# A question of Erdős

In the following paper (pages 122-23), Erdős asks if there is a constant $c > 0$ such that every subset $A$ of plane of area more than $c$ contains the vertices of a triangle of unit area.

Is this still open? Has anyone discovered interesting lower bounds for $c$?

As a "motivational" puzzle you can show that if $c = \infty$ then $A$ contains vertices of triangles of all possible areas.

Addendum: Monroe has suggested looking at outer measure: First note that there is a subset of plane of full outer measure avoiding vertices of unit area triangles. Why? Just use transfinite induction to construct a unit area triangle free set X which meets every compact positive area subset of plane. This is doable because at any stage we just need to avoid less than continuum many "heights" corresponding to the base lengths that we have accumulated by that stage. Similarly for category. But what about the following variation?

Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids vertices of triangles of unit area? Of course the answer is yes under CH or MA, but does this hold in ZFC?

To illustrate why this could be hard, let me state a related problem of P. Komjath.

Question: Let X be a bounded subset of plane. Must there exist a full outer measure subset Y of X that avoids rational distances? Can we even avoid unit distance?

There are uncountably many variations on these but they all currently seem very hard to me.

• Does it matter whether $A$ is measurable? Is there a counterexample of $A$ with outer measure $\infty$ and inner measure 0? – Monroe Eskew Jul 24 '15 at 22:13
• Hi Monroe, it does, although your questions leads to some really interesting questions. I will edit to explain. – Ashutosh Jul 24 '15 at 22:46
• @MonroeEskew I may need to verify the details but I am pretty sure a Bernstein set in the plane would work (hmm, with MA perhaps). Use that given any two points $A,B$, there are exactly two lines such that triangle $ABC$ has area $1$ iff $C$ is on one of these two lines. So given any perfect plane non-null set and any $<\mathfrak c$ (where $\mathfrak c=2^{\aleph_0}$) many points, we could remove all lines determined, as above, by pairs of these points ... we might need MA to conclude that the union of all these $<\mathfrak c$ many lines has measure $0$, to be able to carry on the construction. – Mirko Jul 24 '15 at 22:55
• @Mirko, I think we can argue as follows. Suppose $A$ has positive measure and is the union of $< \frak c$ many lines. By Fubini's theorem there must be a positive subset $B$ of the $X$-axis such that for all $x \in B$, the vertical slice $A_x$ has positive linear measure. Choose such $x$ such that the vertical line $L_x$ at $x$ is not among the original lines. Then there is $y$ on $L_x \cap A$ that is not on any of the original lines, contradiction. – Monroe Eskew Jul 25 '15 at 0:17
• @MonroeEskew Thank you! I suspected that taking the union of lines (and not just any null-sets) might help avoid the use of MA, but didn't think of the details till after I read your message. – Mirko Jul 25 '15 at 1:38

According to Daniel R. Mauldin, Some Problems and Ideas of Erdős in Analysis and Geometry. Erdős Centennial 25 (2013): 365-376. this is still open. At least it was still open in 2013. In this chapter it is also stated that we can assume that $A=A_1\cup\cdots\cup A_n$ where the $A_i$ are compact convex sets, and that the statement is true if $n\leqslant 3$, while $n=4$ is the smallest open case.