Or polyominoes with no hollow in $\mathbb{R}^3$?


I created this conjecture and tried to make counterexample, but it doesn't work well. Thank you for any answer or correcting question.



Does one pieces of every kind of connected polyominoes P in $\mathbb{R}^2$ which has no hole cover a plane?
I think one can do this by standard inductive trickery.


First take a square grid in the plane. Index the union of the set of cells and the set of polyominoes by natural numbers $i$.


At each stage $i$, we will have placed $n\_i$ connected polyominos with no holes in the plane so that they form a single connected polyomino. We wish to either place a polyomino covering a given cell (if we haven't already) or to place a particular polyomino (if we haven't already).


If we have already fulfilled the condition, don't place any polyomino and move on to the next stage.


If we haven't yet covered the cell, choose a path in the grid connecting that cell to the union of the $n\_i$ already existing polyominoes, then enlarge the path to fill in any holes, then further enlarge it to ensure it is a polyomino not already used, then place it.


If we haven't yet included the polyomino, place the polyomino somewhere far away from the existing polyominoes, then add a path connecting it to the union of the $n\_i$ already existing polyominoes, enlarge this path to fill in any holes, and further enlarge it to ensure it is a polyomino not already used (including the one just placed), and place it.


If we iterate this procedure over all natural numbers $i$, we will cover all the squares using all the polyominoes, as desired.


