4
$\begingroup$

As a consideration from the post "Prove by "elementary methods": The plane cannot be covered by finitely-many copies of the letter "Y"", on the basis of the remark made in previous post by the user Moishe Cohen, is it still possible to apply elementary methods to prove weaker results, namely:

The plane cannot be covered by countably many copies of the letter Y.

As in the previous post, by "letter Y", it is meant that letter Y consists of 3 closed segments that have common point.

In other words, how to extend the idea from the case where there are finitely-many copies of the letter Y, however still using only elementary methods?

Advices/hints/solutions very appreciated!!!

$\endgroup$
1
4
$\begingroup$

Consider any countable collection $\{L_1,L_2, \dots \}$ of lines in the plane. If $L$ is any line not parallel to one of the $L_k$'s, then $L_k\cap L$ is a single point for each $k.$ Now there are uncountably many such lines $L.$ That's because we have uncountably many slopes available for $L.$ For any such $L,$ an uncountable set, $L$ cannot be covered by $\cup L_k.$

Now any Y in the plane is contained in the union of three lines. Thus the union of any countable collection of Y's is contained in the countable union of lines, and as we've seen, there is a line that is not covered by this collection of lines.

$\endgroup$
9
$\begingroup$

An idea that occured to me:

Let there be some countable covering $G$ of the plane through "Y letters". Pick some circle $S$ in the plane. Clearly $G$ covers $S$. But each "Y letter" has at most 6 intersections with the circle.

So we've shown that $S$ is a countable subset, which is a contradiction.

$\endgroup$
3
  • $\begingroup$ I think your example is easier than mine. +1 $\endgroup$ – zhw. Nov 14 '16 at 22:56
  • $\begingroup$ Why 6 and not 3? $\endgroup$ – Mars Aug 10 '20 at 6:39
  • $\begingroup$ @Mars I don't remember clearly, but I guess I thought showing $3$ formally could require a bit more effort (you'd have to define what exactly an Y is for example)... and since the whole answer was already just trying to be as simple as possible, $6$ seems like a good enough bound (Y consists of "less than" 3 lines, and there are at most 2 intersections with each line). $\endgroup$ – John P Aug 29 '20 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy