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I'm having some trouble proving the first theorem from Hilbert's Foundations of Geometry, which says «two straight lines of a plane have either one point or no point in common; two planes have no point in common or a straight line in common; a plane and a straight line not lying in it have no point or one point in common».

It is said that this follows from just the axioms of connection, but if we consider for example an elliptic geometry then two straight lines can't have no point in common.

I'm having lots of trouble proving this, theorem 2 seems pretty easy though...

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In an elliptic geometry, two straight lines always share one point. In particular, two straight lines always share either one or no points; we're just always in the first case, never in the second. There is no contradiction here.

The three parts of this theorem are just restatements of some of the axioms, stated contrapositively.

For example, to show "two straight lines of a plane have either one point or no point in common", we can proceed by contradiction. Suppose two distinct lines $a$ and $b$ have some other number of points in common. Then they must have at least two common points; call two of those common points $P$ and $Q$. But by I.2, there cannot be more than one line through $P$ and $Q$; contradiction.

(Here, I'm following Wikipedia to figure out what is axiom I.1 and what is axiom I.2. Reading some edition of Hilbert's Foundations of Geometry such as this one, both I.1 and I.2 seem to say "Two distinct points always completely determine a straight line" in slightly different words. It is unclear which of them is supposed to mean "there is at least one line through $P$ and $Q$" and which says "there is at most one line through $P$ and $Q$.)

Similarly, if you suppose that there is a counterexample to "two planes have no point in common or a straight line in common" you will either contradict I.6 (if they have only one point in common) or I.5 (if they have several points on a line in common, but not an entire line) or I.4 (if they have several points, not all on line, in common). This is the most complicated part of the theorem, since there are several cases to consider.

If you suppose that there is a counterexample to "a plane and a straight line not lying in it have no point or one point in common" you will contradict I.5.

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  • $\begingroup$ Thanks, this was very helpful. I tend to avoid proofs by contradiction a bit too much, so I forget they can be very powerful. $\endgroup$ – AstlyDichrar Mar 14 '18 at 16:00

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