contrast more clear).

2. In the paragraph that follows, there is another example:

... the set $\omega$ of non-negative integers are directed by $\geq$. Observe that 0 is a member of $\omega$ which follows every other member in the order $\leq$.

From the context I feel that the $\geq$ in the above text should be interpreted as the usual ordering of natural numbers, i.e., $j \geq i$ means $j$ is bigger than or equal to $i$ in the usual sense. If we follow this interpretation, then I do not see how the property of 0 explains why $\omega$ is directed by $\geq$. In contrast, the property of 0 would explain why $\omega$ is directed by $\leq$.

I've also tried hard with alternative interpretations in multiple places, but could not make Kelley's exposition consistent with itself. Anybody has a clue?

(1) Kelley, General Topology, Dover edition 0-486-81544-7
(2) Munkres, Topology: a first course, Prentice-Hall 1975, 0-13-925495-1
(3) Willard, General Topology, Dover edition, 0-486-43479-6

Attached are the pictures of p.13 and p.65 from the Springer edition of GTM27 0-387-90125-6, which is the same as (1).

Answer 1

A directed set has as its primitive relation $\ge$. If we replace it by $F$ to be neutral and pronounce $nFm$ as $n$ follows $m$, the axioms for a directed set say

1. If $m$ follows $n$ and $n$ follows $p$ then $m$ follows $p$, or $$\forall m,n,p \in D: (mFn \land nFp) \to mFp$$

2. Every $m$ follows itself. Or $$\forall m \in D: mFm$$

3. For any two $m,n$ we can find a $p$ that follows both of them, or $$\forall m,n \in D: \exists p \in D: pFm \land pFn$$

I think this is quite clear, so far. Now it gets a bit confusing because $x \ge y$ already an alternative symbol for $y \le x$, when $\le$ is a partial order. But at least the word "follows" is used the same way; at page 13 if $x \le y$ , $y$ follows $x$, which corresponds to page 65's $y \ge x$ as well.

The remark on $\omega$ is true as it stands. He purposefully inverts the order to illustrate that the symbol doesn't always has the obvious meaning:

So he has $\omega =\{0,1,2,\ldots\}$ and he defines (using my symbol $F$ for clarity) that $mFn$ iff $m \le n$, where $\le$ is the usual ordering on $\omega$. Then axioms 1. and 2. are clear because the usual ordering on $\omega$ is transitive and reflexive. 3. follows from the fact that whatever $m,n$ are we can always take $p=0$: $0Fm$ because $0 \le m$ in the usual order. By reversing the order (so now $1$ "follows" $2$ in the directed set relation, etc.) the directed set in question has a "maximum", which "follows" all elements, to wit $0$. So this example serves (I think) to warn the reader that "follows" in a directed set doesn't always mean the same as an already defined notion "follows" from a partial order. It's meant to make you think and warn that "$\ge$" is not always what you think it is. He still uses $\ge$ in directed sets and not the boring $F$ I used, because it evokes the right idea that the left element in the directed-order relation is "further along". This doesn't have to be literally true, as we saw using $\omega$ or the trivial example that $nFm$ is true for all pairs ("any set is directed.."), where there is no "direction" in an intuitive sense. These examples all illustrate that a notion of directed-ness can be very general. The "direction reversal" is also seen in his example of the "set of finite subsets of a set" where $A \ge B$ ($A$ follows $B$) means that $B \subseteq A$; here the confusion is less as we don't write inclusion as $\le$, so it's not confusing symbols. Axiom 3. shows what the "direction" is, and this axiom distinguishes a directed set from a partial order.

If a set $(A,\le_A)$ is already a partial order, then if it's directed "naturally" in the sense that a directed set $(A,\ge)$ is just defined as $nFm$ or $n \ge m$ iff $m \le_A n$, then Fremlin (in his book Consequences of Martin's Axiom, where posets play a central role) says that $A$ is "upwards-directed", whereas if $nFm$ or $n \ge m$ is defined as $n \le_A m$ (as in Kelley's $\omega$) the set is "downward-directed". I quote from page 4 paragraph 11G(b):

It's not customary to distinguish, for example, between "upwards-ccc" and "downwards-ccc" partially ordered sets; most authors fix on one of these (usually, I think, downwards) and call it simply "ccc". The difficulty with this is that many partially ordered sets come to us with a natural orientation. (Consider, for instance, 11F above; or the cases where our partially ordered set $P$ is a collection of sets ordered by $\subseteq$, as in 32B and 33E below.) There is no logical difficulty in declaring, if necessary, that '$A \le B$ iff $B \subseteq A$'; but I find that following the subsequent arguments is like drinking a glass of water while hanging upside-down .I think it easier to take the trouble to use a language which can itself perform the necessary inversions.[..]

TLDR: I don't think Kelley is inconsistent, he's just giving an illustrative example to clarify. I do think Fremlin has a good point here, though.

Answer 2

Given the length restriction to comments, I hope what follows can help clarify why I think the $\omega$-example on p.65 is inconsistent with itself and/or with the definition earlier on the same page.

I think if $\omega$-example is stated in the opposite way, it would be consistent, and here is my version of the statement, followed by the proof/explanation.

Statement: The set $\omega$ is directed by $\leq$ because $0$ is a member of $\omega$ which follows every other member in the order $\leq$. (Note that in my statement $\omega$ is directed by $\leq$ while the textbook says $\omega$ is directed by $\geq$.)
Proof: we try to follow the definition of directs (directed set) given on p.65. Since we are trying to prove that "$\mathbf{\omega}$" is "directed by $\leq$", the set $D$ and the abstract $\geq$ symbol in the definition (or $F$ in Hennon's terminology) would become $\omega$ and $\leq$ everywhere, respectively.

(a) and (b) are obvious.
(c) we want to show that $(\forall m, n \in \omega)((\exists p \in \omega) (p\leq m \wedge p\leq n))$. Since "$0$ is a member of $\omega$ which follows every other member in the order $\leq$," $0$ is such a $p$:

$$(\forall m, n \in \omega)(0\leq m \wedge 0\leq n)$$

$\blacksquare$

I agree that the "$\geq$" in the definition could be anything, including a concrete $\leq$, as Hennon stated/implied in his/her answer:

He purposefully inverts the order to illustrate that the symbol doesn't always has the obvious meaning:

but when it boils down to a specific example (e.g., the $\omega$-example or the subsequent $\subset$-example), whatever symbol used needs to stand for a concrete order.

In other words, the $\geq$ in the definition (abstract, $F$ in Hennon's terminology, for now written as $\geq_F$) and the $\geq$ in the sentence "... $\omega$ ... (is) directed by $\geq$" (concrete, for now written as $\geq_\omega$) are different and should be different: the former could be anything, while the latter should have a specific meaning and actually needs to be consistent with the $\leq$ in the subsequent sentence "Observe $0$ is ... in the order $\leq$", i.e., $\geq_\omega$ means the opposite of $\leq$. Unfortunately, with this setting the subsequent sentence does not justify the statement "... $\omega$ ... (is) directed by $\geq$" and that's the reason why I think changes are needed, as given at the beginning of this answer.

Answer 3

I now think (on second reading) that Kelley is giving two examples in those two sentences:

The real numbers as well as the set $\omega$ of non-negative integers are directed by $\ge$.

This is true, I think we can agree on that. We have a linear order which is always directed because $\max(x,y) \ge x$ and $\max(x,y) \ge y$ and $\max(x,y)$ always exists in a linear order. The example with $\omega$ is especially important as it shows that a net is a generalisation of a sequence. It's two examples in one go, one for sequences, one for "long sequences" (indexed by reals)

Then the second example:

Observe that $0$ is a member of $\omega$ which follows every other member in the order $\le$.

This is not literally true, of course, but it is true here because the direction is reversed: $\le =\, \le_\omega\, = F$, as I defined $F$ (follows) in my earlier answer. So $0 \le n$ is true in $\omega$ so $0 \ge n$ pronounced as "$0$ follows $n$" is true in the directed set. So we have an example of a linear order with a minimum and when we interpret the natural $\le_\omega$ as the $\ge$ symbol of a directed set (this always has this symbol, as Kelley makes clear, I think) we get the reverse order example, different from the one (or two) in the sentence before.

Note that this interpretation is consistent with the next one:

It is also noteworthy that the family of all neighborhoods of a point in a topological space is directed by $\subset$ (the intersection of two neighborhoods is a neighborhood which follows both (my emphasis) in the ordering $\subset$.

So $N_1 \cap N_2 \ge N_2$ and $N_1 \cap N_2 \ge N_2$ (the intersection follows both) (as $N_1 \cap N_2 \subset N_2$ and $N_1 \cap N_2 \subset N_2$). This speaks for my interpretation in the second example as $\ge$ (direction) is synonymous with $\le$ (order).

The next to last example is again consistent:

The family of all finite subsets of a set is, on the other hand, directed by $\supset$.

This is clear because we can take the union of two finite sets $F$ and $G$ and see $F \cup G \ge F$ and $F \cup G \ge G$ (using the synonyms $\ge$ (direction) = $\supset$ (order)).

And trivially (to see any set can be made a directed set, so that there is no set of all subnets of a net, but a class):

Any set is directed by agreeing that $x \ge y$ for all members $x$ and $y$, so that each element follows both itself and any other element.

IMHO this hammers in the notion that

We always use $\ge$ for the relation in a directed set, no matter what other structure the set we're working on already had, and we always pronounce $x \ge y$ (for a directed set) as "$x$ follows $y$".

There is no contradiction with p.13 because there he is talking about an order ("merely" a transitive relation), and later he's talking about a "directed set", a separate notion. Note that $x \ge y$ is pronounced "$x$ follows $y$" in either case.