If we have a family of squares such that the sum of their areas is infinite, then can we tile the plane with them?

Question

Let $\{S_{i}\}_{i \in \mathbb{N}}$ be a family of squares such that the Sum of the areas of $S_{i}$s is infinite. Can we tile the plane $\mathbb{R}^{2}$ with these squares?

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Note that you can freely move the squares in the plane but they should not overlap.

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let me rephrase my question, you are given $\{Ai\}_{i\in \mathbb{N}}$ a sequence of numbers, where each $A_{i}$ is positive such that $\sum_{i=1}^{\infty}A_{i}=\infty$. Is there a tilling of the plane with the countable number of squares such that the area of the i'th square is $A_{i}$?

Answer 1

No. Example: for $i \in \mathbb N$ let $S_i$ be the open square with the corners $(i-1,0), (i,0),(i,1)$ and $(i-1,1)$.

Answer 2

No, the squares associated with some sequences will not tile $\mathbb{R}^{2}$.

In particular the sequence $A_i$ = $2^{2i}$ fails.

Note that if $a_1$, $a_2$, ..., $a_k$ and $b_1$, $b_2$, ..., $b_h$ are distinct sets of integers, then $\sum_{r=1}^{k}$ $\sqrt{A_{a_r}}$ $\neq$ $\sum_{s=1}^{h} $$\sqrt{A_{b_s}}$. That is, the sum of the sides of any set of squares will not equal the sums of the sides of any other set.

The smallest square must go somewhere and some other square must be next to it. The side of the other square will overlap the side of the first on the top or bottom or both. The best case scenario of a single overlap is shown in the figure below. The squares are not to scale.

Another square must fit on the other side of the smallest square. The best case scenario , overlapping the bottom, is shown below.

Some square, D, must fit on the right side of the square I have labeled B. If it overlaps the bottom of square B, the space between C and D cannot be filled with any combination of squares. So if the tiling is to work, D must overlap the top of B.

Whatever square or squares are fitted on top of B, the leftmost must stick out creating an unfillable gap between A and F.