T Nation

Unusual Math Question (Hilbert's Axioms)


#1

Regarding Hilbert's axioms, I've never understood why I.4 is needed, given I.3. From Wikipedia:

I.1: Two distinct points A and B always completely determine a straight line a. We write AB = a or BA = a. Instead of â??determine,â?? we may also employ other forms of expression; for example, we may say â??A lies upon aâ??, â??A is a point of aâ??, â??a goes through A and through Bâ??, â??a joins A and/or with Bâ??, etc. If A lies upon a and at the same time upon another straight line b, we make use also of the expression: â??The straight lines a and b have the point A in common,â?? etc.

I.2: Any two distinct points of a straight line completely determine that line; that is, if AB = a and AC = a, where B â?  C, then also BC = a.

I.3: Three points A, B, C not situated in the same straight line always completely determine a plane �±. We write ABC = �±. We employ also the expressions: â??A, B, C, lie in �±â??; â??A, B, C are points of �±â??, etc.

I.4: Any three points A, B, C of a plane �±, which do not lie in the same straight line, completely determine that plane.

I.5: If two points A, B of a straight line a lie in a plane �±, then every point of a lies in �±. In this case we say: â??The straight line a lies in the plane �±,â?? etc.

I.6: If two planes �±, �² have a point A in common, then they have at least a second point B in common.

I.7: Upon every straight line there exist at least two points, in every plane at least three points not lying in the same straight line, and in space there exist at least four points not lying in a plane.

II. Order

II.1: If a point B is between points A and C, B is also between C and A, and there exists a line containing the points A,B,C.

II.2: If A and C are two points of a straight line, then there exists at least one point B lying between A and C and at least one point D so situated that C lies between A and D.

II.3: Of any three points situated on a straight line, there is always one and only one which lies between the other two.

II.5: Pasch's Axiom: Let A, B, C be three points not lying in the same straight line and let a be a straight line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the straight line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

III. Parallels

III.1: In a plane �± there can be drawn through any point A, lying outside of a straight line a, one and only one straight line which does not intersect the line a. This straight line is called the parallel to a through the given point A.

IV. Congruence

IV.1: If A, B are two points on a straight line a, and if A�¡ is a point upon the same or another straight line a�¡ , then, upon a given side of A�¡ on the straight line a�¡ , we can always find one and only one point B�¡ so that the segment AB (or BA) is congruent to the segment A�¡ B�¡ . We indicate this relation by writing AB â?? A�¡ B�¡. Every segment is congruent to itself; that is, we always have AB â?? AB.

We can state the above axiom briefly by saying that every segment can be laid off upon a given side of a given point of a given straight line in one and only one way.

IV.2: If a segment AB is congruent to the segment A�¡B�¡ and also to the segment A�¡�¡B�¡�¡, then the segment A�¡B�¡ is congruent to the segment A�¡�¡B�¡�¡; that is, if AB â?? A�¡B and AB â?? A�¡�¡B�¡�¡, then A�¡B�¡ â?? A�¡�¡B�¡�¡

IV.3: Let AB and BC be two segments of a straight line a which have no points in common aside from the point B, and, furthermore, let A�¡B�¡ and B�¡C�¡ be two segments of the same or of another straight line a�¡ having, likewise, no point other than B�¡ in common. Then, if AB â?? A�¡B�¡ and BC â?? B�¡C�¡, we have AC â?? A�¡C�¡.

IV.4: Let an angle (h, k) be given in the plane �± and let a straight line a�¡ be given in a plane �±�¡. Suppose also that, in the plane �±�¡, a definite side of the straight line a�¡ be assigned. Denote by h�¡ a half-ray of the straight line a�¡ emanating from a point O�¡ of this line. Then in the plane �±�¡ there is one and only one half-ray k�¡ such that the angle (h, k), or (k, h), is congruent to the angle (h�¡, k�¡) and at the same time all interior points of the angle (h�¡, k�¡) lie upon the given side of a�¡. We express this relation by means of the notation â? (h, k) â?? (h�¡, k�¡)
Every angle is congruent to itself; that is, â? (h, k) â?? (h, k)
or â? (h, k) â?? (k, h)

IV.5: If the angle (h, k) is congruent to the angle (h�¡, k�¡) and to the angle (h�¡�¡, k�¡�¡), then the angle (h�¡, k�¡) is congruent to the angle (h�¡�¡, k�¡�¡); that is to say, if â? (h, k) â?? (h�¡, k�¡) and â? (h, k) â?? (h�¡�¡, k�¡�¡), then â? (h�¡, k�¡) â?? (h�¡�¡, k�¡�¡).

IV.6: If, in the two triangles ABC and A�¡B�¡C�¡ the congruences AB â?? A�¡B�¡, AC â?? A�¡C�¡, â? BAC â?? â? B�¡A�¡C�¡ hold, then the congruences â? ABC â?? â? A�¡B�¡C�¡ and â? ACB â?? â? A�¡C�¡B�¡ also hold.

V. Continuity

V.1: Axiom of Archimedes. Let A1 be any point upon a straight line between the arbitrarily chosen points A and B. Take the points A2, A3, A4, . . . so that A1 lies between A and A2, A2 between A1 and A3, A3 between A2 and A4 etc. Moreover, let the segments AA1, A1A2, A2A3, A3A4, . . . be equal to one another. Then, among this series of points, there always exists a certain point An such that B lies between A and An.

V.2: Line completeness. To a system of points, straight lines, and planes, it is impossible to add other elements in such a manner that the system thus generalized shall form a new geometry obeying all of the five groups of axioms. In other words, the elements of geometry form a system which is not susceptible of extension, if we regard the five groups of axioms as valid.

Lest I be criticized for citing Wikipedia, I provided the above directly from Townsend's transation, with some small modifications being made since from a German speaker who improved some translation points.

For about a year probably on the Wikipedia discussion page I've had a question about why I.4 is not implied by I.3. No takers.

Can any here advise?

There must be answer. Hilbert was no fool. There must be something I do not see.


#2

Sorry for the gibberish of symbols that didn’t carry over… They appeared correctly when cutting and pasting.

In the case of the axioms in question, the gibberish refers to plane alpha.

For the correct symbols for other axioms, they are at http://en.wikipedia.org/wiki/Talk:Hilbert’s_axioms


#3

It looks to me like I.3 is talking about three random points in space where a plane has not yet been determined, but if you take those points then you can always completely define a plane.

In I.4 a plane has already been defined and if you take any three random points from that plane it will always completely define the said plane.

I know it seems like one should be implied in the other, but this is just what I see.


#4

Thanks. What you wrote got me to finally get it, after all this time.


#5

I see how I.3 and I.4 are different, but it does seem as though I.3 implies I.4. If I.3 were true and I.4 were false, then the three points A, B, and C of the plane Ã??Ã?± that do not lie in the same straight line could determine a different plane, which is not Ã??Ã?±. That would imply that there are two different planes that include the same group of three non-collinear points, which as I understand it would mean that the three points do not “determine” a plane.

Maybe I.4 is included because the process of determining I.4 from I.3 is somewhat indirect? Or not sufficiently rigorous?


#6

What came first the plane or the points? If the points, then I3. If the plane, then I4.


#7

I3 and I4 taken together means that three non-linear points are sufficient to define a plane but are not unique. It means that ABC = alpha there can also be DEF = alpha. ABC and DEF don’t get to argue over who defines alpha.


#8

[quote]FrankNStang wrote:
What came first the plane or the points? If the points, then I3. If the plane, then I4.[/quote]

What Hilbert was working with here was (gross oversimplification I’m sure) coming up with geometric axioms and proofs that truly did not lie on assumptions about what things already were or how they ought to behave.

We’re taught in high school that this is what Euclid did, and it is what Euclid thought that he did and people continued to think for many centuries, but it’s since been found that his proofs and axiom system are not airtight and do rely on pre-existing opinion about the nature of things.

So the answer to your question is, Hilbert has neither the term plane nor point meaning anything at all until assigned the meanings that the axioms assign them. And the subsequent proofs don’t assume any other properties.

Philosophically, I think it is the case that Euclid perceived “lines,” “points,” “planes” and so forth to be natural pre-existing truths, in the sense of Platonic ideals. The axioms were simply clearly true statements regarding these already-existing things.

Whereas Hilbert is doing no such thing.

I actually have a project that I would like to do, but suspect I won’t get to till I’m too senile for it. Namely – for background – work has long since been completed in identifying where Euclid in fact fails to make given proofs, as there are hidden assumptions involved; and all (I think) of Euclid’s proofs have been redone in Hilbert’s system.

But I have an aesthetic as well as historical appreciation for Euclid’s work, and I would like to do a revisiting of it where everything is in fact rigorous to Hilbert’s axioms, but the presentation is as close to Euclid as possible.

It would in fact contribute nothing to mathematical knowledge – I don’t have the gifts for that – but I would like it myself.


#9

Let’s forget about points, lines, and planes for the moment. Let’s talk about three fictitious entities: herberts, nerberts, and sherberts. Let’s also use different herberts for I.3 and I.4.

I.3: Three herberts A, B, C not associated with the same nerbert always completely determine a sherbert alpha.

I.4: Any three herberts D, E, F that are associated with the same sherbert alpha, and which are not associated with the same nerbert, completely determine that sherbert.

Let’s see if starting from I.3 we can get to I.4, without making any assumptions about the nature of herberts, nerberts, or sherberts.

If I understand “determine” correctly, I.3 means that for any group of three herberts that are not associated with the same nerbert, there is one and only one sherbert with which all three of those herberts are associated. That being the case, herberts D, E, F in I.4 cannot possibly all be associated with two different sherberts. Therefore, herberts D, E, F are all associated with sherbert alpha, and there is no other sherbert with which D, E, and F are all associated. Therefore, the sherbert that is determined by D, E, F must be sherbert alpha. Therefore, I.4 necessarily follows from I.3. Therefore, I.4 is not necessary for a truly minimum set of axioms (as far as I can tell).

Here’s a crazy thought: is it possible that Hilbert thought that I.4 necessarily followed from I.3, but was not absolutely sure? And that he included I.4 just in case, to defend his legacy of an airtight system against possible future attack? Roughly analogous to a physician ordering an extra test to make sure he doesn’t get sued.


#10

Are those the correct axioms? Other websites give different ones for I.3 and I.4.

This is from a university math department so it is more likely to be correct:
“Postulate I.3.
There exists at least two points on a line. There exist at least three points that do
not lie on a line.
Postulate I.4.
For any three points A, B, C that do not lie on the same line there exists a plane α
that contains each of the points A, B, C. For every plane there exists a point which
it contains.”

http://www.math.nus.edu.sg/~matlmc/Hilbert_Geometry.pdf


#11

If by correct you mean as Hilbert originally wrote them, that is in German.

Wikipedia (the relevance will be seen shortly) used to have the axioms written up in someone’s own words intended to convey the same things. Perhaps there was concern for copyright infringement, as the translation that I know of is in a published book and it would be a lengthy amount of text being copied. However I found clear errors, and also found that the e-book of the Townsend translation provides license to copy, and so was able to replace everything with that.

Since then someone familiar with German saw some room for improvement with regard to “straight line” and “line” and made these improvements.

So the Wikipedia entry, linked above, is “correct” so far as being the best English representation of Hilbert’s German that seems available.

If you would like to verify for yourself, the Townsend translation of Hilbert’s book is at http://www.gutenberg.org/files/17384/17384-pdf.pdf


#12

I’m going to, below, keep your concept but change the wording because the rhyming confused the heck out of me.

[quote]I.3: Three herberts A, B, C not situated in the same boson always completely determine a river alpha.

I.4: Any three herberts D, E, F of a river, including when this is river alpha, which are not in the same boson, completely determine that river. [/quote]

[quote]

Let’s see if starting from I.3 we can get to I.4, without making any assumptions about the nature of herberts, bosons, or rivers.

If I understand “determine” correctly, I.3 means that for any group of three herberts that are not [of] the same boson, there is one and only one river with which all three of those herberts are associated. [/quote]

Yes

True

Well we have established from this what we stipulated, that D, E, and F are of river alpha.

But as MrToad pointed out, if we do not have I.4, we do not know from I.3 that the three herberts are not a unique definition of the river that they lie in: that the river also may be completely defined by herberts other than these.

Or that the plane may be completely defined by points other than any given three.

It’s true as you say that the “any” three points of I.4 includes the three points of I.3. This is where I was missing it myself. But the thing is, it is not only those three points that completely define the plane.


#13

Willus, btw I found another restatement of Hilbert’s axioms, similar to what you had found. That is to say, similar in being worded and numbered entirely differently.

The way it relates to the original reminds me of how a Professor X rewrite compares to one of my, shall we say, less-than-well-written posts.

Rather than present the entire set, here are the first few:

I. Hilbert’s Axioms of Incidence:

(1) Any two points are on at least one line.
(2) Any two points are on at most one line.
(3) Any three points not all on one line are on at least one plane.
(4) Any three points not all on one line are on at most one plane.
(5) If two points on a line are on a plane, then every point on the line is on the plane.
(6) If a point is on each of two planes, then there is another point on each of the two planes.
(7) There are at least two points on each line; there are at least three points on each plane; and there are [at least?] four points not all on one plane and not all on one line.

The full set is at http://www.math.niu.edu/~rusin/known-math/99/hilbert_axioms

But as long as we are in the business of completely rewriting statements while retaining all the original meaning and adding absolutely no new meaning, or at least so intending, it seems to me the above would be better re-written as:

  1. Any two points are on one and only one line.
  2. There are at least two points on each line.
  3. Any three points not all one one line are on one and only one plane.
  4. There are at least three points on each plane.
  5. If two points on a line are on a plane, then every point on the line is on the plane.
  6. If a point is on each of two planes, then there is another point on each of the two planes.
  7. There are at least four points not all on one plane.

(The “and not all on one line” of the last seems unnecessary, given #5, if we can prove from this set of axioms, combined with the rest of Hilbert’s axioms, that all lines are on at least one plane. If we can’t prove that, then there’s a problem.)

However, as for this business of totally rewriting this man’s axioms, considerable care needs to be given that in fact everything is retained and nothing new is added. I am at this point not clear that either the rewrite cited above, or the rewrite of the rewrite, accomplishes this.

They’re a lot easier to read, but they had better be equivalent or they are wrong.


#14

On the surface 1 & 2 seem to share the same commonality between them as 3 & 4 but in regard to points and the lines (rather than points and planes). Let me explain:

(below is not a rewording of the axioms, but rather just shows the structure of the axioms discussion)

I.1 Points and the X that contains them
I.2 X and the points it contains
(replace X with Line)

I.3 Points and the Y that contains them
I.4 Y and the points it contains
(replace Y with Plane)

So you can see how they appear similar on the surface.

So why is each I.2 and I.4 necessary, given I.1, and I.3. If you know why I.1 and I.2 coexist, you will know why I.3 and I.4 coexist.

Well not exactly. Each point actually says something a bit different.

(below IS a rewording of the axioms)

I.1 says: Any two points in space (completely) define (the geometry of) one and only one line (or if you want to be anal, the two points define infinite lines that share all points in common)

I.2 says: Any line is defined by as few as two points it possesses. The points can be any two points the line possesses.

I.3 says: Any three points in space (completely) define (the geometry of) one and only one plane (or if you want to be anal, the three points define infinite planes that share all points in common)

I.4 says: Any plane is defined by as few as three points it possesses. The points can be any three points the plane possesses.

So the answer to why each I.2 and I.4 necessary, given I.1 and I.3, is that I.1 and I.3 each give the definition of line and plane respectively. That’s all they do is give the definition of the respective geometric element. I.2 and I.4 are different in that each show that the respective geometric element (line and plane) can be defined any of the points they contain (any two for line and any three for plane).

Hilbert knew he must first give the definition before he went ahead and discussed the properties.


#15

Bill, this thread rocks.

I have literally made a career out of applied geometry and trigonometry (land surveying, geodesy, writing mapping software, etc.).

Good to work the mind as well as the body. It’s these kinds of problems that can keep one on a treadmill for an hour.


#16

As you point out it would make sense that a person (such as myself) not seeing why 1.4 was necessary given 1.3 would also not see why I.2 was needed given 1.1.

But oddly enough, I didn’t have any trouble with the first pair but did with the second.


#17

[quote]Bill Roberts wrote:
I’m going to, below, keep your concept but change the wording because the rhyming confused the heck out of me.

I.3: Three herberts A, B, C not situated in the same boson always completely determine a river alpha.

I.4: Any three herberts D, E, F of a river, including when this is river alpha, which are not in the same boson, completely determine that river.

Let’s see if starting from I.3 we can get to I.4, without making any assumptions about the nature of herberts, bosons, or rivers.

If I understand “determine” correctly, I.3 means that for any group of three herberts that are not [of] the same boson, there is one and only one river with which all three of those herberts are associated.

Yes

That being the case, herberts D, E, F in I.4 cannot possibly all be associated with two different rivers.

True

[We know that] herberts D, E, F are all associated with river alpha, and there is no other river with which D, E, and F are all associated. Therefore, the river that is determined by D, E, F must be river alpha.

Well we have established from this what we stipulated, that D, E, and F are of river alpha.

But as MrToad pointed out, if we do not have I.4, we do not know from I.3 that the three herberts are not a unique definition of the river that they lie in: that the river also may be completely defined by herberts other than these.

Or that the plane may be completely defined by points other than any given three.

It’s true as you say that the “any” three points of I.4 includes the three points of I.3. This is where I was missing it myself. But the thing is, it is not only those three points that completely define the plane.[/quote]

Start with I.3: Three herberts A, B, C not associated with the same boson always completely determine a river alpha.

Either there are more herberts in addition to A, B, C that are associated with the same river alpha, or there are not.

If there are not more herberts associated with the same river alpha: then A, B, C are the only herberts associated with river alpha. If A, B, C are the only herberts associated with river alpha, then “any” three herberts not associated with the same boson but associated with river alpha are necessarily A, B, C. In that case it is true that “any” three herberts associated with river alpha but not the same boson determine river alpha, because of I.3.

If there are more herberts associated with the same river alpha, then any three of the herberts that are associated with river alpha but not associated with the same boson constitute a set of three herberts that are not associated with the same boson, for the same reason that any car with black fenders and whitewalls is a car with whitewalls. According to I.3, any three herberts not associated with the same boson determine a river. But if three herberts all associated with river alpha determine a river, the river they determine would have to be alpha because of what it means to “determine”. And so, if there are more herberts associated with the same river alpha in addition to A, B, and C: it is also true that “any” three herberts associated with river alpha but not the same boson determine river alpha.

I hadn’t noticed it before, but now that somebody pointed out the parallel of I.1-I.2 vs. I.3-I.4: it seems to me that I.2 can be derived from I.1 by a similar process as the above three paragraphs.

One way I could be wrong would be if the definition of one or more key German words are slightly different than the definition of the English substitute: e.g. if the German “any” in I.2 and I.4 necessarily implies a proper subset, in which case I.2 and I.4 would introduce a requirement for the existence of additional herberts associated with the boson or with the river, which I.1 and I.3 had not yet introduced.

Or I could be wrong even based on the literal meanings of the words in English, in some way that I am just not seeing.

Edit – another way I could be wrong would be if I am wrong to assume that I.3 necessarily implies that the herberts A, B, C are associated with river alpha. Does it make sense that something could hypothetically determine something else but not be associated with it? Could that be the key to how I.2 and I.4 do not necessarily follow from I.1 and I.3? Going back to the points, lines, and planes: I.1 has only said the points determine the line but has not yet said the points are in the line; I.3 has only said the points determine the plane but has not yet said the points are in the plane?


#18

NR, you’ve made strong points.

I think it is clear that points A, B, and C are associated with alpha, as Hilbert adds in his text, “We write ABC = alpha. We employ also the expressions: A, B, C, lie in alpha; A, B, C are points of alpha, etc.”

On your German language question, I surely don’t know (even though I was forced in graduate to school to learn to read German, not with any great success but enough to at the time squeak by. These days I know nothingk, nothingk.)

Bulletproof Tiger’s argument, if I’m summing it up correctly, that Hilbert wanted to separate minimal definition from further discussion of properties wouldn’t seem to cover the situation of I.4 being provable from I.3. That would just leave I.4 unnecessary.

I suspect that the answer to your derivation of I.4 from I.3 is that I.3 would be clearer if written as “There exist points A, B, C that completely define a plane.”

If written that way, then where in your derivation you write that if additional points in the plane exist – call them A’, B’, and C’ – that they would necessarily have this same property of defining the plane, that wouldn’t be supported by I.3.

In other words, it seems to me that you’re reading an implicit “any” into I.3, thus making it equivalent to I.4. But no implicit “any” was intended, I believe.


#19

[quote]Bill Roberts wrote:
NR, you’ve made strong points.

I think it is clear that points A, B, and C are associated with alpha, as Hilbert adds in his text, “We write ABC = alpha. We employ also the expressions: A, B, C, lie in alpha; A, B, C are points of alpha, etc.”

On your German language question, I surely don’t know (even though I was forced in graduate to school to learn to read German, not with any great success but enough to at the time squeak by. These days I know nothingk, nothingk.)

Bulletproof Tiger’s argument, if I’m summing it up correctly, that Hilbert wanted to separate minimal definition from further discussion of properties wouldn’t seem to cover the situation of I.4 being provable from I.3. That would just leave I.4 unnecessary.

I suspect that the answer to your derivation of I.4 from I.3 is that I.3 would be clearer if written as “There exist points A, B, C that completely define a plane.”

If written that way, then where in your derivation you write that if additional points in the plane exist – call them A’, B’, and C’ – that they would necessarily have this same property of defining the plane, that wouldn’t be supported by I.3.

In other words, it seems to me that you’re reading an implicit “any” into I.3, thus making it equivalent to I.4. But no implicit “any” was intended, I believe.

[/quote]

Bill, I see how you are right if I.1 and I.3 do not have an implicit “any”.

But I think I might now see a way my earlier reasoning was wrong, even if I.1 and I.3 do have any implicit “any”. I.1 and I.3, via “determine”, do imply an association from points to line and from points to plane; but they do not say what kind of association.

My earlier analogy with herberts and other entities was faulty, because I did not take the possibility of different kinds of associations into account. And so, even if I.3 does mean “any”:

I.3 Any three herberts that do not all have association type X with the same boson determine one and only one river.

I.4 Any three herberts that do not have association X with the same boson, and have association type X with river alpha, determine river alpha.

Can we derive I.4 from I.3? I think not. From I.3, I think it is possible that any three herberts A, B, C that do not have association type X with the same boson, have association type Y with only one river while having association type X with ten rivers.

The river with which the herberts have association type Y would be the river they determine, and at least nine of the rivers with which they have association type X would be rivers they do not determine. Let one of the rivers with which the herberts have association type X but not Y be called alpha.

Let the river with which the herberts have association type Y be called beta. The same three herberts A, B, C would not have association type X with the same boson, would have association type X with river alpha, and yet would not determine alpha: instead, these three herberts would determine river beta, which is not river alpha. Therefore, we have a scenario where I.3 is true and means “any” but nevertheless I.4 is false. Therefore, I.4 cannot be derived from I.3, even if I.3 does mean “any”.

Now lets go back to the points, lines, and planes. Even if I.3 means “any”: it only says that any three points that are not in the same line determine a plane, it does not say by means of what kind of association they do that determination. Maybe three points that are not in the same line are in ten planes, but they “behoove” only one plane.

By “behooving” only one plane, they determine that plane. At least nine of the planes that the points are in are not planes that the points “behoove”. For each plane that these three points are in but do not “behoove”: it would be the case that these same three points are not in the same line; are in the plane in question; and yet do not determine that plane; but instead determine a different plane (the plane that those points behoove).

So again, a scenario where I.3 is true and means “any”, and yet I.4 is false. So again, I.4 cannot be derived from I.3, even if I.3 does mean “any”.


#20

It can be funny how the forum swings back and forth randomly between different versions of posts :slight_smile:

Shortly after writing the above, I edited it to include the part you’re talking about as a quote, but saying that I had to be wrong because Hilbert wrote (as translated, anyway) “always determine.”

That would seem equivalent to leading the sentence off with “any.”

So I concluded that I was still in puzzlement.

Let’s bring back I.3 and I.4 (per the Townsend translation) for clarity:

[b]I.3: Three points A, B, C not situated in the same straight line always completely determine a plane alpha. We write ABC = alpha. We employ also the expressions: A, B, C, lie in alpha; A, B, C are points of alpha, etc.

I.4: Any three points A, B, C of a plane alpha, which do not lie in the same straight line, completely determine that plane. [/b]

This wouldn’t seem to leave open the avenue of differing types of associations.

The only differences I find between the statements are:

I.3 is “three points always” while I.4 is “Any three points,” and

I.3 identifies these points by virtue of not being in the same straight line and states that they completely determine a plane, while I.4 identifies them as being of a plane and not being in the same straight line and says they completely determine the plane.

However, in this second difference it seems to me that I.3 is the universal case – where we don’t first know that they must lie in a plane but learn it from the axiom – while is more limited in that it does not answer the question whether there might exist three points not in a straight line that aren’t in the same plane.

I don’t see why the second difference is needed; and I don’t understand what any consequence could be for the first difference (“three points… always” vs “Any three points…”)

So I’m still puzzled.