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We have one square with area 1/n for each positive integer n.

Is it possible to place these squares in the xy-plane in such a way that they completely cover the entire plane. If Yes, can you describe how this can be done (you might also want to draw a picture). If No, explain why this cannot be done.

The sum of their areas correspond to the harmonic series which is divergent. That is the 'total' area is 'infinite'. The 'total' area of the entire plane is also 'infinite'. Yet, infinity has some levels, how can we compare them?

Edit: We could consider both cases :

the first case : overlap of squares is allowed

the second case : overlap of squares is not allowed. I am particularly interested in that case.

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    $\begingroup$ since you can fill any square $C \times C$ with infinitely many small enough squares, and if it is allowed not using them all, then yes, whenever the area $a_n \to 0$ and that $\sum_n a_n$ diverges (and it will cover the plane in the sense that for every fixed square of the plane, the un-covered area $\to 0$) $\endgroup$
    – reuns
    May 4, 2016 at 12:04
  • $\begingroup$ @user1952009 sure the non-covered area converges to 0, because (a) the sum diverges and (b) we have arbitrarily small tiles, but the question is whether the plane can be filled exactly somehow (preferably constructively), and that's not at all obvious, because each square has sides of length $\frac{1}{\sqrt{n}}$, so the proportions between the tiles are usually not rational. $\endgroup$
    – Lieuwe Vinkhuijzen
    May 4, 2016 at 12:15
  • $\begingroup$ @LieuweVinkhuijzen : my proof wasn't so un-constructive (at least it is definable, it doesn't require the axiom of choice or whatever) : split the plane into $C \times C$ squares, choose an ordering on those squares, and everytime $a_n$ can be put somewhere, choose the first place where it is possible (on the leftmost and upper most part of the $C \times C$ square), and so on. now if you want instead a construction with finitely many squares per $C \times C$ square, I'm not sure it is possible. $\endgroup$
    – reuns
    May 4, 2016 at 12:17
  • $\begingroup$ I got an idea, let me finish it and I'll post it $\endgroup$
    – cxz
    May 4, 2016 at 12:28
  • $\begingroup$ I believe that we first need to define what "covering" the entire plane means. If an area is $\lt \infty$ this makes sense in an intuitive manner. But for an infinite plane what does it mean? $\endgroup$
    – MathematicianByMistake
    May 4, 2016 at 12:29

2 Answers 2

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I'm not quite familiar with real analysis, after all, my major is in physics. So forgive me for some tiny mistakes, but I believe the idea is right.

Lemma. When you have finitely many squares with their total area to be 3. Then you can cover a unit square with them parallelly. By parallelly we mean each edge of the individual squares to be parallel to the unit square.

Proof:Denote the unit square by $M$. Firstly, let us sort the small squares from large to small parallelly along the bottom side of $M$,until the sum of the edges to be no less than $1$. Suppose the edge of the last square is $h_i$.

Accordingly, let us perform the same operation. This time, we sort the rest of the squares just above the first line of squares, where the bottom edges are right $h_1$ above $M$. Suppose the edge of the last square is $h_2$.

After several operations stated above. we have a series $h_i$. and the largest edge of the $(i+1) th$ line is less than $h_1$. and the largest edge of the first square is less than $1$.

We have: $$ Sum\ of\ the\ squares \leq 1\times (1+h_1) +h_1\times (1+h_2)+...\leq 1+2h_1+2h_2+... $$

That is to say: $$ 1+2h_1+2h_2+...\geq 3 $$

Which implies $h_1+h_2+...\geq 1$.

Back to the problem:

We have: $$ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\geq 1+ \frac{1}{2} + 2\times \frac{1}{4} + 4\times \frac{1}{8} + 8\times \frac{1}{16}=3\\ ... $$ so we take the first 16 squares out the cover a unit square.

Accordingly we can take the squares of area $\frac{1}{2^{6k+4}+1}...\frac{1}{2^{6k+10}}$ to cover a unit square.

Let us now construct a map form $k$ to $(x,y)$, where $x,y$ are integers. Then let the $k th$ unit square to be centered at $(x,y)$, thus complete coverage.

PS. I'm not a native English user, hope you can forgive my poor English.

the schematic of the unit coverage

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  • $\begingroup$ I don't get why this should give a tiling of $\mathbb{R}^2$ using every square with area $\frac{1}{n}$ exactly once. $\endgroup$
    – Jack D'Aurizio
    May 4, 2016 at 13:25
  • $\begingroup$ @JackD'Aurizio We use the squares of area from $\frac{1}{2^{6k+4}+1}$ to $\frac{1}{2^{6k+10}}$ once to cover a unit square.Then use the unit squares to cover the space. Of cause each square is used once. $\endgroup$
    – cxz
    May 4, 2016 at 13:31
  • $\begingroup$ Edit to remove: sorry, I thought it had to be a tiling. $\endgroup$
    – Julien Clancy
    May 4, 2016 at 13:41
  • $\begingroup$ Oh, sorry, I bet there was a language issue on my side, since I was looking for tilings, but the OP seems to be fine with those squares massively overlapping, so your solution is fine. $\endgroup$
    – Jack D'Aurizio
    May 4, 2016 at 13:41
  • $\begingroup$ @JulienClancy $4\times 0.75 = 2 <3$ $\endgroup$
    – cxz
    May 4, 2016 at 13:43
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The argument of @user1952009 can be improved in order to give a tiling. Since the harmonic series is divergent, we may fill $\mathbb{R}^+\times \mathbb{R}^+$ with some of the squares whose area is $\frac{1}{4k+1}$. We may fill $\mathbb{R}^+\times \mathbb{R}^+$ also with some of the squares whose area is $\frac{1}{4k+2}$, or $\frac{1}{4k+3}$, or $\frac{1}{4k+4}$. So we may fill $\mathbb{R}^2$ by using some of the given squares, by just glueing the four copies of $\mathbb{R}^+\times\mathbb{R}^+$, each of them tiled with some squares with area $\frac{1}{4k+j}$, for $j\in\{1,2,3,4\}$. Assume that the largest unused square, $Q_1$, has area $\frac{1}{n_1}$.

enter image description here

By the same argument as above we may fill $\mathbb{R}^2\setminus Q_1$ by using exactly the same squares as before, then place $Q_1$ in the hole. By moving a bit those four corners we may make room for the other unused squares. So it is possible to tile $\mathbb{R}^2$ with squares having areas $\frac{1}{1},\frac{1}{2},\frac{1}{3},\ldots$, each of them used once, but the tiling procedure is quite non-constructive. Also because how we move the four corners above depends on the unused squared having a finite total area or not.

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